I want to investigate the convergence or divergence of the following series:$$\sum _{n=0}^{\infty }\:\frac{\left(n!\right)^2}{\left(2n\right)!}(-4)^n$$
However, the alternating series test fails because the absolute value of the ratio of successive is
$$\frac{4n+4}{4n+2}$$
which means the sequence $\{|a_n|\}$ is increasing and so the alternating series test does not apply.
I have proven that this series does not converge absolutely, because
$$n\frac{a_n}{a_{n+1}}-n=\frac{-2n}{4n+4}\leq 0$$
so $\sum^\infty|a_n|$ diverges by Raabe's test.
But how do I know if the series might still converge conditionally?
I also tried the root test, and it turns out that
$$ \limsup_{n\rightarrow\infty}\sqrt[n]{|a_n|}=\lim_{n\rightarrow\infty}\sqrt[n]{\frac{4^n\left(n!\right)^2}{\left(2n!\right)}}=1$$ so that test was also inconclusive. What else can I try?
Best Answer
If $\lvert a_n\rvert$ is increasing and nonzero, then $a_n\not \to 0$, and therefore $\sum_{n=0}^\infty a_n$ does not converge.