Do the convergent sum


posses a closed form? ($a \in \mathbb{R}$)

Special case is known, for $a=0$ one recalls well known alternating harmonic series :

$$\sum_{n=1}^{\infty}\frac{(-1)^n}{n}=-\ln 2$$

  • $\begingroup$ I highly doubt it and that is in part because the square root of a quadratic term is linear and so we have a conditionally convergent series that can have many sums depending on the arrangement of terms. That includes your stated example. But it is a good post though, +1 $\endgroup$ – imranfat Jul 21 '16 at 2:34
  • $\begingroup$ the sum can be written as $$\frac{1}{a\Gamma(1/2)}\int_0^\infty \frac{e^{-x}}{\sqrt{x}} F(\frac{x}{a^2}),$$ with $$F(x) = \sum_{n=1}^{\infty} \exp(-x n^2) (-1)^n$$. The graph of $F(x)$ is: i.stack.imgur.com/enPeb.png. I think a nice asymptotic expansion can be derived from the above, that could provide high accuracy with a small number of terms. Hope this helps. $\endgroup$ – Chip Jul 21 '16 at 3:08
  • $\begingroup$ According to Wolfram Alpha, in my comment above $F(x)=(\theta_4(0,e^{-x})-1)/2$, where $\theta_4$ is the Jacobi theta function of order $4$. $\endgroup$ – Chip Jul 21 '16 at 5:49
  • $\begingroup$ @imranfat why would that ever stop us? $\endgroup$ – Simply Beautiful Art Dec 28 '16 at 23:36
  • $\begingroup$ @SimpleArt (late response,back from holidays). Stop us from what? $\endgroup$ – imranfat Jan 9 '17 at 2:08

As we are considering a function of $a$, it is always useful to build a Taylor series for it.

Unfortunately, in this case it's possible only when $|a|<1$, which we are going to assume here.

Using the binomial series and some identities, we can expand the radical as:

$$\frac{1}{\sqrt{n^2+a^2}}=\frac{1}{n} \sum_{k=0}^\infty \frac{(-1)^k (2k)!}{k!^2} \left( \frac{a}{2n} \right)^{2k}$$

Now we need to find a closed form for the following series, which is well known:

$$\sum_{n=1}^\infty \frac{(-1)^n}{n^{2k+1}}=-\left(1-\frac{1}{2^{2k}} \right) \zeta(2k+1)$$

We need to carefully separate the case with $k=0$ so we don't need to deal with divergencies. Finally we have:

$$\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n^2+a^2}}=- \log 2 -\sum_{k=1}^\infty \frac{(-1)^k (2k)!}{k!^2}\left(1-\frac{1}{2^{2k}} \right) \zeta(2k+1) \left( \frac{a}{2} \right)^{2k}$$

Introducing a new function:

$$g(y)=\sum_{k=1}^\infty \frac{(-1)^k (2k)!}{k!^2} \zeta(2k+1) y^{2k}$$

We can write:

$$\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n^2+a^2}}=- \log 2 -g \left( \frac{a}{2} \right)+g \left( \frac{a}{4} \right)$$


Using the integral form for the zeta function we can write:

$$\zeta(2k+1)=\frac{1}{(2k)!} \int_0^\infty \frac{x^{2k}}{e^x-1}dx$$

Now after summation the function under the integral has a closed form in terms of the Bessel function:

$$g(y)=\int_0^\infty \frac{J_0 (2 yx)-1}{e^x-1}dx$$

Which makes it possible to write the original series neatly as:

$$\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n^2+a^2}}=- \log 2 -\int_0^\infty \frac{J_0 (a x)-J_0 (a x/2)}{e^x-1}dx$$

What's more important, this formula works for $|a|>1$ as well, which can be justified by analytic continuation of the Taylor series.


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