(05/31/2011, 03:02 AM)JmsNxn Wrote: I do not have a proof of its convergence
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\( \sum_{n=0}^{\infty} \frac{x^n}{n!} \psi_0(n+1) \)
Hm, lets see, we make a very rough estimation:
\( \psi_0(n+1) \le n-\gamma < n \)
Then the radius of convergence must be bigger than the one for replacing \( \psi_0(n+1) \) with n:
\( \sum_{n=0}^\infty x^n \frac{n}{n!} = \sum_{n=1}^\infty \frac{x^n}{(n-1)!}= x\sum_{n=1}^\infty \frac{x^{n-1}}{(n-1)!}= x\sum_{n=0}^\infty \frac{x^n}{n!} \)
But this series has infinite convergence radius, hence the same is true for the series \( \sum_{n=0}^\infty x^n \frac{\psi_0(n+1)}{n!} \).
