Summation problem

  • #1

Main Question or Discussion Point

Hello, i have been trying to solve a summation question for a while, and i'm not too much of an expert at the subject, so i couldn't figure it out. it is the sum of n/n! in which n takes the value from one to infinity. In other words, just 1/1! + 2/2! + 3/3! + 4/4!....

Well, firstly, does the series diverge or converge? i think it converges and has a limit. In the end i came up with this


lim 1/((n-1)!)
n->infin

PS Could someone tell me how to write limits here?
 

Answers and Replies

  • #2
22,097
3,282
Hi Ashwin Kumar! :smile:

Limit can be written as \lim_{n\rightarrow +\infty}{ ... } between [ itex ] and [ /itex] tags (without spaces).

Anyway, do you know the Taylor series of the exponential function?
 
  • #3
79
0
\sum_{n=0}^\infty \frac{n}{n!}=\sum_{n=1}^\infty\frac{1}{(n-1)!}
Gives:
[tex]
\sum_{n=0}^\infty \frac{n}{n!}=\sum_{n=1}^\infty\frac{1}{(n-1)!}
[/tex]
when used with tex, use the advance button to see all math things you can do

So far it is correct and it does converge, how to solve it I don't know. Maybe use Z-transform to solve it.

Also [tex](n-1)!=\prod_{k=1}^{n-1}k[/tex] might help
 
Last edited:
  • #4
Mute
Homework Helper
1,388
10
\sum_{n=0}^\infty \frac{n}{n!}=\sum_{n=0}^\infty\frac{1}{(n-1)!}
Gives:
[tex]
\sum_{n=0}^\infty \frac{n}{n!}=\sum_{n=0}^\infty\frac{1}{(n-1)!}
[/tex]
when used with tex, use the advance button to see all math things you can do

So far it is correct and it does converge, how to solve it I don't know. Maybe use Z-transform to solve it.

Also [tex](n-1)!=\prod_{k=1}^{n-1}k[/tex] might help
The sum should start at n=1, not n=0. The sum is also quite easy, once you recognize it. As micromass suggested, taking a look at the Taylor series expansion for e^x is quite helpful.
 
  • #5
79
0
The sum should start at n=1, not n=0. The sum is also quite easy, once you recognize it. As micromass suggested, taking a look at the Taylor series expansion for e^x is quite helpful.
it doesn't matter if n starts at 0 since n!=1 when n=0 in the first summation, but it should be n=1 in the other summation
 
Last edited:
  • #6
Ok i'll take a look at the taylor series expansion
 
  • #7
371
0
Hello, i have been trying to solve a summation question for a while, and i'm not too much of an expert at the subject, so i couldn't figure it out. it is the sum of n/n! in which n takes the value from one to infinity. In other words, just 1/1! + 2/2! + 3/3! + 4/4!....

Well, firstly, does the series diverge or converge? i think it converges and has a limit. In the end i came up with this


lim 1/((n-1)!)
n->infin

PS Could someone tell me how to write limits here?
The answer is e.
 
  • #8
371
0
you simply need the e^x expansion (this is also a way to find e to the power of pi i +1but sin x+i cos x is more useful since the sine disapears and the i cos x is -1.).
 

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