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I am trying to find the sum of the series from $n=2$ to infinity of $\frac{ 1}{n(n-2)!}$

I was originally given: 1/2 + 1/3 + 1/8 + 1/30 + 1/144 ... ? and I had to figure out the next term (which is 1/840) and then figure out the sum of the series. I deduced what the formula is to make the series but I have no clue how to find the sum. (Brute forcing the sum on a calculator gave me a number very very close to 1 so I'm guessing that's the sum, but I don't know how to find that sum using calculus) Please help!

imranfat
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Kaiproberts
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4 Answers4

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This is $$\sum_{n=2}^\infty\frac{(n-1)}{n!} =\sum_{n=2}^\infty\frac{n}{n!}-\sum_{n=2}^\infty\frac{1}{n!} =\sum_{n=2}^\infty\frac{1}{(n-1)!}-\sum_{n=2}^\infty\frac{1}{n!} =\sum_{m=1}^\infty\frac{1}{m!}-\sum_{n=2}^\infty\frac{1}{n!} $$ etc.

Angina Seng
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  • That's a brilliant way of writing it! The part I don't get though is how to find the sum. Normally, ∑n=0->∞ of 1/n! = e In this case though, we are starting at 1 and 2 respectively. – Kaiproberts May 05 '18 at 02:41
  • @Kaiproberts These are the series for $e$, but in each case with a few terms missing.... – Angina Seng May 05 '18 at 02:43
  • I was being dumb. I just subtracted out the missing terms and got the answer to be 1, thank you so much! – Kaiproberts May 05 '18 at 02:45
  • Is not there a small erratum in the term $-\sum\limits_2^\infty\frac{n}{n!}$, there should be 1 instead of n in the nominator? – JV.Stalker May 05 '18 at 02:54
  • Yeah, I corrected for that on my whiteboard i'm working with and it worked out. Probably just an error from typing too fast. – Kaiproberts May 05 '18 at 03:10
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You already accepted the answer, which is fine. Here is just another way of looking at it:

$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....=\sum_{n=0}^\infty\frac{x^n}{n!} $$ Now when you divide everything by $x$ and then take derivative on both sides, you end up with: $$\frac{xe^x-e^x}{x^2}=\frac{-1}{x^2}+1+\frac{x}{2!}+\frac{x^2}{3!}+\frac{x^3}{4!}....=\sum_{n=0}^\infty\frac{(n-1)x^{n-2}}{n!}$$ When you plug in $x=1$ and move over $\frac{-1}{x^2}$ to the left side, you arrive at your desired sum, which is indeed $1$ as you suggested.

imranfat
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  • I think what threw me off from going this way was that the sum started at 2 instead of 0 for n. I really appreciate this way of looking at it though. – Kaiproberts May 05 '18 at 03:11
  • When you take the derivative, the "first" term is gone, the $-1/x^2$ is moved to the left, which makes the series start with $n=2$...Hope it helps – imranfat May 05 '18 at 03:13
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    Ohhhh, I didn't know that. I'll have to use this for more problems. The deeper I get into math, the more fun it gets. – Kaiproberts May 05 '18 at 03:18
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$$\dfrac{n-1}{n!}=f(n-1)-f(n)$$ where $f(x)=\dfrac1{(x-1)!}$

$$\sum_{n=2}^m\dfrac{n-1}{n!}=f(1)-f(m)$$

Now $\lim_{m\to\infty}f(m)=?$

See also: How to solve this summation without taylor?

lab bhattacharjee
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$$\frac{1}{3}+\frac{1}{8}+\frac{1}{30}+\frac{1}{144}+\frac{1}{840}+\cdots=\frac{1}{2}$$

$$\frac{1}{2}+\frac{1}{3}+\frac{1}{8}+\frac{1}{30}+\frac{1}{144}+\cdots=\frac{1}{2}+\frac{1}{2}=1$$

Lê Thành Đạt
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LeverDuJour
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    Welcome to Math.SE! Please elaborate with an [edit] how you you have come to this conclusion. At the moment, it is hard to understand your reasoning. – Brian61354270 Feb 10 '20 at 14:24