How can the following infinite series be evaluated to a closed form?:

$$ \sum_{l=1}^\infty \left(1 -\frac{l}{\sqrt{l^2 + a^2}}\right) $$

I tried to evaluate this using Mathematica, but it was not able to find a closed form.

  • 2
    $\begingroup$ Do you have a reason to believe that a closed form exists? $\endgroup$ – Arthur Apr 20 '20 at 17:00
  • $\begingroup$ If yes is the response to @Arthur, do you try integral comparison ? (it seems like hyperbolic functions. $\endgroup$ – EDX Apr 20 '20 at 17:08
  • $\begingroup$ No, I don't have a reason to believe a closed form exists, but I also don't know how to tell if a series has a closed form. $\endgroup$ – Libby Apr 20 '20 at 18:01
  • $\begingroup$ Looks like: $\sum _{l=1}^{\infty } \left(1-\frac{l}{\sqrt{l^2+a^2}}\right)=\int_0^{\infty } \frac{a J_1(a x)}{\exp (x)-1} \, dx$ where: $J_1$ is Bessel function of the first kind. $\endgroup$ – Mariusz Iwaniuk Apr 20 '20 at 18:49
  • $\begingroup$ @MariuszIwaniuk How do you show that? $\endgroup$ – Libby Apr 20 '20 at 21:13

$$ \frac{\ell}{\sqrt{\ell^2+a^2}}=\frac{1}{\sqrt{1+\frac{a^2}{\ell^2}}}=\sum_{n=0}^{+\infty}\frac{(2n)!a^{2n}}{4^n(n!)^2\ell^{2n}} $$ Thus $$ \sum_{\ell=1}^{+\infty}\left(1-\frac{\ell}{\sqrt{\ell^2+a^2}}\right)=\sum_{\ell=1}^{+\infty}\sum_{n=1}^{+\infty}\frac{(2n)!a^{2n}}{4^n(n!)^2\ell^{2n}}=\sum_{n=1}^{+\infty}\sum_{\ell=1}^{+\infty}\frac{(2n)!a^{2n}}{4^n(n!)^2\ell^{2n}}=\sum_{n=1}^{+\infty}\frac{(2n)!a^{2n}}{4^n(n!)^2\zeta(2n)} $$ The convergence is way faster when $a\in(-1,1)$.

  • $\begingroup$ And of course $\zeta(2n)$ can be expressed using Bernoulli numbers. But it's still an infinite sum, so not closed form. $\endgroup$ – Robert Israel Apr 20 '20 at 17:10

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