Convergence and estimation of the rests of $\sum\limits_{k=1} ^\infty \frac{1}{\sqrt{k^3+x}} - \sum\limits_{k=1} ^\infty \frac{1}{\sqrt{k^3 - x}} $ For $x\in ]-1,1[$  is required to evaluate the expression
$$
S(x) = \sum_{k=1} ^\infty \frac{1}{\sqrt{k^3+x}} - \sum_{k=1} ^\infty \frac{1}{\sqrt{k^3 - x}}
$$
with absolute error $\epsilon= 3\times  10^{-10}.$


*

*I need to prove that the series $\displaystyle\sum_{k=1} ^\infty \frac{1}{\sqrt{k^3+x}} $ converges and $\displaystyle \sum_{k=1} ^\infty \frac{1}{\sqrt{k^3 - x}}$ also.

*How many terms are required to evaluate the first series in order to have an absolute error lower than $\epsilon$?

*How (if possible) you can re-write $S(x)$ in order to be evaluated faster.
Sorry if that question is very simple, but for me it is hard for me.
Thanks.
 A: Using cancellations yields smaller errors, since $$\frac{1}{\sqrt{k^3+x}} - \frac{1}{\sqrt{k^3-x}}=\frac{-2x}{\sqrt{k^6-x^2}\left(\sqrt{k^3+x}+\sqrt{k^3-x}\right)}\in\Theta\left(\frac1{k^{9/2}}\right),$$ which shows that, for every $K\geqslant1$, $$\sum_{k=K+1}^\infty\left(\frac{1}{\sqrt{k^3+x}} - \frac{1}{\sqrt{k^3-x}}\right)\in\Theta\left(\frac1{K^{7/2}}\right),$$ with some explicit bounds if one desires so.
Edit: Let us first show that there exists some $c$ such that, for every $|x|\lt1$ and $k\geqslant2$, $$\left|\frac{1}{\sqrt{k^3+x}} - \frac{1}{\sqrt{k^3-x}}\right|\leqslant\frac{c}{(k-1)^{7/2}}-\frac{c}{(k+1)^{7/2}}.$$
It is enough to ask that, for every $k\geqslant2$, $$\frac{1}{\sqrt{k^3-1}} - \frac{1}{\sqrt{k^3+1}}\leqslant\frac{c}{(k-1)^{7/2}}-\frac{c}{(k+1)^{7/2}}.$$ Using the mean value theorem twice, one sees that there exists some $u$ and $v$ in $(-1,1)$, depending on $k$, such that $$\frac{1}{\sqrt{k^3-1}} - \frac{1}{\sqrt{k^3+1}}=\frac1{k^{3/2}}\left(\frac{1}{\sqrt{1-1/k^3}} - \frac{1}{\sqrt{1+1/k^3}}\right)=\frac1{k^{3/2}}\frac2{k^3}\frac{1/2}{(1+u/k^3)^{3/2}}$$ and $$\frac{1}{(k-1)^{7/2}}-\frac{1}{(k+1)^{7/2}}=\frac1{k^{7/2}}\left(\frac{1}{(1-1/k)^{7/2}}-\frac{1}{(1+1/k)^{7/2}}\right)=\frac1{k^{7/2}}\frac2{k}\frac{7/2}{(1+v/k)^{9/2}}.$$ Using, for every $k\geqslant2$, $$1+u/k^3\geqslant7/8,\qquad 1+v/k\leqslant3/2,$$ one gets that some suitable $c$ in the inequality above is $$c=\frac{3^{9/2}}{7^{5/2}}\lt1.1.$$ Using a telescoping series, one gets, for every $K\geqslant1$, $$\left|\sum_{k=K+1}^\infty\left(\frac{1}{\sqrt{k^3+x}} - \frac{1}{\sqrt{k^3-x}}\right)\right|\leqslant\frac{c}{K^{7/2}}+\frac{c}{(K+1)^{7/2}}\leqslant\frac{2.2}{K^{7/2}}.$$
A: Rewrite $$\frac{1}{\sqrt{k^3 \pm x}}=\frac{1}{k^{3/2}}\frac{1}{\sqrt{1 \pm y}}$$ where $y=\frac{x}{k^3}$. Now, consider that $y$ is small compared to $1$ and expand $$\sqrt{1 \pm y}=1\pm\frac{y}{2}+\frac{3 y^2}{8}\pm\frac{5 y^3}{16}+O\left(y^4\right)$$ and replace $y$ by $\frac{x}{k^3}$. You then end, after simplifications, with $$\frac{1}{\sqrt{k^3+x}} - \frac{1}{\sqrt{k^3 - x}}=-\left(\frac{1}{k}\right)^{9/2} x-\frac{5}{8} \left(\frac{1}{k}\right)^{21/2}
   x^3+O\left(\left(\frac{1}{k}\right)^{31/2}\right)$$ which, after summation, write $$S(x) = \sum_{k=1} ^\infty \frac{1}{\sqrt{k^3+x}} - \sum_{k=1} ^\infty \frac{1}{\sqrt{k^3 - x}}\approx-\zeta \left(\frac{9}{2}\right)x-\frac{5 \zeta \left(\frac{21}{2}\right)}{8}x^3$$ $$S(x) \approx -1.05471x-0.625438 x^3$$
Let us try with $x=\frac{1}{4}$; the approximation leads to $-0.273449$ while the whole summation would lead to $-0.273957$; for sure, if you increase the value of $x$, more terms should be required.
A: HINTS
Ignoring the first term, we can set $x=-1$ (worst case) and use
$$\sqrt{k^3-1}>\sqrt{(k-1)^3},$$
so that the series is bounded by the sum of $-3/2$ powers ($\zeta(3/2)=2.612...$).
The partial sum can be approximated by an integral
$$\sum_1^n k^{-3/2}\approx\int_1^n t^{-3/2}dt=-\frac23t^{-1/2}\Big|_1^n.$$
This gives an estimate of the remainder, proportional to $n^{-1/2}$, showing that the series converges, but that the number of terms required to reach $\epsilon$ will be astronomical.
The second series is just the same, with $-x$ instead of $x$.
