Convergence of $\sum_{n=1}^{\infty}\frac{a^{-n+1}+1}{n^a}$

Question

How do I study the convergence of the series $$\sum_{n=1}^{\infty}\frac{a^{-n+1}+1}{n^a}$$ for $a>0$? I have tried all the tests I know without succeeding: ratio test, comparison test, root test...

Answer 1

We have that

$$\frac{a^{-n+1}+1}{n^a}=\frac{a+a^n}{a^nn^a}$$

then we can distinguish two cases

• $0<a\le 1$

$$\frac{a^{-n+1}+1}{n^a}\sim \frac{1}{a^{n-1}n^a}>\frac{1}{n^a}$$

• $a>1$

$$\frac{a^{-n+1}+1}{n^a}\sim \frac{1}{n^a}$$

Answer 2

Hint:

Write

$$\frac{a^{-n+1}+1}{n^a}=\frac{a^n+a}{a^nn^a}$$

and now apply the (limit or not) comparison test with $\;\frac1{n^a}\;$ . Distinguish cases.

Answer 3

The series converges iff $a>1$: if it converges then so does $\sum \frac 1 {n^{a}}$ so we must have $a>1$. Conversely, if $a>1$ then split the series into two parts and show that each of them converges.