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Do the convergent sum

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

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

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

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

Machinato
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  • 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 – imranfat Jul 21 '16 at 02:34
  • 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: http://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. – Chip Jul 21 '16 at 03:08
  • 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 $4$. – Chip Jul 21 '16 at 05:49
  • @imranfat why would that ever stop us? – Simply Beautiful Art Dec 28 '16 at 23:36
  • @SimpleArt (late response,back from holidays). Stop us from what? – imranfat Jan 09 '17 at 02:08
  • @imranfat the square root... – Simply Beautiful Art Jan 09 '17 at 11:56
  • Also see here: http://math.stackexchange.com/questions/2079389/how-does-a-mathematician-create-a-new-zeta-function/2079454#2079454 – Simply Beautiful Art Jan 09 '17 at 12:18
  • @SimpleArt. OK I see – imranfat Jan 09 '17 at 17:56
  • @SimplyBeautifulArt, in case you have a minute, would you look over my answer? I'm curious, how do I justify the final result $$\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n^2+a^2}}=- \log 2 -\int_0^\infty \frac{J_0 (a x)-J_0 (a x/2)}{e^x-1}dx$$ for $|a|>1$, as my derivation is only valid for $|a|<1$? I'm a little sketchy on the analytic continuation – Yuriy S Apr 10 '18 at 09:22
  • @YuriyS Consider changing the first step from $\frac1{\sqrt{n^2+a^2}}=\frac1n\frac1{\sqrt{1+(a/n)^2}}$ to $\frac1a\frac1{\sqrt{1+(n/a)^2}}$ – Simply Beautiful Art Apr 10 '18 at 17:52
  • @SimplyBeautifulArt, that won't work for large $n$, because for any $a$ there's $n>a$... – Yuriy S Apr 10 '18 at 20:41
  • @YuriyS Well, I don't see anything really immediate, sorry D: – Simply Beautiful Art Apr 13 '18 at 23:31
  • @SimplyBeautifulArt, thank you – Yuriy S Apr 14 '18 at 05:32

1 Answers1

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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:

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

$$|a|<1$$

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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