watchmath
  • watchmath
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.
Mathematics
chestercat
  • chestercat
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watchmath
  • watchmath
I think I got some idea :D
watchmath
  • watchmath
I think the answer is \(\frac{3}{2}\).
anonymous
  • 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.

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watchmath
  • 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 \(\sum_{n=0}^\infty \frac{1}{n!(n+a)(n+b)}\)
anonymous
  • anonymous
do you think this is something nice?
watchmath
  • watchmath
It turns out to be nice satellite. It can make your afternoon beautiful :D.
anonymous
  • anonymous
i mean say \[\frac{e}{2}\] or \[\frac{\pi^2}{6}\]
watchmath
  • watchmath
It is \(3/2\).
anonymous
  • anonymous
get outa dodge.
watchmath
  • watchmath
I made a mistake. You are right satellite. It is \(e/2\).
anonymous
  • anonymous
It's \(\frac{e}{2}\), not \(\frac{3}{2}\)
anonymous
  • anonymous
really? that was a total guess. really.
anonymous
  • anonymous
i just wrote it to say something.
anonymous
  • anonymous
i am going to go play the lottery! what are the chances of that? pick a number. ok \[\frac{e}{2}\]!
anonymous
  • anonymous
commutant how did you get it?

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