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...
We have \begin{align}
&amp;\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}\\
&amp;=\sum_{n=1}^m \frac{n}{2 n-1}-\sum _{n=3}^{m+2} \frac{n}{2 n-1}\\
&amp;=\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}\\
&amp;=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}
&amp;\sum _{n=1}^\infty \left(\frac{n}{2 n-1}-\frac{n+2}{2 (n+2)-1}\right) \\&amp;= \lim_{m\to\infty}\left( 1+\frac{2}{3}-\frac{m+1}{2 (m+1)-1}-\frac{m+2}{2 (m+2)-1}\right)\\
 &amp;= 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 &quot;remainder&quot; terms after &quot;telescoping&quot; doesn't tends to $0$.
It becomes easier if you generalise the problem: You are calculating the sum of f(n) - f(n+2), where f(n) = n / (2n-1).
The partial sums for the sum up to k are f(1)-f(3) when k=1, f(1)+f(2)-f(3)-f(4) = f(1)+f(2)-f(k+1)-f(k+2) when k=2, and from then on the partial sum is f(1)+f(2)-f(k+1)-f(k+2) for all k.
In this case, since f(k+1) and f(k+2) both converge towards 1/2, the limit is f(1)+f(2)-1 = 1+2/3-1 = 2/3. The same method will always work if f(k+1)+f(k+2) has a limit, and otherwise the sum doesn’t converge.