Using Leibniz to prove convergence of $\sum\limits_{n=1}^{\infty}(-1)^{n+1}\frac{3^{n}}{3 \cdot 5 \cdot 7 \cdots (2n+1)}$

Question

$$\sum\limits_{n=1}^{\infty}(-1)^{n+1}\frac{3^{n}}{3 \cdot 5 \cdot 7 \cdots (2n+1)}.$$

I am trying to use Leibniz in order to prove that the series converges. I don't know if I am doing it correctly. Here it is.

I want to prove that it is decreasing.

$$\frac{a_{n+1}}{a_{n}}= \frac{3}{2n+3}<1 ,$$ so it is decreasing.

Then we want the $\lim\limits_{n \to \infty}a_{n}=0$. The problem is that the limit is not zero. So either I am missing something or the series does not converge.

Answer 1

Answering my own question with the help provided.

Using the ratio test we have:

$$a_{n} = (-1)^{n+1}\frac{3^{n}}{3\cdot5\cdot7\cdots(2n+1)} $$

$$a_{n+1} = (-1)^{n+2}\frac{3^{n+1}}{3\cdot5\cdot7\cdots(2n+1)\cdot(2n+3)} $$

$$\frac{a_{n+1}}{a_{n}}=\frac{-3}{2n+3} $$

$$\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_{n}}=\lim_{n\rightarrow\infty}\frac{-3}{2n+3}=\frac{3}{\infty}=0 < 1$$ Which means that the series converge.

Answer 2

As $0<a_n$ and $a_{n+1}<a_n$ then $a_n$ converges to some point $c$. From the equality $a_{n+1}=\frac{3a_n}{2n+1}$, we have the following when $n\to\infty$:

$c=0$. So we get that $a_n\to 0$.