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)?
 A: 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.
