If $\sum_{n=1}^{\infty} a_n$ is a convergent series of positive terms then $\sum_{n=1}^{\infty} \frac{a_n^{1/n}}{n^{4/5}}$ coverges or diverges$?$

Using comparison test

$\lim_{n \to \infty} \frac{(a_n)^{1/n}}{n^{4/5} a_n}$

And I am getting nothing.

I know that any such convergent series of positive terms should be of the form $\sum_{n=1}^{\infty} (1/n^{a})$ where $a>1$. I tried to solve the problem by replacing $a_n$ with $1/n^{a}$, used comparison test, but found no result.

How to proceed?

  • $\begingroup$ it can diverge. $a_n := \frac{1}{n^2} \implies a_n^{1/n} \to 1$ $\endgroup$ Dec 30 '18 at 13:36
  • $\begingroup$ I tried with this example but since $n^{4/5}$ is in denominator so $n^{th}$ term would tend to zero (in comparison test). So we can't say anything about convergence. $\endgroup$
    – Mathaddict
    Dec 30 '18 at 13:42
  • $\begingroup$ no. the sum $\sum_n \frac{1}{n^{4/5}}$ diverges $\endgroup$ Dec 30 '18 at 13:43
  • $\begingroup$ Can we directly find the behaviour of the series $\sum_{n} \frac{1}{n^{2/n} n^{4/5}}$ by applying limit on $n^{2/n}$ in all terms$?$ $\endgroup$
    – Mathaddict
    Dec 30 '18 at 13:56

You can have both situations.

1) If $a_n = q^n$, with $0<q<1$, then $\sum_n a_n$ is convergent but $\sum_n \frac{a_n^{1/n}}{n^{4/5}} = \sum_n \frac{q}{n^{4/5}}$ diverges to $+\infty$.

2) If $a_n = n^{-n}$, then $\sum_n a_n$ is convergent and $\sum_n \frac{a_n^{1/n}}{n^{4/5}} = \sum_n \frac{1}{n^{1+4/5}}$ is convergent.


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