watchmath
  • watchmath
Calculate \[\sum_{n=1}^\infty \frac{n-1}{n!}\]
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
0
anonymous
  • anonymous
it has to be positive though
watchmath
  • watchmath
Can't be zero since it is a positive series starting from \(n=2\)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
I can't prove it, but the sum seems to approach one, in that every successive sum is (n!-1)/n!
watchmath
  • watchmath
Well observation is half of the proof thebestpig :). You just need to push a little bit more :D.
anonymous
  • anonymous
gimme a second - i'll do an induction thing
anonymous
  • anonymous
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.
anonymous
  • anonymous
oh wait one more thing
anonymous
  • anonymous
\[\lim_{n \rightarrow \infty} \frac{n!-1}{n!}=1\]
watchmath
  • watchmath
Awesome! :)
watchmath
  • watchmath
We can also using telescoping sum. Note that \[\frac{n-1}{n!}=\frac{1}{(n-1)!}-\frac{1}{n!}\]
anonymous
  • anonymous
ha, /facepalm/
anonymous
  • anonymous
well, I enjoy using induction :)
anonymous
  • anonymous
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 \[ 1 e^1 -e^1 +1 = 1 = \sum_{n=0}^\infty \frac{1}{(n+2)n!}. \]

Looking for something else?

Not the answer you are looking for? Search for more explanations.