# Why is $\sum\limits_{n=N+1}^{\infty} \frac {1}{2^n!} < \sum\limits_{n=(N+1)!}^{\infty} \frac {1}{2^n} = \frac{1}{2^{(N+1)!-1}}$

I'm reading a proof of a theorem and some applications and in the process the author used the following without any explanation of why it's true. I'm trying to understand why exactly this is true, but I have no idea how one would justify it besides saying it seems true intuitively. Any help would be much appreciated.

$$\sum_{n=N+1}^{\infty} \frac {1}{2^n!} < \sum_{n=(N+1)!}^{\infty} \frac {1}{2^n} = \frac{1}{2^{(N+1)!-1}}$$

• The equality follows since this is a geometric series. – vadim123 Apr 20 '18 at 1:14

We have: $(2^{N+k})! > 2^{(N+k)!}$ for any $k \ge 1$, and this is what you have to prove to deduce the above inequality. It is just a starting point, and could be a hint. In fact,you could prove it by taking log both sides. Thus you have to prove: $\ln\left(\left(2^{N+k}\right)!\right)= \ln 1 + \ln 2 + \cdots + \ln\left(2^{N+k}\right)> ((N+k)!)\ln (2)$. Basically, I rewrite the statement into something that is a bit easier for you to handle. Alternatively, you can prove the original inequality above by induction on $k \ge 1$,and for each of these cases, you do induction on $N\ge 1$. I think it is probably better than taking Log.