Find $\lim_{n\to \infty} \big[\frac{1}{1\cdot3}+\frac{1}{2\cdot5}+...+\frac{1}{n\cdot(2n+1)}\big]$

Question

Find $\displaystyle\lim_{n\to \infty} \bigg[\frac{1}{1\cdot3}+\frac{1}{2\cdot5}+...+\frac{1}{n\cdot(2n+1)}\bigg]$ .

This is a similar question as I've asked an hour ago . Find $\lim_{n\to \infty} (1-\frac{1}{2}+\frac{1}{3}-...-\frac{1}{2n})$

But here, I am not able to rearrange terms to use the fact that $\gamma_{n}=\sum_{n=1}^n \frac{1}{n} -\log(n)$ .

Any ideas ? Or Is there any other way to find limit?

Answer 1

Use that $\frac1{n(2n+1)}=2\left[\frac1{2n}-\frac1{2n+1}\right]$ to reduce the problem to that of finding $1-\frac12+\frac13-\frac14+\dots$ This is a classical series, that adds up to $\ln 2$. (You can see this by considering the power series expansion of $\ln(1+x)$ and evaluating at $x=1$ using Abel's theorem. There are other ways.) I'll leave to you the easy algebra that gives the value you need from this series.

Answer 2

Note: I'm posting this answer to help other users.

Finally, As Martin Sleziak sir stated in chat and Andrés E. Caicedo hinted in his answer ,

$T_n=\frac1{n(2n+1)}=2\left[\frac1{2n}-\frac1{2n+1}\right]$ .

We can relate the limit of this sequence $\langle x_n\rangle$ with the partial sum of $n$ terms of the series $\sum_{n = 1}^\infty 2\left[\frac1{2n}-\frac1{2n+1}\right]$ . And as $n\to \infty$ , the series sums to [$2-\ln(4)$] , which is equal to our limit.