Convergence of a sum $\sum_{n \geq1} \frac{1}{(a^n+n)}$

Question

I'm supposed to study the convergence of this series and consider separately the cases of $a > 1$ and $a \leq 1$ :

Let $a \geq 0$.

$$\sum_{n \geq1} \frac{1}{(a^n+n)}$$

I've tried using the integral test and the simplest divergence test. I've noticed it's a geometric series when $a = 2$ but I'm having trouble getting anything rigorous. Any help is appreciated.

Answer 1

if $a > 1$, then we have

$$\sum_{n=1}^{\infty} \frac{1}{a^n+n} \leq \sum_{n=1}^{\infty} \frac{1}{a^n} $$

which converges.

If $a < 1$ we have, for example,

$$\sum_{n=1}^{\infty} \frac{1}{a^n+n} \geq \sum_{n=1}^{\infty} \frac{1}{1+n} $$ which diverges to infinity.

Answer 2

As an alternative, note that for $a > 1$

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

and for $a \le 1$ we have

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

and refer to limit comparison test.