# Show that a series converge or diverge: $\sum_{n=1}^{\infty}\frac {(-1)^n(2n-1)!}{3^n}$

Check wether this converges or diverges $$\sum_{n=1}^{\infty}\frac {(-1)^n(2n-1)!}{3^n}$$

I have a couple of questions:

Using Abel-Dirichlet criteria when you have $$\sum_{k=1}^{n}b_k$$ bounded and $$a_n\to0$$ when $$n\to\infty$$ it $$\implies\sum_{n=1}^{\infty} a_n$$ converges.

Here if we take $$b_n = (-1)^n$$ we have that $$a_n\to\infty$$ as $$n\to\infty$$, can I or I cannot say that the series diverges?

Also when testing the absolute convergence of this we have that the new series are: $$\sum_{n=1}^{\infty} \frac{(2n-1)!}{3^n}$$ doing the ratio test we get:

$$\lim_{n\to\infty}\frac {(2n+1)}{3}=\infty > 1$$

So it diverges so it's not absolute convergent, here can I or I cannot say that the series diverges? probably not here.

Else what criteria should I use here?

If a serie converges its general term tends to $$0$$ (this is a necessary condition, not sufficient). Here we clearly have :

$$\lim_{n \to \infty} \frac{(-1)^n(2n-1)!}{3^n} \ne 0$$

So it diverges.

• Actually that limit doesn't exist, but it does if you take its absolute value – zhw. Nov 14 '18 at 20:24
• @DonAntonio yes of course. Thank you for noticing ! – Thinking Nov 14 '18 at 20:36

Put

$$a_n=\frac{(2n-1)!}{3^n}\implies\frac{a_{n+1}}{a_n}=\frac{(2n+1)!}{3^{n+1}}\cdot\frac{3^n}{(2n-1)!}=\frac{2n(2n+1)}3\xrightarrow[n\to\infty]{}\infty$$

and thus $$\;\lim\limits_{n\to\infty}\cfrac{(-1)^n((2n-1)!}{3^n}\neq 0\;$$ (In fact the limit doesn't even exist) and the series doesn't converge.