Convergence of $\sum_{n=1}^{\infty}\frac{2n+1}{(n^{2}+n)^{2}}$ The series $\sum_{n=1}^{\infty}\frac{2n+1}{(n^{2}+n)^{2}}$
(a) converges to $1$
(b) converges to a number $>1$
(c) diverges to $\infty$
(d) has an oscillating sequence of partial sum
How to deal with convergence of the series $\sum_{n=1}^{\infty}\frac{2n+1}{(n^{2}+n)^{2}}$.
Here, I can use limit comparision test taking series  $b_{n}=\sum\frac{1}{n^{3}}$ and got $\lim\frac{a_{n}}{b_{n}}=2$, so by limit comparison test, since the series $\frac{1}{n^{3}}$ is convergent, $a_n=\sum_{n=1}^{\infty}\frac{2n+1}{(n^{2}+n)^{2}}$ is also convergent.
But will it converge to $1$ or to a number greater than $1$? How should I proceed?
 A: The most important aspect of this problem is to notice that $2n+1=(n+1)^2-n^2$. From there, we have
$$\sum_{n=1}^\infty\frac{2n+1}{(n^2+n)^2} = \sum_{n=1}^\infty \frac{(n+1)^2-n^2}{n^2(n+1)^2}.$$
Also, note that $\frac{a-b}{ab}=\frac1b-\frac1a$. Then,
\begin{align*}\sum_{n=1}^\infty\frac{(n+1)^2-n^2}{n^2(n+1)^2} &= \sum_{n=1}^\infty\frac1{n^2}-\frac1{(n+1)^2}\\
&=1-\frac1{2^2}+\frac1{2^2}-\frac1{3^2}+\dots\end{align*}
I'm sure you can get the final answer from here. :)
A: Use computer or maually calculation, you can find the limit close to 1. So you need to find the limit = 1 instead of use inequality to show it >1.
Then you should notice that the expression looks like a partial fraction type question.
$\sum\limits^{\infty}_{n=1}\frac{2n+1}{n^2(n+1)^2}=\sum\limits^{\infty}_{n=1}\frac{(n+1)^2-n^2}{n^2(n+1)^2}=...$
You can try to finish the rest and show the limit is 1.
If you have problem. I can finish the rest steps later.
A: Notice that
\begin{align}
\sum_{n=1}^{N}\frac{2n+1}{(n^{2}+n)^{2}}=\sum_{n=1}^{N}\left(\frac{1}{n^{2}}-\frac{1}{(n+1)^{2}} \right),
\end{align}
this sum is telescoping, then taking $N\to+\infty$ we have
\begin{align}
\sum_{n=1}^{+\infty}\frac{2n+1}{(n^{2}+n)^{2}}=\lim_{N\to +\infty}\sum_{j=1}^{N}\left(\frac{1}{n^{2}}-\frac{1}{(n+1)^{2}} \right)=1.
\end{align}
More reference and examples, see here:

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*https://tutorial.math.lamar.edu/Classes/CalcII/Series_Special.aspx
