Convergence or divergence of given series

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.

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

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

Comparison Test.

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

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