Question

\[\sum_{n=1}^{∞}1/n!\]

Answer 1

We know that:\[e^x=\sum_{n=0}^\infty \frac{x^n}{n!}\]

Letting x=1 gives:
\[e^1=\sum_{n=0}^{\infty}\frac{1^n}{n!}\]

Therefore:
\[e-1=\sum_{n=1}^{\infty}\frac{1}{n!}\]

Answer 2

so about 1 .718

Answer 3

yep, thats right.

Answer 4

thanks

Answer 5

\[so, -∑(iπ)^n/n!=1 ?\]

Answer 6

yes that is correct lolol