Summation of series by definite integration Questions in English

(C) Let $l = \lim _{n \rightarrow \infty} \frac{1}{n} \log \left(\frac{(2 n)!}{n^n n!}\right)$.
We can write $\frac{(2n)!}{n! n^n} = \frac{(n+1)(n+2)\dots(n+n)}{n^n} = \prod_{r=1}^n \left(\frac{n+r}{n}\right) = \prod_{r=1}^n \left(1 + \frac{r}{n}\right)$.
Taking the logarithm,we get $\log \left(\prod_{r=1}^n \left(1 + \frac{r}{n}\right)\right) = \sum_{r=1}^n \log \left(1 + \frac{r}{n}\right)$.
Thus,$l = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \log \left(1 + \frac{r}{n}\right)$.
Using the definition of a definite integral as the limit of a sum,$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n g\left(\frac{r}{n}\right) = \int_0^1 g(x) dx$.
Here,$g(x) = \log(1+x)$,so $l = \int_0^1 \log(1+x) dx$.
Let $1+x = t$,then $dx = dt$. When $x=0, t=1$ and when $x=1, t=2$.
So,$l = \int_1^2 \log t dt = \int_1^2 \log x dx$.
Comparing this with $\int_1^2 f(x) dx$,we get $f(x) = \log x$.

(D) Let $y = \lim _{n \rightarrow \infty} \prod_{r=1}^n \left(1+\frac{r^2}{n^2}\right)^{\frac{2 r}{n^2}}$.
Taking the natural logarithm on both sides:
$\ln y = \lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{2 r}{n^2} \ln \left(1+\frac{r^2}{n^2}\right)$.
This is a Riemann sum,which can be written as a definite integral:
$\ln y = \int_0^1 2x \ln(1+x^2) dx$.
Let $t = 1+x^2$,then $dt = 2x dx$. When $x=0, t=1$ and when $x=1, t=2$.
$\ln y = \int_1^2 \ln(t) dt$.
Using the integration formula $\int \ln(t) dt = t \ln(t) - t$:
$\ln y = [t \ln(t) - t]_1^2 = (2 \ln 2 - 2) - (1 \ln 1 - 1) = 2 \ln 2 - 2 + 1 = \ln(4) - 1 = \ln(4) - \ln(e) = \ln \left(\frac{4}{e}\right)$.
Therefore,$y = \frac{4}{e}$.

(A) The given expression is $\lim _{n \rightarrow \infty} \frac{\pi}{2 n}\left[\sin \frac{\pi}{2 n}+\sin \frac{2 \pi}{2 n}+\ldots+\sin \frac{n \pi}{2 n}\right]$.
This can be written as $\lim _{n \rightarrow \infty} \frac{\pi}{2 n} \sum_{r=1}^n \sin \left(\frac{r \pi}{2 n}\right)$.
Using the definition of definite integral as the limit of a sum,$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n f\left(\frac{r}{n}\right) = \int_0^1 f(x) dx$.
Here,we can rewrite the expression as $\int_0^{\pi/2} \sin(x) dx$ by setting $x = \frac{r\pi}{2n}$,where $dx = \frac{\pi}{2n}$.
The limits of integration are $x = 0$ when $r=0$ and $x = \frac{\pi}{2}$ when $r=n$.
Thus,the integral becomes $\int_0^{\pi/2} \sin(x) dx$.
Evaluating the integral: $[-\cos(x)]_0^{\pi/2} = -(\cos(\frac{\pi}{2}) - \cos(0)) = -(0 - 1) = 1$.

(A) The given expression is $S = \lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{n+3r}{n^2+r^2}$.
Dividing numerator and denominator by $n^2$,we get $S = \lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{\frac{1}{n} + 3\frac{r}{n^2}}{1 + (\frac{r}{n})^2} = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \frac{1 + 3(\frac{r}{n})}{1 + (\frac{r}{n})^2}$.
Using the definition of definite integral as the limit of a sum,we have $\int_0^1 \frac{1+3x}{1+x^2} dx$.
This can be split into $\int_0^1 \frac{1}{1+x^2} dx + 3 \int_0^1 \frac{x}{1+x^2} dx$.
Evaluating the integrals,we get $[\tan^{-1} x]_0^1 + \frac{3}{2} [\ln(1+x^2)]_0^1$.
$= (\frac{\pi}{4} - 0) + \frac{3}{2} (\ln 2 - \ln 1) = \frac{\pi}{4} + \frac{3}{2} \ln 2$.

(B) The given limit can be written as a summation:
$\lim _{n \rightarrow \infty} \sum_{r=0}^{n-1} \left[ \frac{n^{3/2}}{(n+2r)^{5/2}} - \frac{n^{1/2}}{(n+3r)^{3/2}} \right]$
Dividing numerator and denominator by $n^{5/2}$ and $n^{3/2}$ respectively:
$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=0}^{n-1} \left[ \frac{1}{(1+2r/n)^{5/2}} - \frac{1}{(1+3r/n)^{3/2}} \right]$
This is a Riemann sum which converts to the definite integral:
$\int_0^1 \left( \frac{1}{(1+2x)^{5/2}} - \frac{1}{(1+3x)^{3/2}} \right) dx$
Integrating term by term:
$= \left[ \frac{(1+2x)^{-3/2}}{-3/2 \times 2} - \frac{(1+3x)^{-1/2}}{-1/2 \times 3} \right]_0^1$
$= \left[ -\frac{1}{3(1+2x)^{3/2}} + \frac{2}{3(1+3x)^{1/2}} \right]_0^1$
Evaluating at limits $0$ and $1$:
$= \left( -\frac{1}{3(3)^{3/2}} + \frac{2}{3(4)^{1/2}} \right) - \left( -\frac{1}{3(1)^{3/2}} + \frac{2}{3(1)^{1/2}} \right)$
$= \left( -\frac{1}{9\sqrt{3}} + \frac{2}{6} \right) - \left( -\frac{1}{3} + \frac{2}{3} \right)$
$= -\frac{1}{9\sqrt{3}} + \frac{1}{3} - \frac{1}{3} = -\frac{1}{9\sqrt{3}}$

(A) We have,
$I = \lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{k}{n^2+k^2} = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \frac{k/n}{1+(k/n)^2}$
By the definition of definite integral as a limit of a sum,we replace $\frac{k}{n}$ with $x$ and $\frac{1}{n}$ with $dx$,where the limits of integration are from $0$ to $1$.
$I = \int_0^1 \frac{x}{1+x^2} dx$
Let $u = 1+x^2$,then $du = 2x dx$,which implies $x dx = \frac{1}{2} du$.
When $x=0, u=1$ and when $x=1, u=2$.
$I = \frac{1}{2} \int_1^2 \frac{1}{u} du = \frac{1}{2} [\log |u|]_1^2 = \frac{1}{2} (\log 2 - \log 1) = \frac{1}{2} \log 2$.

(A) Let $S_n = \sum_{r=1}^n \frac{n}{n^2+r^2}$.
We can rewrite the expression as $S_n = \sum_{r=1}^n \frac{n}{n^2(1 + \frac{r^2}{n^2})} = \sum_{r=1}^n \frac{1}{n} \cdot \frac{1}{1 + (\frac{r}{n})^2}$.
Taking the limit as $n \rightarrow \infty$,we use the definition of a definite integral as the limit of a Riemann sum:
$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n f(\frac{r}{n}) = \int_0^1 f(x) dx$.
Here,$f(x) = \frac{1}{1+x^2}$.
Thus,$S = \int_0^1 \frac{1}{1+x^2} dx$.
Evaluating the integral,we get $S = [\tan^{-1} x]_0^1 = \tan^{-1}(1) - \tan^{-1}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}$.

(C) Let $f(x) = 2+x^2$.
By the definition of a definite integral as the limit of a sum,$\int_a^b f(x) dx = \lim_{n \rightarrow \infty} h \sum_{r=1}^{n} f(a+rh)$,where $h = \frac{b-a}{n}$.
Here,$a=0, b=3$,so $h = \frac{3-0}{n} = \frac{3}{n}$.
Thus,$\int_0^3 (2+x^2) dx = \lim_{n \rightarrow \infty} \frac{3}{n} \sum_{r=1}^{n} f\left(0 + r \cdot \frac{3}{n}\right) = \lim_{n \rightarrow \infty} \frac{3}{n} \sum_{r=1}^{n} \left(2 + \frac{9r^2}{n^2}\right)$.
Alternatively,using the form $\int_0^b f(x) dx = \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{bn} f\left(\frac{r}{n}\right)$,we have:
$\int_0^3 (2+x^2) dx = \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{3n} \left(2 + \left(\frac{r}{n}\right)^2\right)$
$= \lim_{n \rightarrow \infty} \frac{1}{n} \left[ \sum_{r=1}^{3n} 2 + \sum_{r=1}^{3n} \frac{r^2}{n^2} \right]$
$= \lim_{n \rightarrow \infty} \frac{1}{n} \left[ 2(3n) + \frac{1^2+2^2+\ldots+(3n)^2}{n^2} \right]$
$= \lim_{n \rightarrow \infty} \frac{1}{n} \left[ 6n + \frac{1^2+2^2+\ldots+(3n)^2}{n^2} \right]$.

(C) Given the limit expression: $\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{4 r^3}{r^4+n^4}=p$.
We can rewrite the sum as: $\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{4 r^3}{n^4 \left(1+\frac{r^4}{n^4}\right)} = \lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{4 (r/n)^3}{1+(r/n)^4} \cdot \frac{1}{n}$.
Using the definition of a definite integral as the limit of a sum,let $x = \frac{r}{n}$ and $dx = \frac{1}{n}$. As $n \rightarrow \infty$,the sum becomes the integral from $0$ to $1$:
$p = \int_0^1 \frac{4x^3}{1+x^4} dx$.
Let $t = 1+x^4$,then $dt = 4x^3 dx$.
When $x=0$,$t=1$. When $x=1$,$t=2$.
Thus,$p = \int_1^2 \frac{dt}{t} = [\ln t]_1^2 = \ln 2 - \ln 1 = \ln 2$.
Therefore,$e^p = e^{\ln 2} = 2$.

(A) Let $A = \lim_{n \rightarrow \infty} \frac{1}{n} [(n+1)(n+2) \ldots (2n)]^{\frac{1}{n}}$.
We can write this as $A = \lim_{n \rightarrow \infty} \left[ \frac{1}{n^n} (n+1)(n+2) \ldots (2n) \right]^{\frac{1}{n}} = \lim_{n \rightarrow \infty} \left[ \left(1+\frac{1}{n}\right) \left(1+\frac{2}{n}\right) \ldots \left(1+\frac{n}{n}\right) \right]^{\frac{1}{n}}$.
Taking the natural logarithm on both sides:
$\log A = \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \log \left(1+\frac{r}{n}\right)$.
Using the definition of a definite integral as a limit of a sum:
$\log A = \int_{0}^{1} \log(1+x) dx$.
Let $u = 1+x$,then $du = dx$. When $x=0, u=1$; when $x=1, u=2$.
$\log A = \int_{1}^{2} \log u du = [u \log u - u]_{1}^{2} = (2 \log 2 - 2) - (1 \log 1 - 1) = 2 \log 2 - 2 + 1 = \log 4 - 1$.
Since $1 = \log e$,we have $\log A = \log 4 - \log e = \log \left(\frac{4}{e}\right)$.
Therefore,$A = \frac{4}{e}$.

(B) The given limit is $S = \lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{r^4}{r^5+n^5}$.
We can rewrite this as $S = \lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{r^4}{n^5( (r/n)^5 + 1 )} = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \frac{(r/n)^4}{(r/n)^5 + 1}$.
Using the definition of the definite integral as the limit of a sum,$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} f(r/n) = \int_0^1 f(x) dx$.
Here,$f(x) = \frac{x^4}{x^5 + 1}$.
Thus,$S = \int_0^1 \frac{x^4}{x^5 + 1} dx$.
Let $t = x^5 + 1$,then $dt = 5x^4 dx$,or $x^4 dx = \frac{dt}{5}$.
When $x = 0$,$t = 1$. When $x = 1$,$t = 2$.
$S = \int_1^2 \frac{1}{t} \cdot \frac{dt}{5} = \frac{1}{5} [\ln |t|]_1^2 = \frac{1}{5} (\ln 2 - \ln 1) = \frac{1}{5} \ln 2$.

(B) The given expression is $S = \lim _{n \rightarrow \infty} \frac{1}{n^2} \sum_{k=1}^{2n} k e^{k/n}$.
This can be rewritten as $S = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{2n} \left(\frac{k}{n}\right) e^{k/n}$.
Let $x = \frac{k}{n}$,then $dx = \frac{1}{n}$. As $k=1, x \rightarrow 0$ and as $k=2n, x \rightarrow 2$.
The sum becomes a definite integral: $S = \int_{0}^{2} x e^x dx$.
Using integration by parts,$\int u dv = uv - \int v du$,where $u=x$ and $dv=e^x dx$:
$S = [x e^x]_{0}^{2} - \int_{0}^{2} e^x dx$.
$S = (2 e^2 - 0) - [e^x]_{0}^{2}$.
$S = 2 e^2 - (e^2 - e^0) = 2 e^2 - e^2 + 1 = e^2 + 1$.

(B) We know that $(1+x)^{n} = {}^{n}C_{0} + x{}^{n}C_{1} + x^{2}{}^{n}C_{2} + \ldots + x^{n}{}^{n}C_{n}$.
Integrating both sides with respect to $x$ from $0$ to $2$,we get:
$\int_{0}^{2} (1+x)^{n} dx = \int_{0}^{2} \left( {}^{n}C_{0} + x{}^{n}C_{1} + x^{2}{}^{n}C_{2} + \ldots + x^{n}{}^{n}C_{n} \right) dx$.
Evaluating the integral on the left side:
$\left[ \frac{(1+x)^{n+1}}{n+1} \right]_{0}^{2} = \frac{(1+2)^{n+1}}{n+1} - \frac{(1+0)^{n+1}}{n+1} = \frac{3^{n+1}-1}{n+1}$.
Evaluating the integral on the right side:
$\left[ x{}^{n}C_{0} + \frac{x^{2}}{2}{}^{n}C_{1} + \frac{x^{3}}{3}{}^{n}C_{2} + \ldots + \frac{x^{n+1}}{n+1}{}^{n}C_{n} \right]_{0}^{2} = \frac{2}{1}{}^{n}C_{0} + \frac{2^{2}}{2}{}^{n}C_{1} + \frac{2^{3}}{3}{}^{n}C_{2} + \ldots + \frac{2^{n+1}}{n+1}{}^{n}C_{n}$.
Thus,$S = \frac{3^{n+1}-1}{n+1}$.

(A) The given expression can be written as $\lim _{n \rightarrow \infty} \sum_{r=0}^{n-1} \frac{\sqrt{n}}{\sqrt{(n+4 r)^{3}}}$.
We can rewrite the sum as $\sum_{r=0}^{n-1} \frac{1}{n} \left( \frac{n \sqrt{n}}{\sqrt{(n+4 r)^{3}}} \right) = \sum_{r=0}^{n-1} \frac{1}{n} \left( \frac{1}{(1+\frac{4 r}{n})^{3 / 2}} \right)$.
This is a Riemann sum which converges to the definite integral $\int_{0}^{1} \frac{dx}{(1+4x)^{3/2}}$.
Let $z = 1+4x$,then $dz = 4dx$,so $dx = \frac{dz}{4}$.
When $x=0, z=1$ and when $x=1, z=5$.
The integral becomes $\frac{1}{4} \int_{1}^{5} z^{-3/2} dz = \frac{1}{4} \left[ \frac{z^{-1/2}}{-1/2} \right]_{1}^{5} = -\frac{1}{2} \left[ \frac{1}{\sqrt{z}} \right]_{1}^{5}$.
$= -\frac{1}{2} \left( \frac{1}{\sqrt{5}} - 1 \right) = \frac{1}{2} \left( 1 - \frac{1}{\sqrt{5}} \right) = \frac{\sqrt{5}-1}{2\sqrt{5}} = \frac{5-\sqrt{5}}{10}$.

(C) The given limit is $L = \lim _{n \rightarrow \infty} \frac{1}{n^{k+1}} \sum_{r=1}^{n} (2r)^k$.
We can rewrite this as $L = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \left(\frac{2r}{n}\right)^k$.
Using the definition of the definite integral as the limit of a Riemann sum,$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} f\left(\frac{r}{n}\right) = \int_{0}^{1} f(x) dx$.
Here,$f(x) = (2x)^k$.
Thus,$L = \int_{0}^{1} (2x)^k dx = 2^k \int_{0}^{1} x^k dx$.
Evaluating the integral,$L = 2^k \left[ \frac{x^{k+1}}{k+1} \right]_{0}^{1} = 2^k \left( \frac{1}{k+1} - 0 \right) = \frac{2^k}{k+1}$.

(C) We express the sum as a definite integral using the limit of a sum formula: $\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{r^3}{r^4+n^4} = \lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{n^3(r/n)^3}{n^4((r/n)^4+1)} = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \frac{(r/n)^3}{(r/n)^4+1}$.
Let $x = r/n$,then as $n \rightarrow \infty$,the sum becomes the integral $\int_0^1 \frac{x^3}{x^4+1} dx$.
To solve this,let $u = x^4+1$,then $du = 4x^3 dx$,or $x^3 dx = \frac{1}{4} du$.
The integral becomes $\frac{1}{4} \int_1^2 \frac{1}{u} du = \frac{1}{4} [\log_e u]_1^2$.
$= \frac{1}{4} (\log_e 2 - \log_e 1) = \frac{1}{4} \log_e 2$.

(A) The given expression is $\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{n}{n^2+r^2}$.
Dividing the numerator and denominator of the term inside the summation by $n^2$,we get:
$\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{1/n}{1+(r/n)^2}$.
This is a Riemann sum of the form $\int_0^1 f(x) \, dx$ where $f(x) = \frac{1}{1+x^2}$.
Thus,the limit is $\int_0^1 \frac{1}{1+x^2} \, dx$.
Evaluating the integral,we get $[\tan^{-1} x]_0^1 = \tan^{-1}(1) - \tan^{-1}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}$.

(C) The given limit is $L = \lim _{n \rightarrow \infty} \frac{3}{n} \sum_{r=0}^{n-1} \sqrt{\frac{n}{n+3r}}$.
We can rewrite the expression inside the summation as $\sqrt{\frac{1}{1+3(r/n)}} = (1+3(r/n))^{-1/2}$.
Thus,$L = 3 \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=0}^{n-1} (1+3(r/n))^{-1/2}$.
Using the definition of a definite integral $\int_{0}^{1} f(x) dx = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=0}^{n-1} f(r/n)$,we get:
$L = 3 \int_{0}^{1} (1+3x)^{-1/2} dx$.
Integrating $(1+3x)^{-1/2}$ with respect to $x$ gives $\frac{(1+3x)^{1/2}}{3 \times (1/2)} = \frac{2}{3} \sqrt{1+3x}$.
Evaluating the definite integral:
$L = 3 \left[ \frac{2}{3} \sqrt{1+3x} \right]_{0}^{1} = 2 [\sqrt{1+3} - \sqrt{1+0}] = 2 [\sqrt{4} - 1] = 2 [2 - 1] = 2$.

(C) Given the limit: $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \sec ^{2} \left( \frac{r \pi}{4 n} \right)$.
By the definition of a definite integral as the limit of a sum,$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} f \left( \frac{r}{n} \right) = \int_{0}^{1} f(x) dx$.
Here,$f(x) = \sec ^{2} \left( \frac{\pi x}{4} \right)$.
So,the integral becomes $\int_{0}^{1} \sec ^{2} \left( \frac{\pi x}{4} \right) dx$.
Integrating $\sec ^{2} \left( \frac{\pi x}{4} \right)$,we get $\frac{4}{\pi} \tan \left( \frac{\pi x}{4} \right)$.
Evaluating the definite integral from $0$ to $1$: $\left[ \frac{4}{\pi} \tan \left( \frac{\pi x}{4} \right) \right]_{0}^{1} = \frac{4}{\pi} \left( \tan \frac{\pi}{4} - \tan 0 \right)$.
Since $\tan \frac{\pi}{4} = 1$ and $\tan 0 = 0$,the value is $\frac{4}{\pi} (1 - 0) = \frac{4}{\pi}$.

(B) The given limit is $L = \lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{n}{n^{2}+r^{2}}$.
Dividing the numerator and denominator by $n^{2}$,we get:
$L = \lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{1/n}{1+(r/n)^{2}}$.
Using the definition of a definite integral as the limit of a sum,$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} f\left(\frac{r}{n}\right) = \int_{0}^{1} f(x) dx$,where $f(x) = \frac{1}{1+x^{2}}$.
Therefore,$L = \int_{0}^{1} \frac{1}{1+x^{2}} dx$.
Evaluating the integral,we get $L = [\tan ^{-1} x]_{0}^{1}$.
$L = \tan ^{-1}(1) - \tan ^{-1}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}$.

(A) Given the limit: $\lim _{n \rightarrow \infty} \frac{1}{n^{3/2}} \sum_{r=1}^{n-1} \sqrt{n+r}$
Dividing the numerator and denominator by $n$,we get:
$= \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n-1} \sqrt{\frac{n+r}{n}} = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n-1} \sqrt{1+\frac{r}{n}}$
This is a Riemann sum of the form $\int_{0}^{1} f(x) dx$ where $f(x) = \sqrt{1+x}$
$= \int_{0}^{1} (1+x)^{1/2} dx = \left[ \frac{(1+x)^{3/2}}{3/2} \right]_{0}^{1} = \frac{2}{3} \left[ (1+x)^{3/2} \right]_{0}^{1}$
$= \frac{2}{3} (2^{3/2} - 1^{3/2}) = \frac{2}{3} (2\sqrt{2} - 1)$

(C) We are given the limit $L = \lim _{n \rightarrow \infty} \frac{\sum_{r=1}^{n-1} \sqrt{r}}{n \sqrt{n}}$.
To evaluate this,we can write the sum as $\sum_{r=1}^{n} \sqrt{r} - \sqrt{n}$.
Thus,$L = \lim _{n \rightarrow \infty} \left( \frac{\sum_{r=1}^{n} \sqrt{r}}{n \sqrt{n}} - \frac{\sqrt{n}}{n \sqrt{n}} \right)$.
$L = \lim _{n \rightarrow \infty} \left( \frac{1}{n} \sum_{r=1}^{n} \sqrt{\frac{r}{n}} - \frac{1}{n} \right)$.
Using the definition of a definite integral as the limit of a sum,$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} f(\frac{r}{n}) = \int_{0}^{1} f(x) dx$.
Here,$f(x) = \sqrt{x}$.
So,$L = \int_{0}^{1} \sqrt{x} dx - \lim _{n \rightarrow \infty} \frac{1}{n}$.
$L = \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{1} - 0 = \frac{2}{3} (1)^{3/2} = \frac{2}{3}$.

(B) Using the property $x-1 < [x] \leq x$,we have: $\sum_{k=1}^n \left( \frac{k^2}{3^x} - 1 \right) < \sum_{k=1}^n \left[ \frac{k^2}{3^x} \right] \leq \sum_{k=1}^n \frac{k^2}{3^x}$.
Dividing by $n^3$ and taking the limit as $n \to \infty$:
$\lim_{n \to \infty} \frac{1}{n^3} \sum_{k=1}^n \left( \frac{k^2}{3^x} - 1 \right) < f(x) \leq \lim_{n \to \infty} \frac{1}{n^3} \sum_{k=1}^n \frac{k^2}{3^x}$.
Since $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$,we have $\lim_{n \to \infty} \frac{n(n+1)(2n+1)}{6n^3 \cdot 3^x} = \frac{1}{3 \cdot 3^x} = \frac{1}{3^{x+1}}$.
By the squeeze theorem,$f(x) = \frac{1}{3^{x+1}}$.
Now,$12 \sum_{j=1}^{\infty} f(j) = 12 \sum_{j=1}^{\infty} \frac{1}{3^{j+1}} = 12 \left( \frac{1}{9} + \frac{1}{27} + \dots \right)$.
This is a geometric series with first term $a = \frac{1}{9}$ and common ratio $r = \frac{1}{3}$.
Sum $= 12 \left( \frac{1/9}{1 - 1/3} \right) = 12 \left( \frac{1/9}{2/3} \right) = 12 \left( \frac{1}{6} \right) = 2$.