## watchmath 5 years ago For this one I don't know the answer yet for sure Find $\sum_{n=0}\infty \frac{1}{n!(n^4+n^2+1)}$ I will be your fan if you can answer this :D.

1. watchmath

I think I got some idea :D

2. watchmath

I think the answer is $$\frac{3}{2}$$.

3. anonymous

Here's an idea. Suppose that we want to find $\sum_{n=0}^\infty \frac{1}{n!(n+a)}$ for any number $$a$$ not an integer $$\leq 0$$, even a complex one. Start with the infinite series $e^x = \sum_{n=0}^\infty \frac{x^n}{n!},$ multiply by $$x^{a-1}$$, $x^{a-1}e^x = \sum_{n=0}^\infty \frac{x^{n+a-1}}{n!}$ and integrate $\int x^{a-1}e^x = \int \sum_{n=0}^\infty \frac{x^{n+a-1}}{n!} = \sum_{n=0}^\infty \frac{x^{n+a}}{n!(n+a)}$ Find the integral on the left, then you can plug in 1 for x and get the sum (watch your choice of a constant though, should always be zero I think). In a similar manner you could find $\sum_{n=0}^\infty \frac{1}{n!(n+a)(n+b)}$ etc. I'm sure there is a better way to do the problem you posted, but this occurred to me as a possible appoach. I didn't do the computation to get $$n^4+n^2+1$$, but you could factor it over the complex numbers and go for it.

4. watchmath

How can you go from $$\int x^{a-1}e^x = \int \sum_{n=0}^\infty \frac{x^{n+a-1}}{n!} = \sum_{n=0}^\infty \frac{x^{n+a}}{n!(n+a)}$$ to $$\sum_{n=0}^\infty \frac{1}{n!(n+a)(n+b)}$$

5. anonymous

do you think this is something nice?

6. watchmath

It turns out to be nice satellite. It can make your afternoon beautiful :D.

7. anonymous

i mean say $\frac{e}{2}$ or $\frac{\pi^2}{6}$

8. watchmath

It is $$3/2$$.

9. anonymous

get outa dodge.

10. watchmath

I made a mistake. You are right satellite. It is $$e/2$$.

11. anonymous

It's $$\frac{e}{2}$$, not $$\frac{3}{2}$$

12. anonymous

really? that was a total guess. really.

13. anonymous

i just wrote it to say something.

14. anonymous

i am going to go play the lottery! what are the chances of that? pick a number. ok $\frac{e}{2}$!

15. anonymous

commutant how did you get it?