Is $\sum_{n=1}^{\infty} \frac{n!}{2^n+1}$ convergent?

Question

Is the following series convergence? $$ \sum_{n=1}^{\infty} \frac{n!}{2^n+1} $$

I know that $ \sum_{n=1}^{\infty} \frac{n!}{n^n} $ is divergent, can I somehow use this fact? Or should I maybe try to use the ratio test?

yes, that's true, so it doesn't go to zero, so it will be divergent? — Zauberkerl, Mar 21 '17 at 18:02
@Zauberkerl and that is strong enough to conclude that the series diverges:
https://en.wikipedia.org/wiki/Term_test — David, Mar 21 '17 at 18:04
prove that $$\lim_{n \to \infty}\frac{n!}{2^n+1}$$ is not zero — Dr. Sonnhard Graubner, Mar 21 '17 at 18:05
As @Dr.SonnhardGraubner suggests consider (for $n>2$) finding a lower bound for:
$$\frac{n!}{2^n} = \frac{2!}{2^2}\cdot \frac{3}{2}\cdots\frac{n-1}{2}\cdot\frac{n}{2}$$ — David, Mar 21 '17 at 18:11
http://math.stackexchange.com/questions/111146/prove-by-induction-that-n2n it's also some kind of proof for it — Zauberkerl, Mar 21 '17 at 18:12
Your "I know that" is not correct. That sum is nicely convergent to a number somewhat less than $2$ — Ross Millikan, Mar 21 '17 at 18:28

score 2 · Accepted Answer · answered Mar 21 '17 at 18:25

As others have commented, you can use the argument that $$\lim_{n\to \infty}\frac{n!}{2^n+1}\neq 0.$$ Alternatively, using the (incorrect) fact that you've mentioned in the question, you can make the following argument $$\infty = \sum_{n=1}^\infty\frac{n!}{n^n} = 1+\sum_{n=2}^{\infty}\frac{n!}{n^n} \leq 1+\sum_{n=2}^{\infty}\frac{n!}{2^n}$$ Hence, $$\sum_{n=1}^{\infty}\frac{n!}{2^n}$$ diverges. Now since, $$\sum_{n=1}^{\infty}\frac{n!}{2^n}\sim \sum_{n=1}^{\infty}\frac{n!}{2^n+1},$$ we can conclude that the latter series must diverge as well by the limit comparison test (see:https://en.wikipedia.org/wiki/Limit_comparison_test).

Where $`\sim'$ is taken to mean that the limit of the ratio of the terms converges, $$\lim_{n\to \infty}\left\{\frac{n!/2^n}{n!/(2^n+1)}\right\} = \lim_{n \to \infty}\left\{1 + \frac{1}{2^n}\right\} = 1.$$

Now, this argument is not correct, because the 'fact' that you've mentioned is incorrect. But if it were correct, or you had another bounding series, you could use this (type of) argument.

Is $\sum_{n=1}^{\infty} \frac{n!}{2^n+1}$ convergent?

1 Answers1