Answer to Question #346385 in Real Analysis for Mr Alex

Question

find $\lim _{n\to \infty }\left(\frac{1}{\left(2n+1\right)^2}+\frac{2}{\left(2n+2\right)^2}\frac{3}{\left(2n+3\right)^2}\right)+...+\frac{3}{25n}$

Answer 1

ANSWER: $\left [ \ln \frac{5}{3}-\ln \frac{2}{3} +\frac{2}{5}-\frac{2}{2}\right ]=\ln \frac{5}{2}+\frac{2}{5}-1\cong0.3163$

EXPLANATION

Let

$a_{k}^{n}=\frac{3k-2}{\left (2n+1 \right )^{2}}+\frac{3k-1}{\left (2n+2 \right )^{2}}+\frac{3k }{\left (2n+3 \right )^{2}} \, \, (k=1,...,n, n\in\N)$ .

Then $a_{1}^{n}=\frac{1}{\left (2n+1 \right )^{2}}+\frac{2}{\left (2n+2 \right )^{2}}+\frac{3 }{\left (2n+3 \right )^{2}}$ , $a_{n}^{n}=\frac{3n-2}{\left (2n+3n-2 \right )^{2}}+\frac{3n-1}{\left (2n+3n-1 \right )^{2}}+\frac{3 n }{\left (2n+3n \right )^{2}} =\frac{3n-2}{\left (5n -2 \right )^{2}}+\frac{3n-1}{\left (5n-1 \right )^{2}}+\frac{3 n }{\left (5n \right )^{2}}$ and

$S_{n}= \sum_{k=1}^{n}a_{k}^{n}=\sum_{k=1}^{3n}\frac{k}{\left ( 2n+k \right )^{2}} = \sum_{k=1}^{3n}\frac{k}{9n^{2}\left ( \frac{2}{3}+\frac{k}{3n} \right )^{2}}=\frac{1}{3n} \sum_{k=1}^{3n}\frac{\frac{k}{3n}}{\left ( \frac{2}{3}+\frac{k}{3n} \right )^{2}}$ .

$S{n}$ is the Riemann sum on the segment $[0,1]$ , corresponding to the partition $P_{n}=\left \{ 0,\frac{1}{3n } ,\frac{2}{3n},...,\frac{3n}{3n}=1\right \}$ for the function $f(x)=\frac{x}{\left ( \frac{2}{3} +x\right )^{2} }$ .

The Riemann sum is $\sigma\left ( P_{n},f, \right )=\sum_{k=1}^{3n}\left ( x_{k-1}-x_{k} \right )f(x_{k})$ , where $x_{k}=\frac{k}{3n}$ .

Since the function $f$ is continuous on $[0,1]$ then $f$ is integrable and

$\lim_{n\rightarrow\infty}S_{n}=\lim_{n\rightarrow\infty}\sigma\left ( P_{n},f, \right )=\int_{0}^{1}\frac{x}{\left ( \frac{2}{3} +x\right )^{2} }dx$ $=$

$=\int_{0}^{1}\left [\frac{1}{\left ( \frac{2}{3} +x\right ) } -\frac{2}{3}\cdot\frac{1}{\left ( \frac{2}{3} +x\right )^{2}} \right ]dx=\left [ \ln\left ( \frac{2}{3}+x \right )+\frac{2}{3}\cdot \frac{1}{ \left ( \frac{2}{3}+x \right )} \right ]_{0}^{1}= \\=\left [ \ln \frac{5}{3}-\ln \frac{2}{3} +\frac{2}{5}-\frac{2}{2}\right ]=\ln \frac{5}{2}+\frac{2}{5}-1\cong0.3163$