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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.

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

4 Answers 4

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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.

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

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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 $(n-1)^2 \ge 2$.

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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).

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