Logarithmic series Questions in English

(C) The given series is the expansion of $1 - e^{-x}$.
$y = x - \frac{x^2}{2!} + \frac{x^3}{3!} - \frac{x^4}{4!} + \dots = 1 - e^{-x}$
$e^{-x} = 1 - y$
Taking the natural logarithm on both sides:
$-x = \log_e(1 - y)$
$x = -\log_e(1 - y)$
$x = \log_e\left(\frac{1}{1 - y}\right)$

(B) Given the series $y = - \left( {{x^3} + \frac{{{x^6}}}{2} + \frac{{{x^9}}}{3} + \dots} \right)$.
We know the logarithmic expansion $\ln(1 - t) = - \left( {t + \frac{{{t^2}}}{2} + \frac{{{t^3}}}{3} + \dots} \right)$ for $|t| < 1$.
Substituting $t = x^3$,we get $y = \ln(1 - x^3)$.
Taking the exponential of both sides,we have $e^y = 1 - x^3$.
Rearranging for $x^3$,we get $x^3 = 1 - e^y$.
Therefore,$x = (1 - e^y)^{1/3}$.

(C) We know the expansion of the hyperbolic cosine function: $\cosh(x) = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \dots = \frac{e^x + e^{-x}}{2}$.
Substituting $x = \log_e n$,we get:
$1 + \frac{(\log_e n)^2}{2!} + \frac{(\log_e n)^4}{4!} + \dots = \frac{e^{\log_e n} + e^{-\log_e n}}{2}$.
Since $e^{\log_e n} = n$ and $e^{-\log_e n} = e^{\log_e(n^{-1})} = n^{-1}$,the expression becomes:
$\frac{n + n^{-1}}{2}$.

(B) Given the series $S = \frac{1}{1 \times 2} - \frac{1}{2 \times 3} + \frac{1}{3 \times 4} - \frac{1}{4 \times 5} + \dots + \infty$.
Using partial fractions,$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$.
So,$S = (1 - \frac{1}{2}) - (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) - (\frac{1}{4} - \frac{1}{5}) + \dots$.
$S = 1 - 2(\frac{1}{2}) + 2(\frac{1}{3}) - 2(\frac{1}{4}) + 2(\frac{1}{5}) - \dots$.
$S = 1 - 2(\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \dots)$.
We know that $\log_e(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$.
For $x=1$,$\log_e 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$,which implies $\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \dots = 1 - \log_e 2$.
Substituting this into $S$: $S = 1 - 2(1 - \log_e 2) = 1 - 2 + 2 \log_e 2 = 2 \log_e 2 - 1 = \log_e 4 - \log_e e = \log_e(\frac{4}{e})$.
Therefore,$e^S = e^{\log_e(4/e)} = \frac{4}{e}$.

(A) The general term of the series is $T_n = \frac{n+1}{n} \cdot \frac{1}{3^n} = \left( 1 + \frac{1}{n} \right) \frac{1}{3^n} = \frac{1}{3^n} + \frac{1}{n \cdot 3^n}$.
The sum $S = \sum_{n=1}^{\infty} T_n = \sum_{n=1}^{\infty} \frac{1}{3^n} + \sum_{n=1}^{\infty} \frac{1}{n \cdot 3^n}$.
The first part is a geometric series: $\sum_{n=1}^{\infty} \frac{1}{3^n} = \frac{1/3}{1 - 1/3} = \frac{1/3}{2/3} = \frac{1}{2}$.
The second part uses the logarithmic expansion $-\log_e(1-x) = \sum_{n=1}^{\infty} \frac{x^n}{n}$. For $x = \frac{1}{3}$,we get $\sum_{n=1}^{\infty} \frac{1}{n \cdot 3^n} = -\log_e(1 - 1/3) = -\log_e(2/3)$.
Thus,$S = \frac{1}{2} - \log_e \left( \frac{2}{3} \right)$.

(D) Let $S = \sum_{n=1}^{\infty} \frac{n}{n+1} x^{n+1}$.
We can write $\frac{n}{n+1} = 1 - \frac{1}{n+1}$.
So,$S = \sum_{n=1}^{\infty} (1 - \frac{1}{n+1}) x^{n+1} = \sum_{n=1}^{\infty} x^{n+1} - \sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1}$.
$S = (x^2 + x^3 + x^4 + \dots) - (\frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \dots)$.
The first part is a geometric series: $\frac{x^2}{1-x}$.
The second part is the expansion of $-\log_e(1-x) - x$.
Therefore,$S = \frac{x^2}{1-x} - (-\log_e(1-x) - x) = \frac{x^2}{1-x} + x + \log_e(1-x)$.
$S = \frac{x^2 + x(1-x)}{1-x} + \log_e(1-x) = \frac{x}{1-x} + \log_e(1-x)$.

(A) Let $S = \sum_{n=1}^{\infty} \frac{1}{n} \left( \frac{x^n - 1}{(x + 1)^n} \right) = \sum_{n=1}^{\infty} \frac{1}{n} \left( \left( \frac{x}{x + 1} \right)^n - \left( \frac{1}{x + 1} \right)^n \right)$.
Using the logarithmic series expansion $\ln(1 - y) = - \sum_{n=1}^{\infty} \frac{y^n}{n}$,we have:
$S = - \ln\left(1 - \frac{x}{x + 1}\right) - \left( - \ln\left(1 - \frac{1}{x + 1}\right) \right)$.
$S = - \ln\left(\frac{1}{x + 1}\right) + \ln\left(\frac{x}{x + 1}\right)$.
$S = \ln(x + 1) + \ln\left(\frac{x}{x + 1}\right) = \ln\left((x + 1) \cdot \frac{x}{x + 1}\right) = \ln x$.

(B) The general term of the series is $T_n = \frac{1}{(2n - 1)(2n)(2n + 1)}$.
Using partial fractions,we have $T_n = \frac{1}{2} \left[ \frac{1}{2n - 1} - \frac{2}{2n} + \frac{1}{2n + 1} \right] = \frac{1}{2} \left[ \left( \frac{1}{2n - 1} - \frac{1}{2n} \right) - \left( \frac{1}{2n} - \frac{1}{2n + 1} \right) \right]$.
Summing from $n=1$ to $\infty$:
$S = \frac{1}{2} \sum_{n=1}^{\infty} \left( \frac{1}{2n - 1} - \frac{1}{2n} \right) - \frac{1}{2} \sum_{n=1}^{\infty} \left( \frac{1}{2n} - \frac{1}{2n + 1} \right)$.
We know the logarithmic series expansion $\log_{e} 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots = \sum_{n=1}^{\infty} \left( \frac{1}{2n - 1} - \frac{1}{2n} \right)$.
Also,$\sum_{n=1}^{\infty} \left( \frac{1}{2n} - \frac{1}{2n + 1} \right) = \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \dots = 1 - \log_{e} 2$.
Substituting these values:
$S = \frac{1}{2} (\log_{e} 2) - \frac{1}{2} (1 - \log_{e} 2) = \frac{1}{2} \log_{e} 2 - \frac{1}{2} + \frac{1}{2} \log_{e} 2 = \log_{e} 2 - \frac{1}{2}$.

(A) The given series is $S = \sum_{n=1}^{\infty} \frac{2n-1}{n} \cdot \frac{1}{2^n}$.
We can rewrite the general term as $\left( 2 - \frac{1}{n} \right) \frac{1}{2^n} = 2 \cdot \frac{1}{2^n} - \frac{1}{n} \cdot \frac{1}{2^n}$.
Thus,$S = 2 \sum_{n=1}^{\infty} \frac{1}{2^n} - \sum_{n=1}^{\infty} \frac{(1/2)^n}{n}$.
Using the sum of an infinite geometric series $\sum_{n=1}^{\infty} ar^{n-1} = \frac{a}{1-r}$,we have $2 \sum_{n=1}^{\infty} (1/2)^n = 2 \cdot \frac{1/2}{1 - 1/2} = 2(1) = 2$.
Using the logarithmic series expansion $-\ln(1-x) = \sum_{n=1}^{\infty} \frac{x^n}{n}$,for $x = 1/2$,we have $\sum_{n=1}^{\infty} \frac{(1/2)^n}{n} = -\ln(1 - 1/2) = -\ln(1/2) = \ln 2$.
Therefore,$S = 2 - \ln 2$.

(A) We know that $\log_e \sqrt{\frac{1+x}{1-x}} = \frac{1}{2} \log_e \left( \frac{1+x}{1-x} \right)$.
Using the logarithmic expansion $\log_e \left( \frac{1+x}{1-x} \right) = 2 \left( x + \frac{x^3}{3} + \frac{x^5}{5} + \dots \right)$ for $|x| < 1$.
Substituting this into the expression:
$\frac{1}{2} \times 2 \left( x + \frac{x^3}{3} + \frac{x^5}{5} + \dots \right) = x + \frac{x^3}{3} + \frac{x^5}{5} + \dots$
Thus,the correct option is $A$.

(A) The given series is $\frac{1}{x + 1} + \frac{1}{2(x + 1)^2} + \frac{1}{3(x + 1)^3} + \dots \infty$.
We know the logarithmic expansion: $-\log_e(1 - y) = y + \frac{y^2}{2} + \frac{y^3}{3} + \dots \infty$ for $|y| < 1$.
Let $y = \frac{1}{x + 1}$.
Then the series becomes $-\log_e\left(1 - \frac{1}{x + 1}\right)$.
$= -\log_e\left(\frac{x + 1 - 1}{x + 1}\right) = -\log_e\left(\frac{x}{x + 1}\right)$.
$= \log_e\left(\frac{x + 1}{x}\right) = \log_e\left(1 + \frac{1}{x}\right)$.

(C) We know that $\log _e(x + 1) - \log _e(x - 1) = \log _e\left( \frac{x + 1}{x - 1} \right)$.
By dividing the numerator and denominator by $x$,we get:
$\log _e\left( \frac{1 + \frac{1}{x}}{1 - \frac{1}{x}} \right) = \log _e\left( 1 + \frac{1}{x} \right) - \log _e\left( 1 - \frac{1}{x} \right)$.
Using the logarithmic series expansion $\log _e(1 + y) = y - \frac{y^2}{2} + \frac{y^3}{3} - \dots$ and $\log _e(1 - y) = -y - \frac{y^2}{2} - \frac{y^3}{3} - \dots$,where $y = \frac{1}{x}$:
$\log _e\left( 1 + \frac{1}{x} \right) - \log _e\left( 1 - \frac{1}{x} \right) = 2\left( \frac{1}{x} + \frac{1}{3x^3} + \frac{1}{5x^5} + \dots \infty \right)$.

(B) We know the logarithmic series expansion: $-\log_e(1 - x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \dots$ for $|x| < 1$.
Let $x = \frac{a - b}{a}$.
Then the given series is $-\log_e \left( 1 - \frac{a - b}{a} \right)$.
$= -\log_e \left( \frac{a - (a - b)}{a} \right) = -\log_e \left( \frac{b}{a} \right)$.
$= \log_e \left( \frac{a}{b} \right)$.

(A) We know that the logarithmic series expansion is given by $\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots \infty$ for $-1 < x \le 1$.
Substituting $x = a - 1$ in the numerator,we get $(a - 1) - \frac{(a - 1)^2}{2} + \frac{(a - 1)^3}{3} - \dots \infty = \ln(1 + a - 1) = \ln(a)$.
Similarly,substituting $x = b - 1$ in the denominator,we get $(b - 1) - \frac{(b - 1)^2}{2} + \frac{(b - 1)^3}{3} - \dots \infty = \ln(1 + b - 1) = \ln(b)$.
Thus,the expression becomes $\frac{\ln(a)}{\ln(b)}$.
Using the change of base formula $\frac{\log_k a}{\log_k b} = \log_b a$,we get $\frac{\ln(a)}{\ln(b)} = \log_b a$.

(C) The given series is of the form $\frac{x}{1} + \frac{x^2}{2} + \frac{x^3}{3} + \dots = -\ln(1-x)$,where $x = \frac{1}{5}$.
Substituting $x = \frac{1}{5}$ into the series expansion:
$S = -\ln(1 - \frac{1}{5}) = -\ln(\frac{4}{5}) = \ln(\frac{5}{4})$.
We can rewrite $\frac{5}{4}$ as $(\frac{\sqrt{5}}{2})^2$.
Therefore,$S = \ln((\frac{\sqrt{5}}{2})^2) = 2\ln(\frac{\sqrt{5}}{2})$.
Thus,the correct option is $C$.

(C) Let $f(x) = \log_e [(1 + x)^{1 + x} (1 - x)^{1 - x}]$.
Using the property $\log(ab) = \log a + \log b$,we get:
$f(x) = (1 + x) \log_e(1 + x) + (1 - x) \log_e(1 - x)$.
Using the logarithmic series expansions $\log_e(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$ and $\log_e(1 - x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \dots$:
$f(x) = (1 + x)(x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots) + (1 - x)(-x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \dots)$.
Expanding the terms:
$f(x) = (x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots) + (x^2 - \frac{x^3}{2} + \frac{x^4}{3} - \dots) + (-x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \dots) + (x^2 + \frac{x^3}{2} + \frac{x^4}{3} + \dots)$.
Grouping like terms:
$f(x) = 2(x^2(1 - \frac{1}{2}) + x^4(\frac{1}{3} - \frac{1}{4}) + x^6(\frac{1}{5} - \frac{1}{6}) + \dots)$.
$f(x) = 2 [\frac{x^2}{1 \cdot 2} + \frac{x^4}{3 \cdot 4} + \frac{x^6}{5 \cdot 6} + \dots]$.
Thus,the correct option is $C$.

(A) Given expression: $f(x) = 2 \ln x - \ln(x + 1) - \ln(x - 1)$
$= 2 \ln x - \ln((x + 1)(x - 1))$
$= 2 \ln x - \ln(x^2 - 1)$
$= 2 \ln x - \ln(x^2(1 - \frac{1}{x^2}))$
$= 2 \ln x - (\ln x^2 + \ln(1 - \frac{1}{x^2}))$
$= 2 \ln x - 2 \ln x - \ln(1 - \frac{1}{x^2})$
$= - \ln(1 - \frac{1}{x^2})$
Using the expansion $\ln(1 - t) = - (t + \frac{t^2}{2} + \frac{t^3}{3} + \dots)$,where $t = \frac{1}{x^2}$:
$= - [ - (\frac{1}{x^2} + \frac{1}{2(x^2)^2} + \dots) ]$
$= \frac{1}{x^2} + \frac{1}{2x^4} + \dots$
The coefficient of $x^{-4}$ is $\frac{1}{2} = 0.5$.

(B) The given series is $S = \frac{1}{2 \times 3} + \frac{1}{4 \times 5} + \frac{1}{6 \times 7} + \dots$
Using partial fractions,$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$.
Thus,$S = (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{4} - \frac{1}{5}) + (\frac{1}{6} - \frac{1}{7}) + \dots$
We know the logarithmic series expansion $\log_e(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$
For $x=1$,$\log_e(2) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \dots$
Rearranging this,$1 - \log_e(2) = \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \dots$
This matches our series $S$.
Therefore,$S = 1 - \log_e(2) = \log_e(e) - \log_e(2) = \log_e(e/2)$.

(A) We know the logarithmic series expansions:
${\log_e} 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}$
However,the given series is $S = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n(n+1)}$.
Using partial fractions,$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$.
So,$S = \sum_{n=1}^{\infty} (-1)^{n-1} \left( \frac{1}{n} - \frac{1}{n+1} \right)$.
$S = \left( 1 - \frac{1}{2} \right) - \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) - \left( \frac{1}{4} - \frac{1}{5} \right) + \dots$
$S = 1 - 2\left( \frac{1}{2} \right) + 2\left( \frac{1}{3} \right) - 2\left( \frac{1}{4} \right) + \dots$
$S = 1 - 2 \left( \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \dots \right)$.
Since ${\log_e} 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$,we have $\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \dots = 1 - {\log_e} 2$.
Substituting this into $S$:
$S = 1 - 2(1 - {\log_e} 2) = 1 - 2 + 2{\log_e} 2 = 2{\log_e} 2 - 1$.
$S = {\log_e} 4 - {\log_e} e = {\log_e} \left( \frac{4}{e} \right)$.

(A) Let the given series be $S = 1 + \sum_{n=1}^{\infty} \left( \frac{1}{2n} + \frac{1}{2n+1} \right) \frac{1}{4^n}$.
$S = 1 + \sum_{n=1}^{\infty} \frac{1}{2n} \left( \frac{1}{4} \right)^n + \sum_{n=1}^{\infty} \frac{1}{2n+1} \left( \frac{1}{4} \right)^n$.
Using the expansion $\log_e(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$ and $\log_e \left( \frac{1+x}{1-x} \right) = 2 \left( x + \frac{x^3}{3} + \frac{x^5}{5} + \dots \right)$.
Let $x = \frac{1}{2}$. Then $\log_e(1+x) = \log_e(3/2)$ and $\log_e \left( \frac{1+x}{1-x} \right) = \log_e(3)$.
Substituting these values,we get $S = \log_e(2\sqrt{3})$.

(A) The given series is $S = \sum_{n=1}^{\infty} \frac{1}{n(2n-1)(2n+1)}$.
Using partial fractions,$\frac{1}{(2n-1)(2n+1)} = \frac{1}{2} \left( \frac{1}{2n-1} - \frac{1}{2n+1} \right)$.
So,$S = \sum_{n=1}^{\infty} \frac{1}{2n} \left( \frac{1}{2n-1} - \frac{1}{2n+1} \right) = \sum_{n=1}^{\infty} \left( \frac{1}{2n(2n-1)} - \frac{1}{2n(2n+1)} \right)$.
This can be written as $S = \sum_{n=1}^{\infty} \left( \left( \frac{1}{2n-1} - \frac{1}{2n} \right) - \left( \frac{1}{2n} - \frac{1}{2n+1} \right) \right)$.
Expanding the series: $S = (1 - 1/2) - (1/2 - 1/3) + (1/3 - 1/4) - (1/4 - 1/5) + \dots$
$S = 1 - 2(1/2) + 2(1/3) - 2(1/4) + 2(1/5) - \dots$
$S = -1 + 2(1 - 1/2 + 1/3 - 1/4 + 1/5 - \dots) = -1 + 2 \log_e 2$.

(B) The given series is $S = \frac{4}{1 \times 3} - \frac{6}{2 \times 4} + \frac{12}{5 \times 7} - \frac{14}{6 \times 8} + \dots$
We can rewrite the terms as:
$S = \left( \frac{1+3}{1 \times 3} \right) - \left( \frac{2+4}{2 \times 4} \right) + \left( \frac{5+7}{5 \times 7} \right) - \left( \frac{6+8}{6 \times 8} \right) + \dots$
$S = \left( \frac{1}{3} + \frac{1}{1} \right) - \left( \frac{1}{4} + \frac{1}{2} \right) + \left( \frac{1}{7} + \frac{1}{5} \right) - \left( \frac{1}{8} + \frac{1}{6} \right) + \dots$
Rearranging the terms,we get:
$S = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \dots$
This is the standard expansion for $\log_e(1+x)$ at $x=1$,which is $\log_e 2$.

(B) We have $\log_e x - \log_e (x - 1) = \log_e \left( \frac{x}{x - 1} \right)$.
This can be written as $\log_e \left( \frac{1}{1 - \frac{1}{x}} \right) = -\log_e \left( 1 - \frac{1}{x} \right)$.
Using the logarithmic expansion series $-\log_e (1 - y) = y + \frac{y^2}{2} + \frac{y^3}{3} + \dots$ where $y = \frac{1}{x}$,we get:
$-\log_e \left( 1 - \frac{1}{x} \right) = \frac{1}{x} + \frac{1}{2x^2} + \frac{1}{3x^3} + \dots \infty$.

(C) The given series is $S = \frac{1}{n^2} + \frac{1}{2n^4} + \frac{1}{3n^6} + \dots \infty$.
This can be written as $S = \frac{(1/n^2)^1}{1} + \frac{(1/n^2)^2}{2} + \frac{(1/n^2)^3}{3} + \dots$.
Using the logarithmic expansion $-\log_e(1 - x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \dots$ for $|x| < 1$,where $x = \frac{1}{n^2}$,we get:
$S = -\log_e(1 - \frac{1}{n^2}) = -\log_e(\frac{n^2 - 1}{n^2}) = \log_e(\frac{n^2}{n^2 - 1})$.
Thus,the correct option is $C$.

(D) We use the logarithmic series expansion: $\frac{1}{2}\log_e\left( \frac{1+x}{1-x} \right) = x + \frac{x^3}{3} + \frac{x^5}{5} + \dots$ where $x = \frac{m-n}{m+n}$.
Substituting $x$ into the formula:
$= \frac{1}{2}\log_e\left( \frac{1 + \frac{m-n}{m+n}}{1 - \frac{m-n}{m+n}} \right)$
$= \frac{1}{2}\log_e\left( \frac{\frac{m+n+m-n}{m+n}}{\frac{m+n-(m-n)}{m+n}} \right)$
$= \frac{1}{2}\log_e\left( \frac{2m}{2n} \right)$
$= \frac{1}{2}\log_e\left( \frac{m}{n} \right)$.

(B) The given series is $S = \frac{1}{2} + \frac{1}{3} \cdot \frac{1}{2^3} + \frac{1}{5} \cdot \frac{1}{2^5} + \dots \infty$.
We know the logarithmic series expansion: $\log_e \left( \frac{1+x}{1-x} \right) = 2 \left( x + \frac{x^3}{3} + \frac{x^5}{5} + \dots \infty \right)$ for $|x| < 1$.
Let $x = \frac{1}{2}$. Then,$\log_e \left( \frac{1 + 1/2}{1 - 1/2} \right) = 2 \left( \frac{1}{2} + \frac{1}{3} \cdot \frac{1}{2^3} + \frac{1}{5} \cdot \frac{1}{2^5} + \dots \infty \right)$.
$\log_e \left( \frac{3/2}{1/2} \right) = 2S$.
$\log_e(3) = 2S$.
$S = \frac{1}{2} \log_e(3) = \log_e(3^{1/2}) = \log_e \sqrt{3}$.

(A) The given equation is $4\left[ {{x^2} + \frac{{{x^6}}}{3} + \frac{{{x^{10}}}}{5} + \dots} \right] = {y^2} + \frac{{{y^4}}}{2} + \frac{{{y^6}}}{3} + \dots$
Using the logarithmic series expansions $\ln(1+t) = t - \frac{t^2}{2} + \frac{t^3}{3} - \dots$ and $\ln(1-t) = -t - \frac{t^2}{2} - \frac{t^3}{3} - \dots$,we have:
$2 \ln\left( \frac{1+x^2}{1-x^2} \right) = -\ln(1-y^2)$
This simplifies to $\ln\left( \frac{1+x^2}{1-x^2} \right)^2 = \ln\left( \frac{1}{1-y^2} \right)$
Taking the exponential of both sides:
$\left( \frac{1+x^2}{1-x^2} \right)^2 = \frac{1}{1-y^2}$
Expanding and rearranging terms leads to the relation ${x^2}y = 2x - y$.

(B) The given series is $S = \log_{e} 2 + \log_{e} \left( \frac{3}{2} \right) + \log_{e} \left( \frac{4}{3} \right) + \dots + \log_{e} \left( \frac{n}{n - 1} \right)$.
Using the property $\log a + \log b = \log(ab)$,we get:
$S = \log_{e} \left( 2 \times \frac{3}{2} \times \frac{4}{3} \times \dots \times \frac{n}{n - 1} \right)$.
All intermediate terms cancel out:
$S = \log_{e} (n)$.
Thus,the correct option is $B$.

(A) The given series is $\frac{1}{n} - \frac{1}{2n^2} + \frac{1}{3n^3} - \frac{1}{4n^4} + \dots$
This can be rewritten as $\frac{1}{n} - \frac{(1/n)^2}{2} + \frac{(1/n)^3}{3} - \frac{(1/n)^4}{4} + \dots$
We know the logarithmic series expansion $\log_e(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$ for $|x| < 1$.
Substituting $x = \frac{1}{n}$,we get $\log_e(1 + \frac{1}{n}) = \frac{1}{n} - \frac{(1/n)^2}{2} + \frac{(1/n)^3}{3} - \dots$
Therefore,the sum is $\log_e\left(\frac{n+1}{n}\right)$.

(D) Given the series $S = \log_{4} 2 - \log_{8} 2 + \log_{16} 2 - \dots$
Using the property $\log_{y^n} x^m = \frac{m}{n} \log_{y} x$,we have $\log_{2^2} 2 = \frac{1}{2} \log_{2} 2 = \frac{1}{2}$,$\log_{2^3} 2 = \frac{1}{3}$,and so on.
Thus,$S = \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \dots$
We know the logarithmic expansion $\log_{e}(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$
For $x=1$,$\log_{e}(2) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$
Rearranging this,$\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \dots = 1 - \log_{e} 2$.
Therefore,$S = 1 - \log_{e} 2$.

(A) Let the given series be $S = \log_{3} e - \log_{9} e + \log_{27} e - \dots$
Using the property $\log_{a^n} b = \frac{1}{n} \log_{a} b$,we can write:
$S = \log_{3} e - \frac{1}{2} \log_{3} e + \frac{1}{3} \log_{3} e - \dots$
Factor out $\log_{3} e$:
$S = (\log_{3} e) \left( 1 - \frac{1}{2} + \frac{1}{3} - \dots \right)$
We know the logarithmic series expansion $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$,so for $x=1$,$\ln(2) = 1 - \frac{1}{2} + \frac{1}{3} - \dots$
Thus,$S = (\log_{3} e) \ln(2) = \frac{\ln e}{\ln 3} \cdot \ln 2 = \frac{1}{\ln 3} \cdot \ln 2 = \log_{3} 2$.

(A) The given series is of the form $x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$ where $x = 0.5$.
We know that the expansion of $\log_e(1 + x)$ is given by $\log_e(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$ for $-1 < x \leq 1$.
Substituting $x = 0.5$ into the series,we get:
$0.5 - \frac{(0.5)^2}{2} + \frac{(0.5)^3}{3} - \frac{(0.5)^4}{4} + \dots = \log_e(1 + 0.5)$.
$= \log_e(1.5) = \log_e(\frac{3}{2})$.

(B) We know that $\log_a(1 + x) = \log_e(1 + x) \cdot \log_a e$.
Using the standard logarithmic series expansion,$\log_e(1 + x) = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^n}{n}$.
Substituting this into the expression,we get:
$\log_a(1 + x) = \log_a e \left( \sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^n}{n} \right)$.
Therefore,the coefficient of $x^n$ is $\frac{(-1)^{n-1}}{n} \log_a e$.

(A) We have $\log_{10}\left(\frac{n}{n-1}\right) = \log_e\left(\frac{n}{n-1}\right) \cdot \log_{10}e$.
$= -\log_e\left(\frac{n-1}{n}\right) \cdot \log_{10}e = -\log_{10}e \cdot \log_e\left(1 - \frac{1}{n}\right)$.
Using the logarithmic series expansion $\log_e(1-x) = -\sum_{r=1}^{\infty} \frac{x^r}{r}$ for $|x| < 1$,we substitute $x = \frac{1}{n}$:
$= -\log_{10}e \cdot \left(-\sum_{r=1}^{\infty} \frac{(1/n)^r}{r}\right) = \log_{10}e \cdot \sum_{r=1}^{\infty} \frac{n^{-r}}{r}$.
Thus,the coefficient of $n^{-r}$ is $\frac{\log_{10}e}{r} = \frac{1}{r \log_e 10}$.

(B) We have $\log_e(1 + 3x + 2x^2) = \log_e((1 + x)(1 + 2x)) = \log_e(1 + x) + \log_e(1 + 2x)$.
Using the expansion $\log_e(1 + y) = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{y^n}{n}$,we get:
$\log_e(1 + x) = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^n}{n}$
$\log_e(1 + 2x) = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{(2x)^n}{n} = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{2^n x^n}{n}$
Adding these,the coefficient of $x^n$ is $(-1)^{n-1} \left( \frac{1}{n} + \frac{2^n}{n} \right) = (-1)^{n-1} \frac{2^n + 1}{n}$.
Since $(-1)^{n-1} = (-1)^{n+1}$,the coefficient is $\frac{(-1)^{n+1}(2^n + 1)}{n}$.

(D) The given expression is $E = e^{\left( {x - \frac{1}{2}{(x - 1)}^2 + \frac{1}{3}{(x - 1)}^3 - \frac{1}{4}{(x - 1)}^4 + \dots} \right)}$.
We can rewrite the exponent by adding and subtracting $1$:
$E = e^{\left( {(x - 1) - \frac{1}{2}{(x - 1)}^2 + \frac{1}{3}{(x - 1)}^3 - \dots} \right) + 1}$.
Using the logarithmic series expansion $\log(1 + t) = t - \frac{t^2}{2} + \frac{t^3}{3} - \dots$,where $t = x - 1$,we get:
$E = e^{\log(1 + (x - 1)) + 1} = e^{\log x + 1}$.
Using the property $e^{a+b} = e^a \cdot e^b$,we have:
$E = e^{\log x} \cdot e^1 = x \cdot e = xe$.

(C) We know that the Taylor series expansion for $\cosh(y)$ is given by $\sum\limits_{n=0}^\infty \frac{y^{2n}}{(2n)!} = \frac{e^y + e^{-y}}{2}$.
Given $S = \sum\limits_{n=0}^\infty \frac{(\log x)^{2n}}{(2n)!}$,we substitute $y = \log x$.
Thus,$S = \frac{e^{\log x} + e^{-\log x}}{2}$.
Since $e^{\log x} = x$ and $e^{-\log x} = e^{\log(x^{-1})} = x^{-1}$,we get $S = \frac{x + x^{-1}}{2}$.

(B) The given series is $S = \frac{1}{3} + \frac{1}{2 \cdot 3^2} + \frac{1}{3 \cdot 3^3} + \frac{1}{4 \cdot 3^4} + \dots \infty$.
We know the logarithmic expansion: $-\log_e(1 - x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \dots \infty$ for $|x| < 1$.
Comparing the given series with the expansion,we set $x = \frac{1}{3}$.
Thus,$S = -\log_e(1 - \frac{1}{3}) = -\log_e(\frac{2}{3})$.
Using the property $-\log_e(\frac{a}{b}) = \log_e(\frac{b}{a})$,we get $S = \log_e(\frac{3}{2})$.
Using the property $\log_e(\frac{a}{b}) = \log_e a - \log_e b$,we get $S = \log_e 3 - \log_e 2$.

(C) The expansion of $\ln(1 - x)$ for $|x| < 1$ is given by $\ln(1 - x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \frac{x^5}{5} - \dots$
Now,consider the expression $(1 - x) \ln(1 - x)$:
$(1 - x) \left( -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \frac{x^5}{5} - \dots \right)$
To find the coefficient of $x^5$,we multiply the terms:
$1 \times (\text{coefficient of } x^5 \text{ in } \ln(1 - x)) - x \times (\text{coefficient of } x^4 \text{ in } \ln(1 - x))$
$= 1 \times \left( -\frac{1}{5} \right) - 1 \times \left( -\frac{1}{4} \right)$
$= -\frac{1}{5} + \frac{1}{4}$
$= \frac{-4 + 5}{20} = \frac{1}{20}$
$= 0.05$

(B) Given expression: $f(x) = \log_e \left[ \frac{1}{1 - x - x^2 + x^3} \right]$
Factorizing the denominator: $1 - x - x^2 + x^3 = (1 - x) - x^2(1 - x) = (1 - x)(1 - x^2) = (1 - x)(1 - x)(1 + x) = (1 - x)^2(1 + x)$
So,$f(x) = \log_e \left[ (1 - x)^{-2}(1 + x)^{-1} \right]$
Using logarithmic properties: $f(x) = -2 \log_e(1 - x) - \log_e(1 + x)$
Using the series expansions $\log_e(1 - x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \dots$ and $\log_e(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$
$f(x) = -2(-x - \frac{x^2}{2} - \dots) - (x - \frac{x^2}{2} + \dots)$
$f(x) = (2x + x^2 + \dots) - (x - \frac{x^2}{2} + \dots)$
$f(x) = (2 - 1)x + (1 + 0.5)x^2 + \dots = x + 1.5x^2 + \dots$
The coefficient of $x$ is $1$.

(B) The given series is $S = 1 + \frac{2}{3} - \frac{2}{4} + \frac{2}{5} - \dots \infty$.
We can rewrite this as $S = 1 + 2 \left( \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \dots \right)$.
We know the logarithmic expansion $\log_e(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$ for $|x| < 1$.
For $x=1$,$\log_e 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$.
Thus,$\frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \dots = \log_e 2 - (1 - \frac{1}{2}) = \log_e 2 - \frac{1}{2}$.
Substituting this back into the expression for $S$:
$S = 1 + 2 \left( \log_e 2 - \frac{1}{2} \right) = 1 + 2 \log_e 2 - 1 = 2 \log_e 2 = \log_e 2^2 = \log_e 4$.

(B) We are given the expression $\log_e \left( 1 + ax^2 + a^2 + \frac{a}{x^2} \right)$.
First,factor the expression inside the logarithm:
$1 + ax^2 + a^2 + \frac{a}{x^2} = (1 + ax^2) + \frac{a}{x^2}(1 + ax^2) = (1 + ax^2)(1 + \frac{a}{x^2})$.
Now,use the property $\log(mn) = \log m + \log n$:
$\log_e \left( (1 + ax^2)(1 + \frac{a}{x^2}) \right) = \log_e(1 + ax^2) + \log_e(1 + \frac{a}{x^2})$.
Using the logarithmic series expansion $\log_e(1 + y) = y - \frac{y^2}{2} + \frac{y^3}{3} - \dots$:
$= \left( ax^2 - \frac{(ax^2)^2}{2} + \frac{(ax^2)^3}{3} - \dots \right) + \left( \frac{a}{x^2} - \frac{(a/x^2)^2}{2} + \frac{(a/x^2)^3}{3} - \dots \right)$.
Grouping the terms by powers of $a$:
$= a(x^2 + \frac{1}{x^2}) - \frac{a^2}{2}(x^4 + \frac{1}{x^4}) + \frac{a^3}{3}(x^6 + \frac{1}{x^6}) - \dots$

(C) The logarithmic series expansion is given by $\log_e(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots \infty$.
This power series converges for $-1 < x \le 1$.
Therefore,the expansion is defined for $x \in (-1, 1]$.

(B) We know the logarithmic series expansion: $\frac{1}{2} \ln \left( \frac{1+z}{1-z} \right) = z + \frac{z^3}{3} + \frac{z^5}{5} + \dots$ for $|z| < 1$.
Let $z = \frac{1}{y}$. Then the given expression is $\frac{1}{2} \ln \left( \frac{1 + 1/y}{1 - 1/y} \right) = \frac{1}{2} \ln \left( \frac{y+1}{y-1} \right)$.
Substitute $y = 2x^2 - 1$:
$\frac{1}{2} \ln \left( \frac{2x^2 - 1 + 1}{2x^2 - 1 - 1} \right) = \frac{1}{2} \ln \left( \frac{2x^2}{2x^2 - 2} \right) = \frac{1}{2} \ln \left( \frac{x^2}{x^2 - 1} \right) = -\frac{1}{2} \ln \left( \frac{x^2 - 1}{x^2} \right) = -\frac{1}{2} \ln \left( 1 - \frac{1}{x^2} \right)$.
Using the expansion $\ln(1-u) = -(u + \frac{u^2}{2} + \frac{u^3}{3} + \dots)$ where $u = \frac{1}{x^2}$:
$-\frac{1}{2} [ -(\frac{1}{x^2} + \frac{1}{2x^4} + \frac{1}{3x^6} + \dots) ] = \frac{1}{2} [ \frac{1}{x^2} + \frac{1}{2x^4} + \frac{1}{3x^6} + \dots ]$.

(B) Since $x, y, z$ are three consecutive positive integers,we have $y = x + 1$ and $z = x + 2$,so $z - x = 2$.
Also,$2y = x + z$,which implies $y^2 = xz + 1$ (since $y^2 - xz = (x+1)^2 - x(x+2) = x^2 + 2x + 1 - x^2 - 2x = 1$).
Let $S = \frac{1}{2}\log_e x + \frac{1}{2}\log_e z + \sum_{n=1, 3, 5, \dots} \frac{1}{n} \left( \frac{1}{2xz + 1} \right)^n$.
Using the logarithmic series expansion $\frac{1}{2} \log_e \left( \frac{1+u}{1-u} \right) = u + \frac{u^3}{3} + \frac{u^5}{5} + \dots$,where $u = \frac{1}{2xz+1}$.
$S = \frac{1}{2} \log_e (xz) + \log_e \left( \frac{1 + \frac{1}{2xz+1}}{1 - \frac{1}{2xz+1}} \right) = \frac{1}{2} \log_e (xz) + \log_e \left( \frac{2xz+2}{2xz} \right) = \frac{1}{2} \log_e (xz) + \log_e \left( \frac{xz+1}{xz} \right)$.
$S = \frac{1}{2} \log_e (xz) + \log_e (xz+1) - \log_e (xz) = \log_e (xz+1) - \frac{1}{2} \log_e (xz) = \log_e (y^2) - \frac{1}{2} \log_e (xz)$.
Since $xz = y^2 - 1$,this simplifies to $\log_e y$.

(D) The $n^{th}$ term of the series is $T_n = \frac{(-1)^{n+1}}{n(n+1)}$.
Using partial fractions,$T_n = (-1)^{n+1} \left( \frac{1}{n} - \frac{1}{n+1} \right)$.
The sum $S = \sum_{n=1}^{\infty} (-1)^{n+1} \left( \frac{1}{n} - \frac{1}{n+1} \right)$.
Expanding the terms:
$S = \left( 1 - \frac{1}{2} \right) - \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) - \dots$
$S = 1 - 2 \left( \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \dots \right)$.
We know the expansion $\log_e(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$. For $x=1$,$\log_e 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$.
Thus,$\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \dots = 1 - \log_e 2$.
Substituting this back: $S = 1 - 2(1 - \log_e 2) = 1 - 2 + 2 \log_e 2 = 2 \log_e 2 - 1 = \log_e 4 - \log_e e = \log_e \left( \frac{4}{e} \right)$.

(B) The given series is the expansion of the logarithmic function:
$y = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots = \ln(1 + x)$
Taking the exponential of both sides:
$e^y = 1 + x$
Rearranging for $x$:
$x = e^y - 1$
Using the exponential series expansion $e^y = 1 + \frac{y}{1!} + \frac{y^2}{2!} + \frac{y^3}{3!} + \dots$:
$x = (1 + \frac{y}{1!} + \frac{y^2}{2!} + \frac{y^3}{3!} + \dots) - 1$
$x = y + \frac{y^2}{2!} + \frac{y^3}{3!} + \dots$

(A) Given that $\alpha$ and $\beta$ are the roots of $x^2 - px + q = 0$,we have $\alpha + \beta = p$ and $\alpha \beta = q$.
Now,consider the expression $\log_e(1 + px + qx^2)$.
Substituting $p = \alpha + \beta$ and $q = \alpha \beta$,we get:
$\log_e(1 + (\alpha + \beta)x + \alpha \beta x^2) = \log_e((1 + \alpha x)(1 + \beta x))$.
Using the property $\log(ab) = \log a + \log b$,we have:
$\log_e(1 + \alpha x) + \log_e(1 + \beta x)$.
Using the logarithmic series expansion $\log_e(1 + t) = t - \frac{t^2}{2} + \frac{t^3}{3} - \dots$,we get:
$\left( \alpha x - \frac{(\alpha x)^2}{2} + \frac{(\alpha x)^3}{3} - \dots \right) + \left( \beta x - \frac{(\beta x)^2}{2} + \frac{(\beta x)^3}{3} - \dots \right)$.
Grouping the terms by powers of $x$:
$= (\alpha + \beta)x - \frac{\alpha^2 + \beta^2}{2}x^2 + \frac{\alpha^3 + \beta^3}{3}x^3 - \dots \infty$.

(B) We know that $\log(1+z) = z - \frac{z^2}{2} + \frac{z^3}{3} - \frac{z^4}{4} + \dots$ for $|z| < 1$.
Given $\log(1 - x + x^2) = \log((1+x^3)/(1+x)) = \log(1+x^3) - \log(1+x)$.
Expanding both terms:
$\log(1+x^3) = x^3 - \frac{x^6}{2} + \frac{x^9}{3} - \frac{x^{12}}{4} + \dots$
$\log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \frac{x^6}{6} + \dots$
Subtracting these,the coefficients of $x^{3k}$ are given by the terms in $\log(1+x^3)$ minus the corresponding terms in $\log(1+x)$.
Specifically,$a_{3k} = \frac{(-1)^{k-1}}{k} - \frac{(-1)^{3k-1}}{3k} = \frac{(-1)^{k-1}}{k} - \frac{(-1)^{k-1}}{3k} = \frac{(-1)^{k-1}}{k} (1 - 1/3) = \frac{2}{3} \frac{(-1)^{k-1}}{k}$.
Thus,$\sum_{k=1}^{\infty} a_{3k} = \sum_{k=1}^{\infty} \frac{2}{3} \frac{(-1)^{k-1}}{k} = \frac{2}{3} (1 - 1/2 + 1/3 - 1/4 + \dots) = \frac{2}{3} \log(2)$.