Does this sum have an analytic form?
$$\sum_{n=1}^{\infty}\frac{a^{n}\left(-1\right)^{n}}{n\sqrt{n+1}}$$
Numerically, it appears to only converge on $-1 < a \le 1$
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Not trying to sound like a challenger, but what leads you to believe the sum may have a closed-form expression? – Alann Rosas Oct 02 '20 at 06:45
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Mathematica throws its hands up even when $a=1$. If you replace $n$ with $n+1$, it's $a \mathrm{LerchPhi}[a, 3/2, 2]$. – Joshua P. Swanson Oct 02 '20 at 06:51
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@A.E.Rosas Just hope. To paraphrase Dazed and Confused, "You got an analytic form? It'd be a lot cooler if you did..." – Jerry Guern Oct 02 '20 at 20:09
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@JerryGuern I see. That’s always an acceptable reason :) – Alann Rosas Oct 03 '20 at 04:29
2 Answers
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For analytic continuation purposes, we have (for $|a|\color{red}{\leqslant}1$) $$\sum_{n=1}^\infty\frac{(-a)^n}{n\sqrt{n+1}}=-a\int_0^\infty\frac{\operatorname{erf}\sqrt{t}}{a+e^t}\,dt,$$ with the RHS being analytic on $\color{blue}{a\in\mathbb{C}\setminus\mathbb{R}_{\leqslant-1}}$. The equality is obtained using $$\int_0^\infty e^{-st}\operatorname{erf}\sqrt{t}\,dt=\frac{1}{s\sqrt{s+1}}\quad(\Re s>0)$$ (this Laplace transform may be proved for $\Re s>1$ using the power series for $\operatorname{erf} z$ and termwise integration, and extended to $\Re s>0$ analytically; another, perhaps better, alternative is just integration by parts) and the geometric series $-a/(a+e^t)=\sum_{n=1}^\infty(-ae^{-t})^n$.
metamorphy
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As noted, Mathematica does not provide a closed form.
Plotting the numerical results for different values of $a,$ it seems very close to
$$
k(1-\sqrt[3]{1+a})
$$
where $k=2.18401...$ is the value of the sum for $a=-1:$
Vincenzo Tibullo
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I like your approach to math. But I need to analytically continue this thing, so an in-domain approximation doesn't quite do the job. Thanks though! – Jerry Guern Oct 02 '20 at 20:06