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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.

Sasha
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Jon
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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
  • 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.

     – Jon Aug 23 '12 at 21:05
  • 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.

     – Jon Aug 29 '12 at 14:17

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