Limit $\lim_{n\to\infty} \frac{n!}{n+2^n}$

Question

In an exercise I'm asked to evaluate the following limit:

$$\lim_{n\to\infty} \frac{n!}{n + 2^n}$$

I arrived at the conclusion that:

$$\frac{n!}{n + 2^n} \geq \frac{n!}{2^n}$$

So my plan is to prove that $\frac{n!}{2^n}$ diverges and as a consequence show that $\frac{n!}{n + 2^n}$ also diverges, But I'm not being able to do so.

How can I prove that $\lim n! / 2^n$ is infinity?

Surely a duplicate somewhere on this site, if you looked a little harder. But more pedagogically: how would you explain $n!$ and $2^n$ to your bright 12-year-old cousin? What does this imply about their behavior? — Integrand, Nov 24 '20 at 14:56
Also your inequality points the wrong way: bigger denominators mean smaller terms. — Integrand, Nov 24 '20 at 14:57
$n!>1\cdot 2\cdot 3\cdot 4^{n-3}=6\cdot 2^{2n-6}$. Therefore, $n!/2^n>2^{n-6}\cdot 6=\infty$ as $n\rightarrow\infty$ — W. Wongcharoenbhorn, Nov 24 '20 at 15:02
This question is partially or fully (depending on how you want to view it) answered here. — Aldoggen, Nov 24 '20 at 15:08
As pointed out already your inequality is wrong. Miguel's answer let's you know how to prove $\frac{n!}{2^n}$ diverges. Then you can use the limit comparison test to show that $\frac{n!}{n+2^n}$ also diverges. — Ameet Sharma, Nov 24 '20 at 15:45
Specifically with $2^n$ rather than $x^n$: Show that $ \lim_{n \rightarrow \infty} \frac{n!}{2^{n}} = \infty$ or How can I calculate the limit of exponential divided by factorial? — Martin Sleziak, Nov 24 '20 at 17:27

score 6 · Accepted Answer · answered Nov 24 '20 at 14:59

Notice that for $n \ge 2$, we have $\frac{n!}{2^n} = \frac{n}{2} \cdot \frac{n-1}{2} \cdot \dots \frac {2}{2} \cdot \frac{1}{2} \ge \frac{n}{2} \cdot \frac{1}{2} = \frac{n}{4}$, which diverges to $\infty$.

I'll point out that your inequality is wrong, since $n +2^n \ge 2^n$, though. You could fix this by establishing something like $n + 2^n \le 2^{n+1}$.

Neat Math · Answer 2 · 2020-11-25T00:26:44.640

2

Hint:

$$\forall n>3, \frac{n!}{n+2^n} \geqslant \frac{n(n-1)2^{n-3}}{n+2^n} = \frac{n-1}{\frac{1}{2^{n-3}}+\frac 8n} > \frac{n-1}{1+8}.$$

edited Nov 25 '20 at 00:26

answered Nov 24 '20 at 16:36

Neat Math

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Limit $\lim_{n\to\infty} \frac{n!}{n+2^n}$

2 Answers2