What is the sum of n^2/n from 1 to infinity?

[upsidedowntop]

Homework Statement

$\sum_1^\infty \frac{n^2}{n!} =$

The Attempt at a Solution

Context: practice Math GRE question I don't know how to answer.

Well, it's bigger than e and converges by the ratio test. Adding up the first 5 or 6 terms suggests that it converges to 2e. That's good enough for a standardized test, but I'd like to know how to handle this series legitimately.

Thanks in advance for any help.

 

[dalcde]

It does converge to 2e (by wolfram alpha), but I'm not sure how to derive it.

 

[SammyS]

Is $\displaystyle \sum_{n=1}^\infty \frac{n}{(n-1)!}$ any easier ?

 

[Ray Vickson]

Homework Statement

$\sum_1^\infty \frac{n^2}{n!} =$

The Attempt at a Solution

Context: practice Math GRE question I don't know how to answer.

Well, it's bigger than e and converges by the ratio test. Adding up the first 5 or 6 terms suggests that it converges to 2e. That's good enough for a standardized test, but I'd like to know how to handle this series legitimately.

Thanks in advance for any help.

n^2/n! = n/(n-1)!, so your sum = sum_{n=1..inf} n/(n-1)! = sum_{k=0..inf} (k+1)/k!. Look at (d/dx) sum_{k=0..inf} x^(k+1)/k! .

RGV

 

[upsidedowntop]

Thank you Sammy and Ray, I have it now. In case anyone is curious:

$\sum_1^\infty \frac{n^2}{n!} <br /> <br /> = \sum_0^\infty \frac{k+1}{k!}<br /> <br /> = lim_{m \to \infty} (p_m + p_{m-1})$
where $p_k = \sum_0^m 1/k!$ .
And that limit is 2e.

Ray, I see that that derivative, evaluated at 1, is equal to our series. But what good does that do?

 

[dynamicsolo]

You can look at this in stages:

What is $\frac{n}{n!}$ equal to?

Then what is $\sum_1^\infty \frac{n}{n!}$ equal to? (It will help to keep in mind the infinite series for e .)

Next, what is $\frac{n (n-1)}{n!}$ equal to? (Write it two different ways: by cancelling terms in numerator and denominator and by simply multiplying out the numerator.)

Then what would $\sum_2^\infty\frac{n (n-1)}{n!}$ equal? (It may help at any of these stages to write out the first few terms of the series.)

You can now add your result for $\sum_1^\infty \frac{n}{n!}$ to your new result to find $\sum_1^\infty \frac{n^2}{n!}$ .

(You could also use this iteratively to find a general result for $\sum_1^\infty \frac{n^p}{n!}$ .)

 

[SammyS]

Thank you Sammy and Ray, I have it now. In case anyone is curious:
...

And that limit is 2e.

Ray, I see that that derivative, evaluated at 1, is equal to our series. But what good does that do?

Ray's sum (to be differentiated) is: $\displaystyle \sum_{k=0}^\infty\frac{x^{k+1}}{k!}$

$\displaystyle =x\sum_{k=0}^\infty\frac{x^{k}}{k!}$

$=xe^x$
​

d/dx(xe^x), evaluated at x=1 is 2e .

 

[upsidedowntop]

Thanks a lot everyone, this was really helpful.

Regarding the question I just asked, which Sammy just answered, thanks. I was asking myself what function
$\displaystyle \sum_{k=0}^\infty\frac{x^{k+1}}{k!}$
was a power series for, but for some reason it wasn't clicking. I guess I'm pretty rusty with this sort of thing.