# The series $\sum\limits_{k=1}^{\infty} \frac1{\sqrt{{k}{(k^2+1)}}}$

Given the series

$$\sum_{k=1}^{\infty} \frac1{\sqrt{{k}{(k^2+1)}}}$$

How can I calculate its exact limit (if that is possible)?

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I think $\zeta(3/2) - \sum_{k = 1}^N \left( \frac{1}{k^{3/2}} - \frac{1}{\sqrt{k(k^2 + 1)}} \right)$ for sufficiently large N is a reasonable estimate. It gives me something along the lines of 2.2640623991416056. – TenaliRaman Jun 17 '12 at 10:53
I don't think this has a nice closed form. Is this a problem you created, or is it from a problem book? – Potato Jun 17 '12 at 10:56
@Potato: this problem comes from one of my colleagues. I don't know yet where it comes from. – user 1618033 Jun 17 '12 at 10:59
Do you know if it has a closed form solution? This seems like a pretty random sum, and we may be chasing unicorns here. – Potato Jun 17 '12 at 11:00
@Chris you can also use integral to estimate the sum: $$\int_{\frac{1}{2}}^{+\infty} \frac{dx}{\sqrt{x(x^2+1)}}$$ Mathematica gives the answer in terms of hypergeometric function: $2 \sqrt{2} \, _2 F_1 \left( \frac{1}{4}, \frac{1}{2} ; \frac{5}{4}, -4 \right) \approx 2.3261$ while the sum is around $2.2641$. – qoqosz Jun 17 '12 at 12:01

Rewrite the sum in terms of the zeta function. The resulting sum is alternating, $$\begin{eqnarray*} \sum_{k=1}^{\infty} \frac{1}{\sqrt{{k}{(k^2+1)}}} &=& \sum_{k=1}^{\infty} \frac{1}{k^{3/2}} \frac{1}{\left(1+\frac{1}{k^2}\right)^{1/2}} \\ &=& \sum_{k=1}^{\infty} \frac{1}{k^{3/2}} \sum_{j=0}^\infty (-1)^j \frac{\Gamma(j+\frac{1}{2})}{j!\Gamma(\frac{1}{2})} \left(\frac{1}{k^2}\right)^j \\ &=& \sum_{j=0}^\infty (-1)^j \frac{\Gamma(j+\frac{1}{2})}{j!\Gamma(\frac{1}{2})} \sum_{k=1}^\infty \frac{1}{k^{2j+3/2}} \\ &=& \sum_{j=0}^\infty (-1)^j \underbrace{\frac{(2j)!}{(j!)^2 2^{2j}} \textstyle\zeta(2j+\frac{3}{2})}_{a_j}. \end{eqnarray*}$$ In the second step we expanded $(1+1/k^2)^{-1/2}$ in a binomial series. The last line suggests that there is no nice closed form for the sum. Unfortunately, the convergence properties of this sum are not much better than the original. For large $j$, $a_j \sim 1/\sqrt{j}$, so the sum goes like $\sum_j(-1)^j/\sqrt{j}$.

Let's accelerate the sum using the Euler transform. Then, $$\begin{eqnarray*} \sum_{k=1}^{\infty} \frac{1}{\sqrt{{k}{(k^2+1)}}} &=& \sum_{n=0}^\infty \underbrace{(-1)^n \frac{1}{2^{n+1}} \sum_{k=0}^n (-1)^k {n\choose k} a_{n-k}}_{b_n}. \end{eqnarray*}$$ The sum $\sum_{n=0}^\infty b_n$ is not alternating, but has much better convergence properties than the original. Numerically we find $b_n \sim 1/2^n$.

Below we give the partial sums to $25$ digits. $$\begin{array}{ll} N & \sum_{n=0}^N b_n \\ \hline 1 & 1.818439778099484176707949 \\ 2 & 2.052201427955914030757063 \\ 4 & 2.214850676717845386847939 \\ 8 & 2.261274490824097920144642 \\ 16 & 2.264052732086094056012626 \\ 32 & 2.264062399012245290957591 \\ 64 & 2.264062399141221028566328 \\ 128 & 2.264062399141221028592305 \end{array}$$

Another good approach suggested by @qoqosz in the comments is to estimate the sum with an integral. That estimate can be improved somewhat by applying the Euler-Maclaurin formula. The resulting asymptotic series gets us four digits before it starts to diverge.

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that's very impressive! Thank you! :-) – user 1618033 Jun 20 '12 at 6:03
You might want to expand the argument leading to $b_n\sim1/2^n$ (at present, the invocation of de Moivre is rather mysterious). – Did Jun 20 '12 at 6:04
@Chris: Glad to help. – user26872 Jun 20 '12 at 6:05
@did: Thanks for the suggestion, I'll do that. – user26872 Jun 20 '12 at 6:06
@oen: You can also take a hybrid path that starts like yours (expanding $\frac1{\sqrt{k(k^2+1}}$ using power series) and then uses the Euler-Maclaurin Sum Formula to sum that. I did this here. – robjohn Jul 22 '12 at 21:44

Using the binomial theorem to expand \begin{align} {\small\frac1{\sqrt{k(k^2+1)}}} &=k^{-3/2}\left(1+\frac1{k^2}\right)^{-1/2}\\ &\sim k^{-3/2}-\frac12k^{-7/2}+\frac38k^{-11/2}-\frac5{16}k^{-15/2}+\frac{35}{128}k^{-19/2}-\frac{63}{256}k^{-23/2}\tag{1} \end{align} with an error of approximately $\frac14k^{-27/2}$.

Applying the Euler Maclaurin Sum Formula to $(1)$ yields \begin{align} \sum_{k=1}^n{\small\frac1{\sqrt{k(k^2+1)}}} &\sim C-2n^{-1/2}+\frac12n^{-3/2}+\frac3{40}n^{-5/2}-\frac14n^{-7/2}+\frac{31}{384}n^{-9/2}+\frac3{16}n^{-11/2}\\ &-\frac{12959}{66560}n^{-13/2}-\frac5{32}n^{-15/2}+\frac{214861}{557056}n^{-17/2}+\frac{35}{256}n^{-19/2}\\[6pt] &-\frac{24553421}{27525120}n^{-21/2}-\frac{63}{512}n^{-23/2}\tag{2} \end{align} with an error of approximately $3n^{-25/2}$. Since the series is convergent, $C$ represents the sum.

Computing the partial sum to $n=100$ and using $(2)$ to compute $C$, we get $$C\stackrel{.}{=}2.264062399141221028592305$$ Just as a check, with $n=1000$, $(2)$ gives $$C\stackrel{.}{=}2.264062399141221028592304976971183194$$

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The powerful Euler Maclaurin Sum Formula again at work! :-) – user 1618033 Jul 23 '12 at 8:37

Here is another possible approach for numerically summing this series. Through Laplace transform methods, we find that

$$\frac1{\sqrt{k(k^2+1)}}=\sqrt{\frac{\pi}{2}}\int_0^\infty \exp(-ku)\sqrt{u}\,J_{-\frac14}\left(\frac{u}{2}\right)J_{\frac14}\left(\frac{u}{2}\right)\mathrm du$$

where $J_\nu(u)$ is the Bessel function of the first kind. Substituting this into the summation expression, interchanging summation and integration, and some finagling yields the integral expression

$$\sum_{k=1}^\infty \frac1{\sqrt{k(k^2+1)}}=2\sqrt\pi\int_0^\infty \frac{\sqrt{t}}{\exp(2t)-1} J_{-\frac14}(t) J_{\frac14}(t)\mathrm dt$$

Oscillatory integrals such as the one given above can be treated as alternating series; if you look at a plot of the integrand, you can see that the integral across the zeroes of the function alternate between positive and negative signs. Thus, letting $t_k$ be the $k$-th zero of the function $J_{-\frac14}(t) J_{\frac14}(t)$, we consider the series

$$2\sqrt\pi\left(\int_0^{t_1} \frac{\sqrt{t}}{\exp(2t)-1} J_{-\frac14}(t) J_{\frac14}(t)\mathrm dt+\sum_{k=1}^\infty \int_{t_k}^{t_{k+1}} \frac{\sqrt{t}}{\exp(2t)-1} J_{-\frac14}(t) J_{\frac14}(t)\mathrm dt\right)$$

(This strategy for numerically evaluating oscillatory integrals is due to Longman.)

We can use a computing environment like Mathematica to generate the $t_k$:

\begin{array}{c|c}k&t_k\\\hline 1&2.006299672\\2&2.780887724\\3&5.123062743\\ 4&5.906142699\\5&8.257951176\\6&9.042383664\end{array}

One could use the Euler transform as in oen's answer to sum the alternating series, but the terms decrease rather quickly in magnitude that even direct summation works fine. Letting $S_k$ be the $k$-th partial sum, we have

\begin{array}{c|c}k&S_k\\\hline 1&2.2642181990016\\2&2.2632428613646\\3&2.2640625078551\\ 4&2.2640612519867\\5&2.2640623992549\\6&2.2640623973568\end{array}

The last approximation is already good to eight digits. Taking more terms of the series, we have, to twenty digits

$$\sum_{k=1}^\infty \frac1{\sqrt{k(k^2+1)}}\approx 2.26406239914122102859$$

Thirteen terms were needed to obtain this approximation.

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Very nice. I didn't think of going this way. (+1) – user26872 Jul 22 '12 at 18:32
It's nice to have so many ways to sum things numerically. Using Laplace Tranforms is neat (+1) :-) – robjohn Jul 22 '12 at 21:30
By the way, Longman's way was a new way for me, but now it's not new anymore :) – user 1618033 Jul 23 '12 at 8:40

I might as well... as mentioned by oen in his answer, the series

$$\mathscr{S}=\sum_{j=0}^\infty \left(-\frac14\right)^j\binom{2j}{j}\zeta\left(2j+\frac32\right)$$

is alternating, but rather slowly convergent. oen used the Euler transformation for accelerating the convergence of this series in his answer; for this answer, I will be using the Levin $t$-transform:

$$\mathcal{L}_n=\frac{\sum\limits_{j=0}^n (-1)^j\binom{n}{j}(j+1)^{n-1}\frac{S_j}{a_j}}{\sum\limits_{j=0}^n (-1)^j\binom{n}{j}(j+1)^{n-1}\frac1{a_j}}$$

where $a_j$ is the $j$-th term of the series, and $S_j$ is the $j$-th partial sum.

To demonstrate the superior convergence properties of the $t$-transform, consider the following evaluations:

\begin{array}{c|cc}k&\mathcal{L}_k&|\mathscr{S}-\mathcal{L}_k|\\\hline 2&2.2593704308006952815&4.692\times10^{-3}\\5&2.2640560757360687322&6.323\times10^{-6}\\10&2.2640623990222550236&1.189\times10^{-10}\\15&2.2640623991414400190&2.190\times10^{-13}\\20&2.2640623991412210238&4.828\times10^{-18}\\25&2.2640623991412210286&5.938\times10^{-21}\end{array}

Thirty terms of the series along with the $t$-transform gives a result good to twenty-five digits, compared to the $128$ terms required by the Euler transform.

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I like a lot to see so many different ways to go. However, some things here are completely new for me (Levin t-transform). Thanks! – user 1618033 Jul 23 '12 at 14:31
That's the thing about the good math problems; there's always more than one way to solve them. – J. M. Jul 23 '12 at 14:32
@J.M.: The other day I happened to be reading one of your answers that used the Levin transform and wondered if it might show up here! A very powerful method. (+1) – user26872 Jul 23 '12 at 17:47
Thanks @oen, I've been actually looking for an excuse to use the Levin transform for a while... – J. M. Jul 23 '12 at 18:06