Question

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.

Answer 1

I think I got some idea :D

Answer 2

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

Answer 3

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.

Answer 4

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
\(\sum_{n=0}^\infty \frac{1}{n!(n+a)(n+b)}\)

Answer 5

do you think this is something nice?

Answer 6

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

Answer 7

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

Answer 8

It is \(3/2\).

Answer 9

get outa dodge.

Answer 10

I made a mistake. You are right satellite. It is \(e/2\).

Answer 11

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

Answer 12

really? that was a total guess. really.

Answer 13

i just wrote it to say something.

Answer 14

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

Answer 15

commutant how did you get it?