Explore the convergence of a series I have to explore the convergence of a series. At this picture I used radical Cauchy indication. But I don't now what to do with a denominator to find a limit. Help me please !

Thank You so much :)
 A: The root test in this case is inconclusive. However, the series converges. Indeed, already the series $\sum \frac{1}{9^{\sqrt{n}}}$ converges. We can show this by (limit) comparison with $\sum \frac{1}{n^2}$.
The intuition here is that in the long run, $9^{\sqrt{n}}$ is (much) bigger than $n^2$. More formally, $\lim_{x\to\infty} \frac{x^2}{9^{\sqrt{x}}}=0$. 
This becomes obvious if we let $\sqrt{x}=t$, and note that $9\gt e$. So 
$\frac{x^2}{9^{\sqrt{x}}}\lt \frac{t^4}{e^t}$, and we know that $\lim_{t\to\infty} \frac{t^4}{e^t}=0$.
Remark: There is a "cleverer" way to solve the problem. Consider the integral
$$\int_1^\infty \frac{1}{\sqrt{x}9^{\sqrt{x}}}\,dx.$$
This integral obviously converges (make the substitution $u=\sqrt{x}$), so by the Integral Test our original series converges.
Cute, but I do not like it as much as the cruder argument of the answer. That is because the answer confronts the issue of the "size" of the $n$-th term in a more direct way.  
A: $$9^{1/\sqrt n}=e^{\frac1{\sqrt n}\log 9}\xrightarrow[n\to\infty]{}e^0=1\implies$$
$$\sqrt[n]{\frac1{\sqrt n\, 9^{\sqrt n}}}=\frac1{n^{1/2n}\,9^{1/\sqrt n}}\xrightarrow[n\to\infty]{}1$$
so the $\;n$-th root test doesn't help here.
