Answer to Question #302004 in Real Analysis for Sarita bartwal

Question

Find

lim [1/(2n+1)^2

+2/(2n+2)^2+3/(2n+3)^2+..3/25n] n→ ∞

Answer 1

$\lim\limits_{n\to+\infty} [\frac{1}{(2n+1)^2}+\frac{2}{(2n+2)^2}+\frac{3}{(2n+3)^2}+..\frac{3n}{(5n)^2}]=?$

Note that

$[\frac{1}{(2n+1)^2}+\frac{2}{(2n+2)^2}+\dots+\frac{3n}{(5n)^2}]+2n[\frac{1}{(2n+1)^2}+\frac{1}{(2n+2)^2}+\dots+\frac{1}{(5n)^2}]$

$=\frac{2n+1}{(2n+1)^2}+\frac{2n+2}{(2n+2)^2}+\dots+\frac{5n}{(5n)^2}=\frac{1}{2n+1}+\frac{1}{2n+2}+\dots+\frac{1}{5n}$

Therefore,

$\frac{1}{(2n+1)^2}+\frac{2}{(2n+2)^2}+\dots+\frac{3n}{(5n)^2}=\sum\limits_{k=2n+1}^{5n}\frac{1}{k}-2n\sum\limits_{k=2n+1}^{5n}\frac{1}{k^2}$ (*)

We will get an integral estimate of both terms on the right-hand side of this equality.

For all $x\in[k-1,k]$ we have $\frac{1}{x+1}\leq\frac{1}{k}\leq \frac{1}{x}$. Thus,

$\sum\limits_{k=2n+1}^{5n}\frac{1}{k} \leq \sum\limits_{k=2n+1}^{5n}\int\limits_{k-1}^{k}\frac{dx}{x}=\int\limits_{2n}^{5n}\frac{dx}{x}=\log\frac{5n}{2n}=\log\frac{5}{2}$ ,

$\sum\limits_{k=2n+1}^{5n}\frac{1}{k}\geq \sum\limits_{k=2n+1}^{5n}\int\limits_{k-1}^{k}\frac{dx}{x+1}=\int\limits_{2n}^{5n}\frac{dx}{x+1}=\log\frac{5n+1}{2n+1}=\log\frac{5}{2}+O(\frac{1}{n})$ ,

Therefore,

$\sum\limits_{k=2n+1}^{5n}\frac{1}{k}=\log\frac{5}{2}+O(\frac{1}{n})$

Similarly,

$\sum\limits_{k=2n+1}^{5n}\frac{1}{k^2} \leq \sum\limits_{k=2n+1}^{5n}\int\limits_{k-1}^{k}\frac{dx}{x^2}=\int\limits_{2n}^{5n}\frac{dx}{x^2}=\frac{1}{2n}-\frac{1}{5n}=\frac{3}{10n}$

$\sum\limits_{k=2n+1}^{5n}\frac{1}{k^2} \geq \sum\limits_{k=2n+1}^{5n}\int\limits_{k-1}^{k}\frac{dx}{(x+1)^2}=\int\limits_{2n}^{5n}\frac{dx}{(x+1)^2}=\frac{3n}{(2n+1)(5n+1)}$ $=\frac{3}{10n}+O(\frac{1}{n^2})$

Therefore,

$\sum\limits_{k=2n+1}^{5n}\frac{1}{k^2}=\frac{3}{10n}+O(\frac{1}{n^2})$

Let's apply the obtained asymptotic estimates to equality (*).

$\frac{1}{(2n+1)^2}+\frac{2}{(2n+2)^2}+\dots+\frac{3n}{(5n)^2}$ $=\log\frac{5}{2}+O(\frac{1}{n})-2n(\frac{3}{10n}+O(\frac{1}{n^2}))$ $=\log\frac{5}{2}-\frac{3}{5}+O(\frac{1}{n})$

$\lim\limits_{n\to+\infty} [\frac{1}{(2n+1)^2}+\frac{2}{(2n+2)^2}+\frac{3}{(2n+3)^2}+..\frac{3n}{(5n)^2}]$ $=\log\frac{5}{2}-\frac{3}{5}$

Answer. $\log\frac{5}{2}-\frac{3}{5}$