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Examine the convergence of the series $\sum a_n$, where: $$\sqrt[n]{a_n}\leq 1-\frac{1}{n^\alpha}$$ for all $n$ ($0<\alpha<1$).

Attempt. Since $$\limsup \sqrt[n]{a_n}\leq \limsup\left(1-\frac{1}{n^\alpha}\right)=1$$ we can not use the root test. Comparison test also doesn't work, since $\sum(1-n^{-\alpha})$ diverges to $+\infty$.

Thanks in advance.

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    $\begingroup$ Hint: What happens when $a_n = 0$? And what when $a_n = (1-\frac{1}{n^{\alpha}})^n$? Does the inequality tell us anything about convergence of the series? $\endgroup$
    – Jakobian
    Oct 28, 2018 at 10:27
  • $\begingroup$ Ιn the first case we have convergence and in the second i know that $(1-n^{-a})^n\to 0$. (i am not sure about the series) $\endgroup$ Oct 28, 2018 at 10:41

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Hint:$$\left(1-\frac{1}{n^{\alpha}}\right)^n=e^{n\log \left(1-\frac{1}{n^{\alpha}}\right)};$$ and $$\log \left(1-\frac{1}{n^{\alpha}}\right)<\frac{-1}{n^{\alpha}}.$$ So when $n$ large enough $$n\log \left(1-\frac{1}{n^{\alpha}}\right)<-n^{1-\alpha}.$$ Thus $$0<\left(1-\frac{1}{n^{\alpha}}\right)^n<e^{-n^{1-\alpha}}.$$ Combining the convergent of series $\sum e^{-n^{1-\alpha}}$ for $0<\alpha<1$, we know $\sum \left(1-\frac{1}{n^{\alpha}}\right)^n$ is convergent by Comparison test. Also Comparison test implies $\sum a_n$ is convergent.

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  • $\begingroup$ Very nice. One remark: why do you need $\ln(1+x)>\frac{x}{x+1}$, since $\ln (1+x)<x$ for $x>-1$ is enough? Also, at the last line you meant $e^{-2n^{1-a}}$ and $e^{-n^{1-a}}.$ $\endgroup$ Oct 28, 2018 at 12:12
  • $\begingroup$ Yes, the right inequality is enough. @ Nikolaos Skout $\endgroup$
    – Riemann
    Oct 28, 2018 at 12:16

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