# 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$?$

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?

• it can diverge. $a_n := \frac{1}{n^2} \implies a_n^{1/n} \to 1$ – mathworker21 Dec 30 '18 at 13:36
• 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. – Mathsaddict Dec 30 '18 at 13:42
• no. the sum $\sum_n \frac{1}{n^{4/5}}$ diverges – mathworker21 Dec 30 '18 at 13:43
• 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$?$ – Mathsaddict Dec 30 '18 at 13:56

## 1 Answer

You can have both situations.

1) If $$a_n = q^n$$, with $$0, then $$\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.