Answer to Question #348806 in Calculus for Sarita bartwal

Question

Check whether the sequence (an), where

an = 1/ (n+1) + 1/(n+2) +....+1/(2n) is convergent or not

Answer 1

Consider

$1+\dfrac{1}{2}+...+\dfrac{1}{n}+\dfrac{1}{n+1}+...+\dfrac{1}{2n}$ $=(1+\dfrac{1}{2}+...+\dfrac{1}{n})+(\dfrac{1}{n+1}+...+\dfrac{1}{2n})$

Then

$\dfrac{1}{n+1}+...+\dfrac{1}{2n}=(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{n}$ $+\dfrac{1}{n+1}+...+\dfrac{1}{2n-1}+\dfrac{1}{2n})-2(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2n})$ $=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n}$ $=\displaystyle\sum_{k=1}^{2n}\dfrac{(-1)^{k+1}}{k}$

Consider $\ln(1+x)$ for $x\in(-1, x]$

$\ln(1+x)=\displaystyle\sum_{k=1}^{\infin}\dfrac{(-1)^{k+1}x^k}{k}$

Then for $x=1$

$\ln2=\ln(1+1)=\displaystyle\sum_{k=1}^{\infin}\dfrac{(-1)^{k+1}(1)^k}{k}$ $=\displaystyle\sum_{k=1}^{\infin}\dfrac{(-1)^{k+1}}{k}$

We see that

$\lim\limits_{n\to\infin}a_n=\lim\limits_{n\to\infin}(\dfrac{1}{n+1}+...+\dfrac{1}{2n})$ $=\lim\limits_{n\to\infin}\displaystyle\sum_{k=1}^{2n}\dfrac{(-1)^{k+1}}{k}=\displaystyle\sum_{k=1}^{\infin}\dfrac{(-1)^{k+1}}{k}=\ln2$

Therefore the sequence $(a_n)$ is convergent.