Logarithmic series Questions in English

(A) The general term of the series after the first term $1$ is $T_n = \frac{2}{(2n-1)(2n)(2n+1)}$ for $n \ge 1$.
Using partial fractions,$\frac{2}{(2n-1)(2n)(2n+1)} = \frac{1}{(2n-1)(2n)} - \frac{1}{2n(2n+1)} = \left( \frac{1}{2n-1} - \frac{1}{2n} \right) - \left( \frac{1}{2n} - \frac{1}{2n+1} \right) = \frac{1}{2n-1} - \frac{2}{2n} + \frac{1}{2n+1}$.
Sum $S = 1 + \sum_{n=1}^{\infty} \left( \frac{1}{2n-1} - \frac{2}{2n} + \frac{1}{2n+1} \right)$.
Expanding the series: $S = 1 + (1 - 1 + 1/3) + (1/3 - 1/2 + 1/5) + (1/5 - 1/3 + 1/7) + \dots$
Alternatively,recognize the series as $S = 1 + \sum_{n=1}^{\infty} \left( \frac{1}{2n-1} - \frac{1}{2n} \right) - \sum_{n=1}^{\infty} \left( \frac{1}{2n} - \frac{1}{2n+1} \right)$.
This simplifies to $2 \times (1 - 1/2 + 1/3 - 1/4 + \dots) = 2 \ln 2$.

(A) The given series is $S = \log _4 2 - \log _8 2 + \log _{16} 2 - \dots \infty$.
Using the property $\log _{a^n} b = \frac{1}{n} \log _a b$,we can write:
$\log _4 2 = \log _{2^2} 2 = \frac{1}{2} \log _2 2 = \frac{1}{2}$
$\log _8 2 = \log _{2^3} 2 = \frac{1}{3} \log _2 2 = \frac{1}{3}$
$\log _{16} 2 = \log _{2^4} 2 = \frac{1}{4} \log _2 2 = \frac{1}{4}$
Substituting these into the series:
$S = \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \dots$
Recall the Taylor series expansion for $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$
For $x=1$,$\ln(2) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$
Rearranging this,we get $\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \dots = 1 - \ln 2$.
Thus,the sum of the series is $1 - \ln 2$.

(B) Given the quadratic equation $5x^2 - 3x - 1 = 0$,the sum of roots is $\alpha + \beta = \frac{3}{5}$ and the product of roots is $\alpha\beta = -\frac{1}{5}$.
The given series is $S = (\alpha + \beta)x - \left( \frac{\alpha^2 + \beta^2}{2} \right)x^2 + \left( \frac{\alpha^3 + \beta^3}{3} \right)x^3 - \dots$
We know the expansion $\ln(1 + t) = t - \frac{t^2}{2} + \frac{t^3}{3} - \dots$
Rewriting the series as: $S = (\alpha x - \frac{(\alpha x)^2}{2} + \frac{(\alpha x)^3}{3} - \dots) + (\beta x - \frac{(\beta x)^2}{2} + \frac{(\beta x)^3}{3} - \dots)$
This simplifies to: $S = \ln(1 + \alpha x) + \ln(1 + \beta x) = \ln((1 + \alpha x)(1 + \beta x))$
Expanding the product: $(1 + \alpha x)(1 + \beta x) = 1 + x(\alpha + \beta) + \alpha\beta x^2$
Substituting the values $\alpha + \beta = \frac{3}{5}$ and $\alpha\beta = -\frac{1}{5}$,we get:
$S = \ln(1 + \frac{3}{5}x - \frac{1}{5}x^2)$

(A) Given $\tan^{-1} a + \tan^{-1} b = \frac{\pi}{4}$ where $0 < a, b < 1$.
Using the formula $\tan^{-1} a + \tan^{-1} b = \tan^{-1} \left(\frac{a+b}{1-ab}\right) = \frac{\pi}{4}$.
Taking $\tan$ on both sides,we get $\frac{a+b}{1-ab} = \tan \frac{\pi}{4} = 1$.
Thus,$a+b = 1-ab$,which implies $a+b+ab = 1$.
Adding $1$ to both sides,we get $1+a+b+ab = 2$,which factors as $(1+a)(1+b) = 2$.
The given series is $S = \left(a - \frac{a^2}{2} + \frac{a^3}{3} - \dots\right) + \left(b - \frac{b^2}{2} + \frac{b^3}{3} - \dots\right)$.
Using the logarithmic expansion $\log_e(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$ for $|x| < 1$,we have:
$S = \log_e(1+a) + \log_e(1+b) = \log_e((1+a)(1+b))$.
Substituting $(1+a)(1+b) = 2$,we get $S = \log_e 2$.

(A) Let $S = \frac{3}{2} x^{2} + \frac{5}{3} x^{3} + \frac{7}{4} x^{4} + \ldots \infty$
We can rewrite the general term as $\frac{2n+1}{n} x^{n} = (2 + \frac{1}{n}) x^{n} = 2x^{n} + \frac{x^{n}}{n}$ for $n \geq 2$.
Thus,$S = \sum_{n=2}^{\infty} (2x^{n} + \frac{x^{n}}{n}) = 2 \sum_{n=2}^{\infty} x^{n} + \sum_{n=2}^{\infty} \frac{x^{n}}{n}$.
Using the geometric series sum $\sum_{n=2}^{\infty} x^{n} = \frac{x^{2}}{1-x}$ and the logarithmic expansion $-\log_{e}(1-x) = x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + \ldots$,we have $\sum_{n=2}^{\infty} \frac{x^{n}}{n} = -\log_{e}(1-x) - x$.
Substituting these back: $S = 2 \left( \frac{x^{2}}{1-x} \right) - \log_{e}(1-x) - x$.
$S = \frac{2x^{2} - x(1-x)}{1-x} - \log_{e}(1-x) = \frac{2x^{2} - x + x^{2}}{1-x} - \log_{e}(1-x) = \frac{3x^{2}-x}{1-x} - \log_{e}(1-x)$.
Wait,re-evaluating the series: $S = \sum_{n=1}^{\infty} \frac{2n+1}{n+1} x^{n+1} = \sum_{n=1}^{\infty} (2 - \frac{1}{n+1}) x^{n+1} = 2 \sum_{n=1}^{\infty} x^{n+1} - \sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1}$.
$S = 2 \left( \frac{x^{2}}{1-x} \right) - (-\log_{e}(1-x) - x) = \frac{2x^{2}}{1-x} + x + \log_{e}(1-x) = \frac{2x^{2} + x - x^{2}}{1-x} + \log_{e}(1-x) = \frac{x^{2}+x}{1-x} + \log_{e}(1-x) = x \left( \frac{1+x}{1-x} \right) - \log_{e}(1-x)$.

(A) Given $y = \sum_{n=1}^{\infty} \frac{n}{n+1} x^{n+1} = \sum_{n=1}^{\infty} (1 - \frac{1}{n+1}) x^{n+1}$.
$y = \sum_{n=1}^{\infty} x^{n+1} - \sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1}$.
$y = (x^2 + x^3 + x^4 + \dots) - (\frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \dots)$.
Using the sum of infinite geometric series $\frac{x^2}{1-x}$ and the expansion $-\ln(1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \dots$,we have:
$y = \frac{x^2}{1-x} - (-\ln(1-x) - x) = \frac{x^2}{1-x} + \ln(1-x) + x$.
$y = \frac{x^2 + x(1-x)}{1-x} + \ln(1-x) = \frac{x}{1-x} + \ln(1-x)$.
At $x = \frac{1}{2}$,$y = \frac{1/2}{1-1/2} + \ln(1-1/2) = 1 + \ln(1/2) = 1 - \ln 2$.
Then $e^{1+y} = e^{1 + 1 - \ln 2} = e^{2 - \ln 2} = e^2 \cdot e^{-\ln 2} = e^2 \cdot \frac{1}{2} = \frac{e^2}{2}$.

(A) Let the given series be $S = 1 + \frac{x}{2 \sqrt{3}} + \frac{x^2}{18} + \frac{x^3}{36 \sqrt{3}} + \frac{x^4}{180} + \ldots \infty$,where $x = \sqrt{3} - \sqrt{2}$.
Substitute $t = \frac{x}{\sqrt{3}} = \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}} = 1 - \sqrt{\frac{2}{3}}$.
The series becomes $S = 1 + \frac{t}{2} + \frac{t^2}{6} + \frac{t^3}{12} + \frac{t^4}{20} + \ldots = 1 + \sum_{n=1}^{\infty} \frac{t^n}{n(n+1)}$.
Using partial fractions,$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$.
$S = 1 + \sum_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{n+1}\right) t^n = 1 + \sum_{n=1}^{\infty} \frac{t^n}{n} - \sum_{n=1}^{\infty} \frac{t^n}{n+1}$.
We know $-\log_e(1-t) = \sum_{n=1}^{\infty} \frac{t^n}{n}$.
Also,$\sum_{n=1}^{\infty} \frac{t^n}{n+1} = \frac{1}{t} \sum_{n=1}^{\infty} \frac{t^{n+1}}{n+1} = \frac{1}{t} \left(-\log_e(1-t) - t\right)$.
Thus,$S = 1 - \log_e(1-t) - \frac{1}{t} (-\log_e(1-t) - t) = 1 - \log_e(1-t) + \frac{1}{t} \log_e(1-t) + 1 = 2 + \left(\frac{1}{t} - 1\right) \log_e(1-t)$.
Since $1-t = \sqrt{\frac{2}{3}}$,we have $S = 2 + \left(\frac{1}{1-\sqrt{2/3}} - 1\right) \log_e\left(\sqrt{\frac{2}{3}}\right) = 2 + \left(\frac{\sqrt{3}}{\sqrt{3}-\sqrt{2}} - 1\right) \log_e\left(\sqrt{\frac{2}{3}}\right)$.
$S = 2 + \left(\frac{\sqrt{2}}{\sqrt{3}-\sqrt{2}}\right) \log_e\left(\sqrt{\frac{2}{3}}\right) = 2 + \sqrt{2}(\sqrt{3}+\sqrt{2}) \cdot \frac{1}{2} \log_e\left(\frac{2}{3}\right) = 2 + (\sqrt{6}+2) \cdot \frac{1}{2} \log_e\left(\frac{2}{3}\right) = 2 + \left(\sqrt{\frac{6}{4}} + 1\right) \log_e\left(\frac{2}{3}\right) = 2 + \left(\sqrt{\frac{3}{2}} + 1\right) \log_e\left(\frac{2}{3}\right)$.
Comparing with $2\left(\sqrt{\frac{b}{a}}+1\right) \log_e\left(\frac{a}{b}\right)$,we get $a=2, b=3$.
$11a + 18b = 11(2) + 18(3) = 22 + 54 = 76$.

(B) The given series is $S = \sum_{n=1}^{\infty} \frac{1}{(2n)(2n+1)}$.
Using partial fractions,$\frac{1}{(2n)(2n+1)} = \frac{1}{2n} - \frac{1}{2n+1}$.
Thus,$S = (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{4} - \frac{1}{5}) + (\frac{1}{6} - \frac{1}{7}) + \dots$
This can be rewritten as $S = 1 - (1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \dots)$.
We know the Taylor series expansion for $\log_e(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$,so for $x=1$,$\log_e 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$.
Therefore,$S = 1 - \log_e 2$.
Since $1 = \log_e e$,we have $S = \log_e e - \log_e 2 = \log_e \left(\frac{e}{2}\right)$.

(D) The given series is $\log _4 2 - \log _8 2 + \log _{16} 2 - \ldots$
Using the property $\log _b a = \frac{1}{\log _a b}$,we get:
$= \frac{1}{\log _2 4} - \frac{1}{\log _2 8} + \frac{1}{\log _2 16} - \ldots$
$= \frac{1}{\log _2 2^2} - \frac{1}{\log _2 2^3} + \frac{1}{\log _2 2^4} - \ldots$
$= \frac{1}{2 \log _2 2} - \frac{1}{3 \log _2 2} + \frac{1}{4 \log _2 2} - \ldots$
Since $\log _2 2 = 1$,the series becomes:
$= \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \ldots$
We know the Taylor series expansion $\log _e(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots$
For $x = 1$,$\log _e(1 + 1) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots$
$\Rightarrow \log _e 2 = 1 - (\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \ldots)$
$\Rightarrow \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \ldots = 1 - \log _e 2$

(B) Let $S = \sum_{n=1}^{\infty} \frac{1}{n(2n+1)}$.
Using partial fractions,$\frac{1}{n(2n+1)} = \frac{1}{n} - \frac{2}{2n+1}$.
Thus,$S = \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{2}{2n+1} \right) = \left( 1 - \frac{2}{3} \right) + \left( \frac{1}{2} - \frac{2}{5} \right) + \left( \frac{1}{3} - \frac{2}{7} \right) + \dots$
$S = 1 + \frac{1}{2} - \frac{2}{3} + \frac{1}{3} - \frac{2}{5} + \frac{1}{4} - \frac{2}{7} + \dots$
$S = 1 + \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \dots$
Recall the expansion $\log_e(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$. For $x=1$,$\log_e 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$.
Therefore,$S = 1 - (\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \dots) = 1 - (1 - \log_e 2) = 2 - \log_e 2$.

(C) Let $y = x \log _e a$. The given series is $y + \frac{y^3}{3!} + \frac{y^5}{5!} + \dots$
We know that the Taylor series expansion for the hyperbolic sine function is $\sinh(y) = y + \frac{y^3}{3!} + \frac{y^5}{5!} + \dots$
Substituting $y = x \log _e a$ back into the series,we get $\sinh(x \log _e a)$.
Thus,the value of the series is $\sinh(x \log _e a)$.

(B) Let $x = \log \left(\tan \left(\frac{\pi}{4}+\frac{\theta}{2}\right)\right)$.
Then,$e^x = \tan \left(\frac{\pi}{4}+\frac{\theta}{2}\right) = \frac{1+\tan \frac{\theta}{2}}{1-\tan \frac{\theta}{2}}$.
Applying the componendo and dividendo rule:
$\frac{e^x-1}{e^x+1} = \frac{(1+\tan \frac{\theta}{2}) - (1-\tan \frac{\theta}{2})}{(1+\tan \frac{\theta}{2}) + (1-\tan \frac{\theta}{2})} = \frac{2 \tan \frac{\theta}{2}}{2} = \tan \frac{\theta}{2}$.
We know that $\tanh \left(\frac{x}{2}\right) = \frac{e^{x/2} - e^{-x/2}}{e^{x/2} + e^{-x/2}} = \frac{e^x - 1}{e^x + 1}$.
Therefore,$\tan \frac{\theta}{2} = \tanh \left(\frac{x}{2}\right)$.
Taking the inverse hyperbolic tangent on both sides:
$\frac{x}{2} = \tanh ^{-1}\left(\tan \frac{\theta}{2}\right)$.
Thus,$x = 2 \tanh ^{-1}\left(\tan \frac{\theta}{2}\right)$.
Hence,option $(B)$ is correct.

(C) We know that $1+x+x^2 = \frac{1-x^3}{1-x}$.
Thus,$\log(1+x+x^2) = \log(1-x^3) - \log(1-x)$.
Using the expansion $\log(1-t) = -(t + \frac{t^2}{2} + \frac{t^3}{3} + \dots)$,we get:
$\log(1-x^3) = -(x^3 + \frac{x^6}{2} + \dots)$
$-\log(1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \dots$
Adding these,the expansion is $x + \frac{x^2}{2} + (\frac{1}{3} - 1)x^3 + \dots$
The coefficient of $x^3$ is $\frac{1}{3} - 1 = -\frac{2}{3}$.

(A) Let $y = \tanh^{-1} x$.
Then $x = \tanh y$.
Using the definition of the hyperbolic tangent function,we have $x = \frac{e^y - e^{-y}}{e^y + e^{-y}}$.
Multiplying both sides by $(e^y + e^{-y})$,we get $x(e^y + e^{-y}) = e^y - e^{-y}$.
$xe^y + xe^{-y} = e^y - e^{-y}$.
Rearranging the terms,we get $e^y(1 - x) = e^{-y}(1 + x)$.
Dividing both sides by $(1 - x)$,we get $e^y = e^{-y} \frac{1 + x}{1 - x}$.
Multiplying both sides by $e^y$,we get $e^{2y} = \frac{1 + x}{1 - x}$.
Taking the natural logarithm on both sides,we get $2y = \log \left(\frac{1 + x}{1 - x}\right)$.
Therefore,$y = \frac{1}{2} \log \left(\frac{1 + x}{1 - x}\right)$.

(A) We know that the formula for the inverse hyperbolic cosine function is given by:
$\cosh^{-1}(x) = \log(x + \sqrt{x^2 - 1})$
Substituting $x = 2$ into the formula:
$\cosh^{-1}(2) = \log(2 + \sqrt{2^2 - 1})$
$\cosh^{-1}(2) = \log(2 + \sqrt{4 - 1})$
$\cosh^{-1}(2) = \log(2 + \sqrt{3})$

(C) The given series is $\frac{1}{2} - \frac{1}{2 \cdot 2^2} + \frac{1}{3 \cdot 2^3} - \frac{1}{4 \cdot 2^4} + \ldots$
We know the logarithmic expansion: $\log _e(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots$
Comparing the given series with the expansion,we set $x = \frac{1}{2}$.
Substituting $x = \frac{1}{2}$ into the expansion,we get:
$\log _e\left(1 + \frac{1}{2}\right) = \frac{1}{2} - \frac{(1/2)^2}{2} + \frac{(1/2)^3}{3} - \frac{(1/2)^4}{4} + \ldots$
$= \log _e\left(\frac{3}{2}\right)$.

(B) Let the given series be $S = 1 + \frac{1}{3 \cdot 2^2} + \frac{1}{5 \cdot 2^4} + \frac{1}{7 \cdot 2^6} + \ldots$
We know the logarithmic expansion: $\log _e \left( \frac{1+x}{1-x} \right) = 2 \left( x + \frac{x^3}{3} + \frac{x^5}{5} + \ldots \right)$ for $|x| < 1$.
Multiplying the series by $2$,we get:
$2S = 2 + \frac{2}{3 \cdot 2^2} + \frac{2}{5 \cdot 2^4} + \frac{2}{7 \cdot 2^6} + \ldots = 2 + \frac{1}{3 \cdot 2} + \frac{1}{5 \cdot 2^3} + \frac{1}{7 \cdot 2^5} + \ldots$
This does not match directly. Let us rewrite the series as:
$S = 2 \left[ \frac{1}{2} + \frac{(1/2)^3}{3} + \frac{(1/2)^5}{5} + \ldots \right]$
Using the expansion with $x = 1/2$:
$S = \log _e \left( \frac{1 + 1/2}{1 - 1/2} \right) = \log _e \left( \frac{3/2}{1/2} \right) = \log _e 3$.

(B) Given that $b = \sum_{k=1}^{\infty} \frac{a^k}{k}$.
Using the logarithmic series expansion,we know that $-\ln(1-a) = \sum_{k=1}^{\infty} \frac{a^k}{k}$ for $|a| < 1$.
Therefore,$b = -\ln(1-a)$.
This implies $e^{-b} = 1-a$,so $a = 1 - e^{-b}$.
Using the Taylor series expansion for $e^{-b} = 1 - \frac{b}{1!} + \frac{b^2}{2!} - \frac{b^3}{3!} + \dots$,we get:
$a = 1 - (1 - \frac{b}{1!} + \frac{b^2}{2!} - \frac{b^3}{3!} + \dots)$
$a = \frac{b}{1!} - \frac{b^2}{2!} + \frac{b^3}{3!} - \dots$
$a = \sum_{k=1}^{\infty} \frac{(-1)^{k-1} b^k}{k!}$.

(C) We know that $1+x+x^2 = \frac{1-x^3}{1-x}$.
Thus,$\log(1+x+x^2) = \log(1-x^3) - \log(1-x)$.
Using the expansion $\log(1-u) = -(u + \frac{u^2}{2} + \frac{u^3}{3} + \dots)$,we get:
$\log(1-x^3) = -(x^3 + \frac{x^6}{2} + \dots)$
$-\log(1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \dots$
Adding these,the expansion is $x + \frac{x^2}{2} + (\frac{1}{3} - 1)x^3 + \dots$
The coefficient of $x^3$ is $\frac{1}{3} - 1 = -\frac{2}{3}$.

(A) Given the series expansion: $y = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots$
This is the standard logarithmic series expansion for $\log(1+x)$ where $|x| < 1$.
So,$y = \log(1+x)$.
Taking the exponential of both sides: $e^y = 1+x$.
Therefore,$x = e^y - 1$.
The Taylor series expansion for $e^y$ is $e^y = 1 + y + \frac{y^2}{2!} + \frac{y^3}{3!} + \ldots$.
Substituting this into the expression for $x$:
$x = (1 + y + \frac{y^2}{2!} + \frac{y^3}{3!} + \ldots) - 1$.
$x = y + \frac{y^2}{2!} + \frac{y^3}{3!} + \ldots$.

(A) Given $x(y^3 - 1) = 1$,we have $x = \frac{1}{y^3 - 1}$.
Let $k = \frac{1}{x} = y^3 - 1$. Note that $x = \frac{1}{k}$.
The series is $S = 2(\frac{1}{x}) + \frac{2}{3}(\frac{1}{x})^3 + \frac{2}{5}(\frac{1}{x})^5 + \dots = 2k + \frac{2}{3}k^3 + \frac{2}{5}k^5 + \dots$
We know the logarithmic expansion $\log \left( \frac{1+k}{1-k} \right) = 2(k + \frac{k^3}{3} + \frac{k^5}{5} + \dots) = 2k + \frac{2}{3}k^3 + \frac{2}{5}k^5 + \dots$
Substituting $k = y^3 - 1$:
$S = \log \left( \frac{1 + (y^3 - 1)}{1 - (y^3 - 1)} \right) = \log \left( \frac{y^3}{1 - y^3 + 1} \right) = \log \left( \frac{y^3}{2 - y^3} \right)$.

(C) Given that,$x = \operatorname{sech}^{-1} \frac{1}{2} + \tanh^{-1} \frac{1}{2}$.
Using the logarithmic forms: $\operatorname{sech}^{-1} z = \ln \left( \frac{1 + \sqrt{1 - z^2}}{z} \right)$ and $\tanh^{-1} z = \frac{1}{2} \ln \left( \frac{1 + z}{1 - z} \right)$.
For $z = \frac{1}{2}$,$\operatorname{sech}^{-1} \frac{1}{2} = \ln \left( \frac{1 + \sqrt{1 - 1/4}}{1/2} \right) = \ln \left( 2(1 + \frac{\sqrt{3}}{2}) \right) = \ln (2 + \sqrt{3})$.
And $\tanh^{-1} \frac{1}{2} = \frac{1}{2} \ln \left( \frac{1 + 1/2}{1 - 1/2} \right) = \frac{1}{2} \ln (3) = \ln \sqrt{3}$.
Thus,$x = \ln (2 + \sqrt{3}) + \ln \sqrt{3} = \ln (\sqrt{3}(2 + \sqrt{3})) = \ln (2\sqrt{3} + 3)$.
Now,$\cosh x = \frac{e^x + e^{-x}}{2}$.
Since $e^x = 2\sqrt{3} + 3$,then $e^{-x} = \frac{1}{2\sqrt{3} + 3} = \frac{2\sqrt{3} - 3}{(2\sqrt{3})^2 - 3^2} = \frac{2\sqrt{3} - 3}{12 - 9} = \frac{2\sqrt{3} - 3}{3}$.
Therefore,$\cosh x = \frac{(2\sqrt{3} + 3) + \frac{2\sqrt{3} - 3}{3}}{2} = \frac{6\sqrt{3} + 9 + 2\sqrt{3} - 3}{6} = \frac{8\sqrt{3} + 6}{6} = \frac{4\sqrt{3} + 3}{3}$.

(C) We are given the equation $\tanh ^{-1} x = a \log \left(\frac{1+x}{1-x}\right)$ for $|x| < 1$ ...$(i)$
We know the standard logarithmic definition of the inverse hyperbolic tangent function is $\tanh ^{-1} x = \frac{1}{2} \log \left(\frac{1+x}{1-x}\right)$ ...(ii)
Comparing equation $(i)$ and equation (ii),we can equate the coefficients of the logarithmic term.
Therefore,$a = \frac{1}{2}$.

(D) The sum of an infinite $GP$ with first term $a=k$ and common ratio $r=\frac{k}{k+1}$ is given by $S_{k} = \frac{a}{1-r} = \frac{k}{1-\frac{k}{k+1}} = \frac{k}{\frac{1}{k+1}} = k(k+1)$.
We need to calculate $\sum_{k=1}^{\infty} \frac{(-1)^{k}}{S_{k}} = \sum_{k=1}^{\infty} \frac{(-1)^{k}}{k(k+1)}$.
Using partial fractions,$\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$.
So,the sum is $\sum_{k=1}^{\infty} (-1)^{k} \left( \frac{1}{k} - \frac{1}{k+1} \right) = -(\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) - (\frac{1}{3} - \frac{1}{4}) + (\frac{1}{4} - \frac{1}{5}) - \dots$
$= -1 + \frac{1}{2} + \frac{1}{2} - \frac{1}{3} - \frac{1}{3} + \frac{1}{4} + \frac{1}{4} - \dots$
$= -1 + 2 \left( \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \dots \right)$.
Recall the expansion $\log_{e}(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$. For $x=1$,$\log_{e} 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$,so $\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \dots = 1 - \log_{e} 2$.
Substituting this back,the sum is $-1 + 2(1 - \log_{e} 2) = -1 + 2 - 2 \log_{e} 2 = 1 - 2 \log_{e} 2 = 1 - \log_{e} 4$.

(C) Given,$P = 1 + \frac{1}{2 \times 2} + \frac{1}{3 \times 2^{2}} + \dots = \sum_{n=1}^{\infty} \frac{(1/2)^{n-1}}{n}$.
Using the expansion $-\ln(1-x) = \sum_{n=1}^{\infty} \frac{x^n}{n}$,we have $P = 2 \sum_{n=1}^{\infty} \frac{(1/2)^n}{n} = 2 [-\ln(1 - 1/2)] = 2 \ln 2$.
Given,$Q = \frac{1}{1 \times 2} + \frac{1}{3 \times 4} + \frac{1}{5 \times 6} + \dots = \sum_{n=1}^{\infty} \frac{1}{(2n-1)(2n)}$.
Using partial fractions,$\frac{1}{(2n-1)(2n)} = \frac{1}{2n-1} - \frac{1}{2n}$.
Thus,$Q = (1 - 1/2) + (1/3 - 1/4) + (1/5 - 1/6) + \dots = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} = \ln(1+1) = \ln 2$.
Comparing $P = 2 \ln 2$ and $Q = \ln 2$,we get $P = 2Q$.

(C) Let the series be $S = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{n(n+1)}$.
Using partial fractions,$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$.
Thus,$S = (\frac{1}{1} - \frac{1}{2}) - (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) - (\frac{1}{4} - \frac{1}{5}) + \dots$
$S = 1 - \frac{1}{2} - \frac{1}{2} + \frac{1}{3} + \frac{1}{3} - \frac{1}{4} - \frac{1}{4} + \dots$
$S = 1 - 2(\frac{1}{2}) + 2(\frac{1}{3}) - 2(\frac{1}{4}) + 2(\frac{1}{5}) - \dots$
$S = 1 - 2[\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \dots]$
We know the expansion $\log_{e}(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$
For $x=1$,$\log_{e} 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$
So,$\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \dots = 1 - \log_{e} 2$.
Substituting this back,$S = 1 - 2(1 - \log_{e} 2) = 1 - 2 + 2 \log_{e} 2 = 2 \log_{e} 2 - 1$.