Choose for what constants the series converges or diverges

Question

I have a series $$ \sum_{n=1}^\infty \frac{1}{a^n + 1} $$ where $a > 0$.

I need to find for which values $a$ the series is convergent and divergent.

I have no idea what to do.

I don't think I can apply neither ratio test, integral test, comparison tests.

I'm unsure if I can use alternating series test since $a^n$ potentially could alternate; however, I have the condition $a > n$, so that doesn't apply neither.

I know that it is easier to decide if I first look at $0 < a \leq 1$ and then for $a > 1$.

Answer 1

If $a=1$ then the general term equals $\frac{1}{2} \nrightarrow 0,\;\;n\to \infty,$ hence the series is divergent. If $0<a<1$ then $0<a^n<1 \Rightarrow 1<a^n +1<2 \Rightarrow \frac{1}{2}<\frac{1}{a^n +1}<1.$ In this case the necessary condition for convergence does not hold too.
If $a>1$ then $\frac{1}{a^n +1}<\frac{1}{a^n}$ and the series is convergent by ratio test.

Answer 2

Use D'Alembert's criterion. Compute $r$ given by $$r=\lim_{n\to\infty}\left|\frac{c_{n+1}}{c_n}\right|,\quad\mbox{where } c_n=\frac{1}{a^n+1}$$ If $r<1$, the series converges. If $r>1$, then it diverges.

Since $a>0$, we have: $$\lim_{n\to\infty}\left|\frac{c_{n+1}}{c_n}\right| = \lim_{n\to\infty}\frac{\frac{1}{a^{n+1}+1}}{\frac{1}{a^n+1}} = \lim_{n\to\infty}\frac{a^n+1}{a^{n+1}+1} $$ Thus, for convergence we need $a^n<a^{n+1}$; that is, $a>1$.

The case $r=1$ has to be checked independently. But, clearly, $\sum\frac{1}{2}$ diverges.