I'm stuck at finding pointwise convergence domain of the following function series $$\sum_{n=1}^\infty \frac{\sqrt[3]{(n+1)}-\sqrt[3]{n}}{n^x+1}$$

I tried to use d'Alembert and Weierstrass tests, but it seems to me they don't work here.

  • $\begingroup$ $\frac{\sqrt[3]{(n+1)}-\sqrt[3]{n}}{n^x+1}=\frac{1}{(n^x+1)((n+1)^{2/3}+n_{}^{1/3}(n+1)^{1/3}+n^{2/3})}.$ Comparing with $\sum_{n=1}^{\infty}\frac 1{n^p},$ we get convergence if and only if $x>1/3.$ $\endgroup$
    – yurius
    Jul 23 '13 at 11:54
  • $\begingroup$ Maple finds the following asymptitics of the n-th term $$1/3\, \left( {n}^{-1} \right) ^{x+2/3}+ \left( {n}^{-1} \right) ^{x}O \left( {\frac { \left( {n}^{-1} \right) ^{2/3}}{n}} \right) .$$ $\endgroup$
    – user64494
    Jul 23 '13 at 12:01

We have $$(n+1)^{1/3}-n^{1/3}=n^{1/3}\left(\left(1+\frac{1}{n}\right)^{1/3}-1\right)\sim_\infty\frac{1}{3}n^{-2/3}$$ so there's two cases:

  • If $x>0$ then $$\frac{\sqrt[3]{(n+1)}-\sqrt[3]{n}}{n^x+1}\sim_\infty \frac{1}{3}\frac{1}{n^{x+2/3}}$$ and hence the given series is convergent if $x+2/3>1\iff x>\frac{1}{3}$
  • If $x\leq 0$ then $$\frac{\sqrt[3]{(n+1)}-\sqrt[3]{n}}{n^x+1}\sim_\infty \frac{C}{n^{2/3}}$$ where $C=1/2$ if $x=0$ and $C=1$ if $x<0$ and the series is divergent so we conclude that we have the convergence of the series if and only if $x>1/3$.

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