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)?


1 Answer 1


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 $$ \frac1{\sqrt 2} \sum_{m=1}^M \frac1{\sqrt m} - \frac1{\sqrt 2} \sum_{m=3}^{M+2} \frac1{\sqrt m} = \frac1{\sqrt 2} \left( 1+\frac1{\sqrt 2}\right) + o(1). $$

  • 3
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.