Question

Calculate
\[\sum_{n=1}^\infty \frac{n-1}{n!}\]

Answer 1

0

Answer 2

it has to be positive though

Answer 3

Can't be zero since it is a positive series starting from \(n=2\)

Answer 4

I can't prove it, but the sum seems to approach one, in that every successive sum is (n!-1)/n!

Answer 5

Well observation is half of the proof thebestpig :). You just need to push a little bit more :D.

Answer 6

gimme a second - i'll do an induction thing

Answer 7

the base step is
\[(1!-1)/1! = 0\]
which is in fact the first term, and
\[\frac{(n!-1)}{n!}+\frac{n}{(n+1)!}=\frac{(n!-1)(n+1)+n}{(n+1)!}=\frac{(n+1)!-n-1+n}{(n+1)!}\]
\[=\frac{(n+1)!-1}{(n+1)!}\]
and it looks like we're done.

Answer 8

oh wait one more thing

Answer 9

\[\lim_{n \rightarrow \infty} \frac{n!-1}{n!}=1\]

Answer 10

Awesome! :)

Answer 11

We can also using telescoping sum. Note that
\[\frac{n-1}{n!}=\frac{1}{(n-1)!}-\frac{1}{n!}\]

Answer 12

ha, /facepalm/

Answer 13

well, I enjoy using induction :)

Answer 14

Here's another way: a little relabeling gives (check it!)
\[
\sum_{n=1}^\infty \frac{n-1}{n!} = \sum_{n=0}^{\infty}\frac{1}{(n+2)n!}
\]
Now
\[
e^x = \sum_{n=0}^\infty \frac{x^n}{n!},
\]
so
\[
xe^x = \sum_{n=0}^\infty \frac{x^{n+1}}{n!},
\]
so
\[
\int xe^x = xe^x-e^x +C = \sum_{n=0}^\infty \frac{x^{n+2}}{(n+2)n!}
\]
Plug in \(x=0\) to find that \(C =1\), so
\[
1 e^1 -e^1 +1 = 1 = \sum_{n=0}^\infty \frac{1}{(n+2)n!}.
\]