Solving Summation Problem: 1/1! + 2/2! + 3/3! + 4/4!...

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Ashwin_Kumar
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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?
 
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\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:
Jaynte said:
\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.
 
Mute said:
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:
Ok i'll take a look at the taylor series expansion
 
Ashwin_Kumar said:
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.
 
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.).