<s>[INST] How do I evaluate $\sum_{n=1}^{\infty}\frac{n}{2n-1} - \frac{n+2}{2n+3}$? $$\sum_{n=1}^{\infty}\frac{n}{2n-1} - \frac{n+2}{2n+3}$$
I've tried combining the sum, telescoping series and even trying to make an Nth partial sum but nothing seems to budge. I'm not sure where to go...
 [/INST] We have \begin{align}
&\sum _{n=1}^m \left(\frac{n}{2 n-1}-\frac{n+2}{2 (n+2)-1}\right)=\sum _{n=1}^m \frac{n}{2 n-1}-\sum _{n=1}^m \frac{n+2}{2 (n+2)-1}\\
&=\sum_{n=1}^m \frac{n}{2 n-1}-\sum _{n=3}^{m+2} \frac{n}{2 n-1}\\
&=\frac{1}{2\cdot 1-1}+\frac{2}{2\cdot 2-1}+\sum _{n=3}^m \frac{n}{2 n-1}-\sum
   _{n=3}^m \frac{n}{2 n-1}-\frac{m+1}{2 (m+1)-1}-\frac{m+2}{2 (m+2)-1}\\
&=1+\frac{2}{3}-\frac{m+1}{2 (m+1)-1}-\frac{m+2}{2 (m+2)-1}\end{align}
Taking the limit when $m \to \infty$ we have
\begin{align}
&\sum _{n=1}^\infty \left(\frac{n}{2 n-1}-\frac{n+2}{2 (n+2)-1}\right) \\&= \lim_{m\to\infty}\left( 1+\frac{2}{3}-\frac{m+1}{2 (m+1)-1}-\frac{m+2}{2 (m+2)-1}\right)\\
 &= 1+\frac{2}{3}-\frac{1}{2}-\frac{1}{2} = \frac{2}{3}\end{align}
The others answer don't take into account that the "remainder" terms after "telescoping" doesn't tends to $0$.
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