Sorry if this is a silly question, but I'm trying these types of questions for the first time today.

I have a series and I'm told to check if it's convergent or divergent. The series is:

$1+\frac{2^2}{2!}+\frac{3^2}{3!}+... = \sum_{k=1}^\infty\frac{k^2}{k!}$

I simplified the nth term as $a_n=n/(n-1)!$

Now I'm not sure how to proceed further. Not allowed to use D'Alembert's ratio test yet.

Any help is appreciated.

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    $\begingroup$ Note that for $n\geq 2$, $\frac{n}{(n-1)!}=\frac{1}{(n-2)!}+\frac{1}{(n-1)!}$ $\endgroup$ – Robert Z Sep 1 '19 at 17:50
  • $\begingroup$ can u use the Stirling formula for n! for large n? $\endgroup$ – trula Sep 1 '19 at 17:56
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    $\begingroup$ Which convergence test do you know? $\endgroup$ – Robert Z Sep 1 '19 at 18:19
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    $\begingroup$ Using ratio test, you must be getting that ratio goes to zero as $n$ gets large. This leads to convergence of the series. $\endgroup$ – Aniruddha Deshmukh Sep 1 '19 at 18:23
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    $\begingroup$ In math there are no silly questions. $\endgroup$ – DanielWainfleet Sep 1 '19 at 19:13

I think your second test is known to me as the Limit Comparison Test. Let's use that. I assume you know already that the series $\sum_{k=1}^\infty \frac 1{k!}$ converges. Let $a_k = \frac{k^2}{k!}$. For $k \ge 2$, let $b_k = \frac{1}{(k-2)!}$. Then we have the following.

$$\lim_{k\to\infty}\frac{a_k}{b_k} = \frac{k^2}{k!}\cdot\frac{(k-2)!}{1} = \lim_{k\to\infty} \frac{k^2}{k(k-1)} = 1.$$

It's easy to see that $\sum b_k$ converges, so by the Limit Comparison Test, $\sum a_k$ does too.

  • $\begingroup$ Thank you. I think I got it now. $\endgroup$ – Arka Seth Sep 2 '19 at 6:40

Comparison Test.

For $n\ge 4$ we have $$0<a_n=\frac {1}{1-\frac {1}{n}}\cdot \frac {1}{(n-2)!}<$$ $$<2\cdot \frac {1}{(n-2)!}\le$$ $$\le 2\cdot \frac {1}{2^{n-3}}=\frac {16}{2^n}.$$

  • $\begingroup$ Thank you for the answer. $\endgroup$ – Arka Seth Sep 2 '19 at 6:40

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