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How can I compute this sum?

$\sum_{k=0}^\infty\left[\frac{(2k+1)!}{(2k+b)}-\frac{1}{(2k)!(2k+b)}-\frac{(2k)! }{(2k+b+1)}+\frac{1}{(2k+1)!(2k+b+1)}\right]$

I see the $k=0$ term is equal to zero, but that's about as far as I've gotten. Thanks! $b$ is not equal to zero. If anyone has software that will do it, I would be interested in the answer, if available!!!

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I am afraid that the sum does not converge. The second and last summand tend to zero. But the first and third tend to infinity. Moreover, the ratio of the first and third summand is of the order of $k$, so the first summand is much larger for large $k$ and therefore their difference will still tend to infinity.

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  • $\begingroup$ This is a good answer, thanks. That "order $k$" statement makes a lot of sense. Also, this answer is a bummer too though :( $\endgroup$ – hodop smith Jun 9 at 15:53

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