# Does the series: $\sum_{n=1}^\infty \frac{2n+1}{n^2(n+1)^2}$ converge?

does $\sum_{n=1}^\infty \frac{2n+1}{n^2(n+1)^2}$ converge? I think yes, it does, because the $a_n$ in the series converges to zero. but I'm trying to prove this by the help of the fact that:

$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$

any suggestions?

• If $n$ is very large, the individual terms are approximately $2n/n^4=2/n^3$. That suggest a suitable converging series for a comparison test. – Jyrki Lahtonen Jun 19 '15 at 13:08
• Is convergence sufficient, or do you need to know the limiting value? That it converges is established easily, since $\frac{2n+1}{(n+1)^2}\le1$, and since the series with $1/n^2$ converges. – Bernhard Jun 19 '15 at 13:09
• The fact, that the $a_n$ tend to $0$ is not enough, see the diverging series $$\sum_{j=1}^{\infty} \frac{1}{j}$$ – Peter Jun 19 '15 at 13:10

$$\frac{2n+1}{n^2(n+1)^2}=\frac{1}{n^2}-\frac{1}{(n+1)^2}$$ $$\sum_{n=1}^\infty \frac{2n+1}{n^2(n+1)^2}=\frac{1}{1}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}+....=1$$

• ...and hence the limiting value of the series is 1. – Bernhard Jun 19 '15 at 13:14

Notice that $$\frac{1}{n^2} - \frac{1}{(n+1)^2} = \frac{n^2+2n+1-n^2}{n^2(n+1)^2},$$ so the series telescopes: $$\sum_{n=1}^m \frac{2n+1}{n^2(n+1)^2} = 1-\frac{1}{2^2} + \frac{1}{2^2} -\dotsb - \frac{1}{m^2} + \frac{1}{m^2} - \frac{1}{(m+1)^2} = 1- \frac{1}{(m+1)^2} \to 1$$ as $m \to \infty$

$$\sum_{n\geq1}\frac{2n+1}{n^{2}\left(n+1\right)^{2}}\leq3\sum_{n\geq1}\frac{1}{n^{3}}.$$

$$\frac{2n+1}{n^2(n+1)^2}=\frac{2n}{n^2(n+1)^2}+\frac{1}{n^2(n+1)^2}$$ $$=\frac{2}{n(n+1)^2}+\frac{1}{n^2(n+1)^2}$$

now it is very easy to compare the first term with $\frac{2}{n^3}$ and the second term with $\frac{1}{n^3}$

$$\sum_{n=1}^\infty \frac{2n+1}{n^2(n+1)^2}<\sum_{n=1}^\infty \frac{2n+1}{n^4}=\sum_{n=1}^\infty \frac{2}{n^3} + \sum_{n=1}^\infty \frac{1}{n^4}$$

Another approach: Here $a_n=(2n+1)/n^2(n+1)^2$.
Take auxiliary series whose nth term $v_n =1/n^3$ ; which is convergent as $p>1$ in the power of n appearing in denominator.
Now, use Ratio test to observe that $a_n/v_n → a$ finite non zero number as $n→\infty$. Hence $a$_n also converges.