real analysis - The series $sumlimits_{k=1}^{infty} frac1{sqrt{{k}{(k^2+1)}}}$

Given the series

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

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

Answer

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.