# Convergence/divergence of the series $\sum\frac{n}{(n+1)^{2}}$

The Question Conclude the convergence/divergence of the series $$\sum\frac{n}{(n+1)^{2}}$$.

My attempt

I used derivative test to show that the function $$\frac{x}{(x+1)^{2}}$$ is a monotonically decreasing function after certain point, say $$p$$ in $$\mathbb{R}$$. Then I did the following integral $$\int_{p}^{\infty}\frac{xdx}{(x+1)^{2}} = ln(x+1)+\frac{1}{x+1}\Big|_{p}^{\infty}$$ which doesn't exist.

Is it correct? Also, is there a more elegant way to do this, the analysis done to show that the function is monotonic took a fair amount of time. Is there a way to do it by comparison test? Also on the same note, what are some series that I should keep in mind to apply comparison test handily to conclude divergence, convergence quickly.

• I woud limit-compare to $\sum_n \frac{1}{n}$. Commented May 1 at 16:08
• What does "limit-compare" mean?
– Debu
Commented May 1 at 16:10
• en.wikipedia.org/wiki/Limit_comparison_test Commented May 1 at 16:10

Your approach works. But it's way simpler to use the fact that$$\lim_{n\to\infty}\frac{\frac n{(n+1)^2}}{\frac1n}=\lim_{n\to\infty}\frac{n^2}{(n+1)^2}=1.$$Therefore, since the harmonic series diverges, your series diverges too.

• So, if two sequences are asymptotically same, the series corresponding to them, diverge/converge together?
– Debu
Commented May 1 at 16:14
• Yes, assuming that we are only dealing with real numbers greater than $0$ here. Commented May 1 at 16:16

Noticing that \begin{aligned} \sum \frac{n}{(n+1)^2} & >\sum \frac{n}{(n+n)^2} =\frac{1}{4} \sum \frac{1}{n} , \end{aligned} and $$\displaystyle \sum \frac{1}{n}$$ is divergent, therefore $$\displaystyle \sum \frac{n}{(n+1)^2}$$ is divergent too.

If you know that $$\sum \dfrac1{n}$$ diverges then $$\dfrac{n}{(n+1)^2} \ge \dfrac1{2n}$$, which is true for $$n \ge 3$$, shows that $$\sum \dfrac{n}{(n+1)^2}$$ also diverges.

To show $$\dfrac{n}{(n+1)^2} \ge \dfrac1{2n}$$ for $$n \ge 3$$, cross-multiplying gives $$2n^2\ge (n+1)^2$$ or $$n^2-2n-1\ge 0$$ or $$(n-1)^2 \ge 2$$.

Assume that $$\sum_{n=1}^\infty\frac{n}{(n+1)^2}$$ converges, so that $$\sum_{n=1}^\infty\left(\frac{n}{(n+1)^2}+\frac{1}{(n+1)^2}\right)$$ converges (this is because $$\sum_{n=1}^\infty\frac{1}{n^2}$$ converges, hence by shifting, $$\sum_{n=1}^\infty\frac{1}{(n+1)^2}$$ converges). We get that $$\sum_{n=1}^\infty\left(\frac{n}{(n+1)^2}+\frac{1}{(n+1)^2}\right)=\sum_{n=1}^\infty\frac{n+1}{(n+1)^2}=\sum_{n=1}^\infty\frac{1}{n+1}$$ converges, which is a contradiction to the divergence of the harmonic series (again, by shifting).