# Convergence of the series $\sum a_n$ when $\sqrt[n]{a_n}\leq 1-\frac{1}{n^\alpha}$ for $0<\alpha<1$.

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 $$+\infty$$.

• 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? – Jakobian Oct 28 '18 at 10:27
• 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) – Nikolaos Skout Oct 28 '18 at 10:41
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 $$0<\left(1-\frac{1}{n^{\alpha}}\right)^n Combining the convergent of series $$\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.
• 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 $e^{-n^{1-a}}.$ – Nikolaos Skout Oct 28 '18 at 12:12