# Find $\sum_{n=1}^{\infty}\frac{1}{n!}$

Find $$\sum_{n=1}^{\infty}\frac{1}{n!}$$

All of the advice I've seen to compute this sum says to use the ratio test, but this is in a chapter BEFORE the ratio test, so the book wants me to solve this without the ratio test. My tools are the comparison test, the limit comparison test, and basic knowledge about geometric series and p-series.

I can't find anything to compare this function to. I wish I could type some more work, but I don't even know how to get started.

EDIT: Yes, I had a typo. It is an infinite sum.

• The sum as it stands is very simple. The summand does not depend on $i$. Thus $$\sum_{i=1}^n \frac{1}{n!} = \frac{1}{n!} \sum_{i=1}^n 1 = \frac{n }{n!} = \frac{1}{(n-1)!}$$ Are you sure you do not have a typo in the question? Commented Nov 14, 2012 at 15:49
• What you write is $$\frac{1}{n!}+...+\frac{1}{n!}=\frac{n}{n!}\,$$Is this what you really meant? Commented Nov 14, 2012 at 15:50
• So do you want to evaluate it, or establish its covergence? Commented Nov 14, 2012 at 15:51
• Determine convergence or divergence. Commented Nov 14, 2012 at 15:55
• "Determine convergence or divergence" is different from "finding" the sum, which is the subject of your question and the first word of your question. Always better when posting problems to post as close to the exact problem. Often learners misinterpret problems and end up asking questions that are very different from the problem they are actually trying to solve because they have paraphrased when they didn't understand the problem. Commented Nov 14, 2012 at 16:07

I will assume that you are taking about the infinite sum

Use induction to show that $1/n!<2^{-n}$ for all n>2

Then use the comparison test

$$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}.$$ Put $x=1$. Therefore, the answer is $e-1$.

• I didn't use any of your tests. But you wanted to know what function it is. It is $e^x$. Commented Nov 14, 2012 at 15:58

You say you want to find $\sum\limits_{n=0}^\infty\dfrac{1}{n!}$. But then in comments you say you want to find out whether it converges or not. So you're not expressing yourself clearly.

If you just want to find out whether it converges or not, you can use this comparison test: $$\frac{1}{n!} \le 2\cdot\frac{1}{2^n}.$$

Since you mention $p$-series and comparison test:

First: $$\sum_{n=1}^{\infty}\frac{1}{n^2} = \sum_{n=2}^{\infty} \frac{1}{(n-1)^2}$$ is convergent because it is a $p$ series with $p=2$.

Second $$0\leq \frac{1}{n!} = \frac{1}{1}\frac{1}{2}\dots \frac{1}{n-1}\frac{1}{n} < \frac{1}{n(n-1)} < \frac{1}{(n-1)^2}$$ for all $n\geq 2$. Hence by the comparison test $\sum_{n=2}^{\infty}\frac{1}{n!}$ is convergent, and so then is $\sum_{n=1}^{\infty}\frac{1}{n!}$.