Show that the series $$\sum\limits_{n=2}^{\infty} \frac {(n^3+1)^{1/3}-n}{\log n}$$ converges.

I showed it using Abel's theorem and limit comparison test. Any other simpler method?

  • 3
    $\begingroup$ $$(n^3+1)^{1/3}\leqslant n+\frac1{3n^2}\implies\frac{(n^3+1)^{1/3}-n}{\log n}\leqslant\frac1{3n^2\log n}\implies\sum_{n=2}^\infty\frac{(n^3+1)^{1/3}-n}{\log n}\ \text{converges}$$ $\endgroup$ – Did Nov 14 '17 at 14:50
  • $\begingroup$ thank you . abel's + limit comparision was messy. $\endgroup$ – Magneto Nov 14 '17 at 15:11
  • $\begingroup$ @anirudhb How does Abel apply here?? $\endgroup$ – Mark Viola Nov 14 '17 at 15:25
  • $\begingroup$ @MarkViola $\log n$ is increasing function and so inverse of it is decreasing function converging to zero. Now we can show $\sum (n^3+1)^{\frac {1}{3}} - n$ convergent series by limit comparision test with $\frac {1}{n^2}$. now we can apply abel's test $\endgroup$ – Magneto Nov 14 '17 at 15:28
  • $\begingroup$ @anirudhb You meant "Abel's Test" not "Abel's Theorem." And why not simply realize that $0< \frac{(n^3+1)^{1/3}-n}{\log(n)}\le \frac{(n^3+1)^{1/3}-n}{\log(2)}$ for $n>1$? $\endgroup$ – Mark Viola Nov 14 '17 at 15:35

Write $$\frac{(n^3+1)^{\frac{1}{3}}-n}{\log n} = \frac{n^3+1-n^3}{\log n((n^3+1)^{2/3}+(n^3+1)^{1/3}n+n^2)} = \frac{1}{\log n((n^3+1)^{2/3}+(n^3+1)^{1/3}n+n^2)}$$ and observe that the denominator goes at $0$ faster than $\frac{1}{n^2}.$

  • $\begingroup$ it should be i think $(n^3 + 1)^{\frac {2}{3}} + (n^3 + 1)^{\frac {1}{3}}n + n^2 $ @Gibbs $\endgroup$ – Magneto Nov 14 '17 at 15:16
  • $\begingroup$ Sure, I forgot the 3's. Thanks. $\endgroup$ – Gibbs Nov 14 '17 at 15:19

Note that

$$(n^3+1)^\frac13=n\left(1+\frac1{n^3}\right)^\frac13\sim n+\frac1{3n^2}$$


$$\frac{(n^3+1)^\frac13-n}{\log n}\sim \frac{1}{3n^2\log n}$$

and thus since $\sum\limits_{n=1}^{\infty} \frac{1}{3n^2\log n}$ converges also $\sum\limits_{n=1}^{\infty} \frac{(n^3+1)^\frac13-n}{\log n}$ converges by comparison with it.

  • $\begingroup$ (Please avoid posting duplicate answers) $\endgroup$ – Jack D'Aurizio Feb 3 '18 at 20:27
  • $\begingroup$ @JackD'Aurizio I've posted the answer also here because the other OP was declared a duplicate, I delete the other one. Thanks! $\endgroup$ – gimusi Feb 3 '18 at 20:29

$$\lim_{n \to \infty}\left(\frac{(n^3+1)^\frac13-n}{\ln n}\right)n^2=\lim_{n\rightarrow+\infty}\frac{n^2}{\left(\sqrt[3]{(n^3+1)^2}+n\sqrt[3]{n^3+1}+n^2\right)\ln{n}}=0,$$

then for $n$ large enough we have $$\frac{(n^3+1)^\frac13-n}{\ln n}\le \frac{1}{ n^2}$$ the result follows by comparison test. which says that it converges because $$\sum_{k=1}^{+\infty}\frac{1}{k^2}=\frac{\pi^2}{6}.$$


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