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

## Regarding this sum $\sum_{n=1}^{\infty}(\frac{3n}{3n-1})^{n!}$ I need to tell if it converges or diverges. Here is my attempt: $\sum_{n=1}^{\infty}(\frac{3n}{3n-1})^{n!} = \sum_{n=1}^{\infty}(\frac{3n-1+1}{3n-1})^{n!} = \sum_{n=1}^{\infty}(1+\frac{1}{3n-1})^{n!}=\sum_{n=1}^{\infty}((1+\frac{1}{3n-1})^{3n-1})^{\frac{n!}{3n-1}}$ So the limit inside is $\to e$, but how do I compute a limit of $n!$? I don't think its possible, so I might have gone lost here. EDIT: I can say that $x(x-1) \le \frac{x!}{3x-1}$, so by toast theorem $\lim_{x \to \infty} \frac{x!}{3x-1}= \infty$, so the limit $\to e^\infty = \infty$ ? Any help? Thanks! Regarding this sum $$\sum_{n=1}^{\infty}(\frac{3n}{3n-1})^{n!}$$

I need to tell if it converges or diverges.

Here is my attempt:

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

So the limit inside is $$\to e$$, but how do I compute a limit of $$n!$$? I don't think its possible, so I might have gone lost here.

EDIT:

I can say that $$x(x-1) \le \frac{x!}{3x-1}$$, so by toast theorem $$\lim_{x \to \infty} \frac{x!}{3x-1}= \infty$$, so the limit $$\to e^\infty = \infty$$ ?

Any help?

Thanks!