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

[Ashwin_Kumar]

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?

 

[micromass]

Hi Ashwin Kumar!

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?

 

[Jaynte]

\sum_{n=0}^\infty \frac{n}{n!}=\sum_{n=1}^\infty\frac{1}{(n-1)!}
Gives:
<br /> \sum_{n=0}^\infty \frac{n}{n!}=\sum_{n=1}^\infty\frac{1}{(n-1)!}<br />
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 (n-1)!=\prod_{k=1}^{n-1}k might help

 

[Mute]

\sum_{n=0}^\infty \frac{n}{n!}=\sum_{n=0}^\infty\frac{1}{(n-1)!}
Gives:
<br /> \sum_{n=0}^\infty \frac{n}{n!}=\sum_{n=0}^\infty\frac{1}{(n-1)!}<br />
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 (n-1)!=\prod_{k=1}^{n-1}k 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.

 

[Jaynte]

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

 

[Ashwin_Kumar]

Ok i'll take a look at the taylor series expansion

 

[dimension10]

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.

 

[dimension10]

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.).