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I know the alternating series $$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^3}$$ converges by doing the alternating series test, but I can't seem to find a way to prove $$\sum_{n=1}^{\infty} \frac{1}{n^3}$$ converges. I tried it with $\lim_{n \to +\infty}\sqrt[n] \frac{1}{n^3} $ and $\lim_{n \to +\infty}\frac{a_{n+1}}{a_n}$, but both of these result in 1, which means you can't conclude anything from them.

Any help would be appreciated!

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    $\begingroup$ The corresponding integral converges. $\endgroup$
    – lulu
    Jan 17, 2021 at 16:04
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    $\begingroup$ Hint: For $n\geq 2$, $$ \frac{1}{{n^3 }} < \frac{1}{{n^2 }} < \frac{1}{{(n - 1)n}} = \frac{1}{{n - 1}} - \frac{1}{n}. $$ $\endgroup$
    – Gary
    Jan 17, 2021 at 16:05
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    $\begingroup$ Now you've learned several solutions, this may interest you. $\endgroup$
    – J.G.
    Jan 17, 2021 at 16:07
  • $\begingroup$ In what textbook of calculus is this an exercise? The geometric series and p-series are standard examples when talking about infinite series. $\endgroup$
    – user9464
    Jan 17, 2021 at 16:17

3 Answers 3

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$\displaystyle \sum_{n=1}^{\infty}{\dfrac{1}{n^3}}$ converges by the P-Series Test.

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Since $\displaystyle\int_1^\infty\frac1{x^3}\,\mathrm dx$ converges, your series converges too, by the integral test.

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I thought it might be instructive to present an approach the relies on straightforward elementary analysis only. To that end, we now proceed.

Note that we can write

$$\begin{align} \frac1{(n-1)^2}-\frac1{n^2}&=\frac1{n^2}\left(\frac1{\left(1-\frac1n\right)^2}-1\right)\\\\ &\ge \frac2{n^3}\tag1 \end{align}$$

Summing the telescoping sum and applying the inequality in $(1)$ reveals

$$\begin{align} 1-\frac1{N^2}&=\sum_{n=2}^N \left(\frac1{(n-1)^2}-\frac1{n^2}\right)\\\\ &\ge 2\sum_{n=2}^N \frac1{n^3}\tag2 \end{align}$$

Therefore, we see that

$$\sum_{n=1}^N\frac1{n^3}\le \frac32-\frac1{2N^2}$$

whereupon application of the squeeze theorem we see that the series of interest converges to a value of less than $3/2$.

TOOLS USED: Telescoping series, the squeeze theorem, elementary inequalities

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  • $\begingroup$ I don't really know how I could improve your answer, it seems fine to me, it's nice to see someone using a different approach than normal. $\endgroup$
    – Fooourier
    Jan 28, 2021 at 1:04
  • $\begingroup$ Pleased to hear! $\endgroup$
    – Mark Viola
    Jan 28, 2021 at 2:09

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