Infinte sum, find value that makes it converge I think that this is pretty basic, but I just can't seem to figure it out. I've tried all of the approaches that I can think of, and I can't seem to find the expression for the Nth sum. The Sum is given by: 
$$\sum_{n=1}^\infty \frac{1}{(a^n+1)}$$
And I have to find the values of $a>0$ that makes this sum converge, but I can't figure out how to do this. I thought about comparing it with the sum for $1/a^n$ because I guessed that this would converge for $|x|<1$, since we have learned that this goes for $x^n$ but I'm not sure whether it goes for $x^{-n}$ too, but that didn't seem to work. And I tried to see if the integral of $f(x)$ would converge but them there appeared a lot of $\ln(a^t)$ that would end up as $\ln(0)$. But I am guessing that I have to split it into the case where $a<1$ and $a>1$. A hint would be really nice :)  
 A: If $a = 0$ or $a = 1$ the series is obviously divergent.
If $a>1$ we write
$$\frac{1}{a^n+1} = \frac{1}{a^n}\frac{1}{1+a^{-n}}\le\frac{1}{a^n}.$$
This shows that the series converges for $a>1$.
If $0<a<1$ we have that the sequence $a^n$ is decreasing, so that $\tfrac{1}{a^n+1}$ is bounded below and the series cannot converge.

For $a<0$, we split the sequence into terms with $n$ even and $n$ odd, and check convergence for both separately:
$$\sum_{n=0}^\infty\frac{1}{a^n+1} = \sum_{k=0}^\infty\frac{1}{b^{2k}+1} + \sum_{\ell=0}^\infty\frac{1}{1-b^{2\ell + 1}}$$
where $b=|a|$. By what seen above, the first term converges if, and only if $b>1$. For the second term, we rewrite
$$\frac{1}{1-b^{2\ell + 1}} = \frac{1}{1-b}\frac{1}{1+b+\ldots+b^{2\ell}}\le\frac{1}{1-b}\frac{1}{1+b^{2\ell}}.$$
Again by what we have seen above, this gives convergence in the case where $b>1$.

In conclusion, we have that the series converges if, and only if $|a|>1$.
A: Hint: Try to split it into two cases: 


*

*$ 0 < a\le 1$. Think of the divergence test. 

*$a >1$. Then $a^n > 1^n = 1$. Then try to use comparison test. 
