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$$\sum_{n=1}^{\infty}(-1)^n\frac{3^{n^2}}{(n!)^3}$$

We first absolutely convergence:

$$\big|(-1)^n\frac{3^{n^2}}{(n!)^3}\big|\leq \frac{3^{n^2}}{(n!)^3}$$

We first check $a_n\to 0$

Can we say that if $$\frac{a_{n+1}}{a_n}>1$$

We can conclude that $a_n\not \to 0$?

Masacroso
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newhere
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  • @Masacroso $$a_n = n^{-2}$$ implies $$\frac{a_{n+1}}{a_n} = \frac{n^2}{(n+1)^2},$$ not $(n+1)^2/n^2$ as you have written. – heropup Feb 17 '20 at 17:40
  • @heropup yes, my bad, I was reading the ratio wrongly – Masacroso Feb 17 '20 at 17:41
  • Yes. There's also a related theorem involving limits. Take a look at https://math.stackexchange.com/questions/2971407/proof-of-dalemberts-ratio-test-for-sequences-tending-to-infinity or at https://math.stackexchange.com/questions/2516157/why-does-the-ratio-test-prove-that-this-particular-sequence-converges-on-0 – bjorn93 Feb 17 '20 at 19:11

3 Answers3

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If you define $$a_n = (-1)^n \frac{3^{n^2}}{(n!)^3},$$ then $$\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{3^{2n+1}}{(n+1)^3}\right|.$$ Then by the ratio test, the series is not absolutely convergent as this exceeds $1$.

heropup
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3

We have since $n!<n^n$, $$\frac{3^{n^2}}{(n!)^3}>\exp(n^2\ln(3)-3 n \ln(n))\to\infty$$ so the terms don't converge to $0$ and hence the series is divergent by the term test.

Maximilian Janisch
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2

Yes, because then$$(\forall n\in\mathbb N):a_n>a_1^{\,n}$$and $a_1>1$.

José Carlos Santos
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