I have to find, for all value of constant $\displaystyle a>0$, the interval of converence of $\displaystyle $\sum_{n=0}^{\infty}\frac{x^{n}}{1+a^{n}}$$.

How do I do? By using the Ratio test I get

$\displaystyle \lim_{n \to \infty}\frac{\left (\frac{x^{n+1}}{1+a^{n+1}} \right )}{\left (\frac{x^{n}}{1+a^{n}} \right )}=\lim_{n \to \infty} \frac{x^{n+1}}{x^{n}}\frac{1+a^{n}}{1+a^{n+1}}= \lim_{n \to \infty}x\frac{1+a^{n}}{1+a^{n+1}}$

Now I will use the rule of Hopital to make this fraction 'simpler'.

$\displaystyle \lim_{n \to \infty}x\frac{1+a^{n}}{1+a^{n+1}}=x\lim_{n \to \infty}\frac{\frac{\mathrm{d} \left (1+a^{n} \right )}{\mathrm{d} n}}{\frac{\mathrm{d} \left (1+a^{n+1} \right )}{\mathrm{d} n}}=...=x\lim_{n \to \infty}a^{-1}=\frac{x}{a}$

According to the Ratio test, it states that this series is convergent if $\displaystyle \frac{x}{a}<1$, that is $\displaystyle x<a$. Divergent if $\displaystyle \frac{x}{a}>1$, that is $\displaystyle x>a$. If $\displaystyle x=a$, it has not result. If we test it by inserting the value of $\displaystyle a = 1$ on this series, it would be divergent according to Wolfram Alpha. I don't understand this part because this problem clearly states the value of a should be greater than 0, so $\displaystyle a = 1 > 0$ should also be convergent.

The other hand if we see it converges if $\displaystyle x<a$ where $\displaystyle a>0$, so I am not really sure I could say it would be $\displaystyle 0<x<a$. If it's true, the interval of convergent would be like $\displaystyle x \in ]0, a]$ as $\displaystyle a > 0$.

If I did it wrong, please tell me what to do.