<s>[INST] 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.
 [/INST] 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.
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