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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?

Cheers.

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    Are you interested in $x\to\infty$?2012-08-23
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    Yes, actually! Anything you can say then?2012-08-23
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    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$2012-08-23
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    It seems that this series is close to $\sqrt{x}\exp(x)$2012-08-23
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    Note that the derivative has the slightly nicer form $$ \begin{align} &\sum_{j=1}^\infty\frac{x^{j-1}}{(j-2)!\sqrt{2j-1}} \\ =&\sum_{k=1}^\infty\frac{x^k}{(k-1)!\sqrt{2k+1}} \\ =&x\sum_{n=0}^\infty\frac{x^n}{n!\sqrt{2n+3}}\;. \end{align} $$2012-08-23
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    Thanks for the comments so far people. Numerically, it seems it's even true that, as $x \rightarrow \infty$: $$\exp(-x) \sum_{j=1}^\infty \frac{x^j (j-1)}{j! \sqrt{2j - 1}} \rightarrow 0$$ Although the weaker result I mentioned above suffices for my purposes. joriki's comment above almost helps. I can use it to show that the original sum is bounded above by $1 + exp(x)(x-1)$ for positive $x$. Unfortunately this is not quite enough for either of the two limiting results.2012-08-23
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    I can now show at least the first limit which is all I need for my purposes: $\frac{\exp(-x)}{x} \sum_{j=1}^\infty \frac{x^j (j-1)}{j! \sqrt{2j - 1}} \rightarrow 0$ as $x \rightarrow \infty$. Note that for $j \geq 1$, $\frac{j-1}{\sqrt{2j - 1}} \leq \sqrt{j}$. So $\frac{\exp(-x)}{x} \sum_{j=1}^\infty \frac{x^j (j-1)}{j! \sqrt{2j - 1}} \leq \frac{1}{x} \mathbb{E}[\sqrt{Y}]$ where $Y$ is a Poisson random variable with mean $x$. By Jensen's inequality this is bounded above by $\sqrt{x} / x = 1 / \sqrt{x}$ and the limit follows.2012-08-29

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