Check the convergence of the following series
$$\sum_{n=1}^\infty \frac{(3n-2)!!!}{3^n n!}$$ and $$\sum_{n=1}^\infty (-1)^n\frac{(3n-2)!!!}{3^n n!}$$
My attempt:
I tried Ratio test. I got
\begin{align} \lim_{n\to \infty} \left| \frac{a_{n+1}}{a_{n}} \right| &= \lim_{n\to \infty} \left| \frac{\dfrac{(3n+1)!!!}{3^{n+1} (n+1)!}}{\dfrac{(3n-2)!!!}{3^n n!}} \right| \\ &= \lim_{n\to \infty} \left| \frac{\dfrac{(3n-2)!!!(3n+1)}{3^{n}\cdot 3 (n)!(n+1)}}{\dfrac{(3n-2)!!!}{3^n n!}} \right| \\ &= \lim_{n\to \infty} \left| \frac{\dfrac{(3n+1)}{3 (n+1)}}1 \right| \\ &=1. \end{align}
The test is inconclusive.
When I took Generalized Raabe’s test,
I got $$\lim_{n\to \infty}n\left(\left|\frac{3 (n+1)}{(3n+1)}\right|-1\right)=2/3.$$
This shows that series is not absolutely convergent.
From the normal Raabe’s test, it is clear that $\sum_{n=1}^\infty \frac{(3n-2)!!!}{3^n n!}$ is divergent. I don’t know how do I conclude for the case $\sum_{n=1}^\infty (-1)^n\frac{(3n-2)!!!}{3^n n!}$. Please help me.
Best Answer
Note that $$ \sum_{n=1}^\infty \frac{(3n-2)!!! x^n}{3^n n!} = -1+\sum_{n=0}^\infty \binom{-1/3}{n} (-1)^n x^n =-1+(1-x)^{-1/3}\qquad\text{for } |x|<1 . $$ When $x \to 1^-$, conclude that the series at $x=1$ diverges to $+\infty$. When $x = -1$, the series converges to $-1+2^{-1/3}$ by Abel's convergence theorem.