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While working on some research, I have encountered an infinite series and its improper integral analogue:

\begin{align}\sum_{m=1}^{\infty}\frac1{\sqrt{m(m+1)(m+2)+\sqrt{m^3(m+2)^3}}}&=\frac12+\frac1{\sqrt{2}}, \\ \int_0^{\infty}\frac{dx}{\sqrt{x(x+1)(x+2)+\sqrt{x^3(x+2)^3}}}&=2.\end{align} The evaluations were guessed using numerical evidence.

Can you provide proofs, or any reference (if available)?

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1 Answer 1

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For the integral, notice that the expression under the square root is $$ x(x+1)(x+2)+x(x+2)\sqrt{x(x+2)} = \frac12\,x(x+2)(\sqrt x+\sqrt{x+2})^2. $$ Consequently, \begin{align*} \frac1{\sqrt{x(x+1)(x+2)+x(x+2)\sqrt{x(x+2)}}} &= \frac{\sqrt 2}{(\sqrt x+\sqrt{x+2}) \sqrt{x(x+2)}} \\ &= \frac1{\sqrt 2}\,\frac{\sqrt{x+2}-\sqrt{x}}{\sqrt{x(x+2)}} \\ &= \frac1{\sqrt 2} \left( \frac1{\sqrt x}-\frac1{\sqrt{x+2}}\right); \end{align*} thus, the indefinite integral is $$ \sqrt{2}\, (\sqrt x-\sqrt{x+2})+C $$ and the result follows easily.

As Antony Quas noticed, this also works for the sum showing that the partial sum over $m\in[1,M]$ is $$ \frac1{\sqrt 2} \sum_{m=1}^M \frac1{\sqrt m} - \frac1{\sqrt 2} \sum_{m=3}^{M+2} \frac1{\sqrt m} = \frac1{\sqrt 2} \left( 1+\frac1{\sqrt 2}\right) + o(1). $$

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    $\begingroup$ Sounds as though this might take care of the sum also... $\endgroup$ Commented Dec 24, 2016 at 20:38
  • $\begingroup$ Nice and clean. $\endgroup$ Commented Dec 24, 2016 at 20:39

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