Summation of factorials

Question

Finding the sum of following series $$\frac{1}{1!}+\frac{1}{2!} +...........+\frac{1}{n!}$$ I tried $$\frac{1}{n!}\Bigg[\frac{n!}{1}+\frac{n!}{2!}+\frac{n!}{3!}+..... .....+1\Bigg]$$ And end up getting $$\frac{1}{n!}\Bigg[(n-1)!\binom{n}{1}+(n-2)!\binom{n}{2}+........+\binom{n}{n}\Bigg]$$ After which I am stuck, But after this I am not able to solve it And hence i further want to find $$\sum_{n=3}^{2016} \frac{1}{n!}\Bigg[1-\frac{1}{n}\Bigg]$$ by using the result But I am not able to reach anywhere so far. It goes to a finite number

Answer 1

As I said, you won't be able to find a closed-form expression for these sums. Let's take $\sum_{n=0}^{2016}{\frac{1}{n!}}$ Factorials get big rather quickly. Let's say you compute the sum out to $\frac{1}{20!}$ How much are you leaving out? The next term is $\frac{1}{21!} = \frac{1}{21}\frac{1}{20!}<\frac{1}{20}\frac{1}{20!}$. The term after that is $\frac{1}{22!} = \frac{1}{21\cdot 22}\frac{1}{20!}<\frac{1}{20^2}\frac{1}{20!},$ and so on. The total error is less than $$\frac{1}{20!}\sum_{n=1}^{\infty}{\frac{1}{20^n}} = \frac{19}{20!} \approx 2.163325 \cdot 10^{-20}$$ Even computing 20 terms is probably overkill.

You can make the same sort of analysis for the other series. Add them up though 20 terms and forget it.