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I have come across the series:

$$\sum_{j=1}^\infty \frac{x^j (j-1)}{j! \sqrt{2j - 1}}$$

which is easily seen to be absolutely convergent everywhere (e.g. ratio test). It seems that it should be very close to $\exp(x)$ and I would like to characterize it exactly in terms of simple functions if possible. Does anyone have ideas?


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Are you interested in $x\to\infty$? – Fabian Aug 23 '12 at 14:42
Yes, actually! Anything you can say then? – Jon Aug 23 '12 at 15:48
More specifically, I would like to show that: $\frac{\exp(-x)}{x} \sum_{j=1}^\infty \frac{x^j (j-1)}{j! \sqrt{2j - 1}} \rightarrow 0$ as $x \rightarrow \infty$ – Jon Aug 23 '12 at 15:50
It seems that this series is close to $\sqrt{x}\exp(x)$ – Norbert Aug 23 '12 at 17:40
Note that the derivative has the slightly nicer form $$ \begin{align} &\sum_{j=1}^\infty\frac{x^{j-1}}{(j-2)!\sqrt{2j-1}} \\ =&\sum_{k=1}^\infty\frac{x^k}{(k-1)!\sqrt{2k+1}} \\ =&x\sum_{n=0}^\infty\frac{x^n}{n!\sqrt{2n+3}}\;. \end{align} $$ – joriki Aug 23 '12 at 18:57

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