Is this series convergent $$\sum_{n=0}^\infty\frac{(-2)^{n^2}}{n!}\,?$$

I am supposed to find out if this series is convergent, absolutely convergent or divergent. No test has given me information. By this point i am aware that I'm supposed to simplify something but i don't know how. How do I get rid of $2^{n^2}$? Below is my attempt.

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closed as off-topic by RRL, Leucippus, user10354138, ancientmathematician, Vidyanshu Mishra Dec 7 at 7:59

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  • Recall that we always need to check the foundamental necessary condition for any convergent $\sum a_n$. – gimusi Dec 6 at 22:25
  • In order to reopen your OP, you should improve your question adding your trial by ratio test using MathJax. Do not hesitate to ask if you need some help. Bye – gimusi Dec 7 at 15:21

From $n!\le n^n$ and $2^n\gt n$ for all $n\ge1$, we have

$${2^{n^2}\over n!}\ge{2^{n^2}\over n^n}=\left(2^n\over n\right)^n\gt1$$

so $(-2)^{n^2}/n!\not\to0$, and therefore the series diverges.

HINT

Let consider

$$\lim_{n\to \infty}\left|\frac{(-2)^{n^2}}{n!}\right|=\lim_{n\to \infty}\frac{2^{n^2}}{n!}$$

  • if the limit of this is not zero then the series diverges, right? I'm not sure how to solve for that limit though, can you be more specific? – YoungDumbBroke Dec 6 at 22:30
  • apply a log to the top and bottom And consider $\ln n! = \ln n + \ln(n-1) + \cdots < n\ln n$ – Doug M Dec 6 at 22:34
  • @YoungDumbBroke Exacly, try by ratio test! – gimusi Dec 6 at 22:35
  • @gimusi i tried but i get 1/0 – YoungDumbBroke Dec 6 at 22:43
  • 1
    @YoungDumbBroke I've seen that but I don't get some step, we simply should have $$\frac{2^{(n+1)^2}}{(n+1)!}\frac{n!}{2^{n^2}}=\frac{2^{(n+1)^2-n^2}}{n+1}=\frac{2^{2n+1}}{n+1}$$ – gimusi Dec 6 at 23:01

$|(-2)^{(n+1)^2}| = |2^{(n+1)^2}| = |2^{n^2 + 2n+ 1}| = (2^{n^2})(2^{2n})(2)$

Now apply the ratio test.

  • why apply n+1 before doing the ratio test? I do get the problem solved that way but i don't understand why? – YoungDumbBroke Dec 6 at 22:40
  • The ratio test looks at $|\frac{a_{n+1}}{a_n}|$ If the ratio is greater than 1 it does not converge (conditionally or absolutely). If the ratio is less than one it converges. If it equals 1 the test is inconclusive and may converge conditionally but not absolutely. I was just trying to show that $a_{n+1}$ can be manipulated to form something that is divisible by $a_n$ – Doug M Dec 6 at 23:10

You seem to have an algebra error, using the Ratio Test, you have

$$ \lim_{n \to \infty} \left| \dfrac{\dfrac{(-2)^{(n+1)^2}}{(n+1)!}}{\dfrac{(-2)^{n^2}}{n!}} \right|= \lim_{n \to \infty} \dfrac{2^{n^2+2n+1}}{2^{n^2}} \cdot \dfrac{n!}{(n+1)!}= \lim_{n \to \infty} \dfrac{2^{n^2} \cdot 2^{2n} \cdot 2}{2^{n^2}} \cdot \dfrac{1}{n+1}= \lim_{n \to \infty} \dfrac{2^{2n+1}}{n+1}\stackrel{L.H.}{=} \lim_{n \to \infty} \dfrac{2^{2n+1} \ln 2 \cdot 2}{1}= \infty $$

  • we did not cover L.H. rule in class, so I can't use it – YoungDumbBroke Dec 6 at 23:12

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