I am trying to solve an exercise, but i am not sure that the result i get at the end is correct...May i kindly ask you for a little help or a remark?

Find the radius of convergence of the following power series: $\sum_{n=0}^{\infty }(n^{2}+a^{n})z^{n}$ for any $z,a\in \mathbb{C}$

I use the quotient ratio: $\lim_{n\rightarrow \infty }\left | \frac{a_{n+1}}{a_n} \right |=\lim_{n\rightarrow \infty } \left | \frac{n^{2}+2n+1+aa^{n}}{n^{2}+a^{n}} \right |=\lim_{n\rightarrow \infty }\left | \frac{1+\frac{2}{n}+\frac{1}{n^{2}}+ \frac{aa^{n}}{n^{2}}}{1+\frac{a^{n}}{n^{2}}} \right | = 1$ and then i get radius of convergence $1$.

With Cauchy-Hadamard i get $\lim sup_{n\rightarrow \infty }\sqrt{n^{2}+a^{n}}=\lim sup_{n\rightarrow \infty }\sqrt{n^{2}(1+\frac{a^{n}}{n^{2}})}=\lim sup_{n\rightarrow \infty }n\sqrt{1+\frac{a^{n}}{n^{2}}}=\infty$ and a radius of convergence $0$.

I am not sure which result is correct if any of them is correct...

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If $|a| > 1$, then your quotient ratio calculation is wrong. Also, in Cauchy-Hadamard you need to take the $n$'th root, not the square root. –  Yoni Rozenshein Mar 15 '13 at 15:18
It is lim sup of an $n$=th root, not a square root. –  André Nicolas Mar 15 '13 at 15:20
@Yoni, you're right, Cauchy-Hadamard i did wrong. I am sorry. Thank you for the correction. –  Lullaby Mar 15 '13 at 15:21

First let us assume that $|a| \le 1$. Then the major term of $n^2 + a^n$ is $n^2$, since $a^n$ is bounded by a constant ($1$). By a calculation similar to what you did, the radius of convergence is $1$.
Second, let us assume that $|a| > 1$. Then the major term of $n^2 + a^n$ is $a^n$, which grows exponentially fast. The ratio test gives us (using the fact that exponential growth is much stronger than polynomial growth)
$$\lim_{n\to\infty} \left| \frac {a_{n+1}} {a_n} \right| = \lim_{n\to\infty} \left| \frac {a^{n+1} + (n+1)^2} {a^n + n^2} \right| = \lim_{n\to\infty} \left| \frac {a + \frac {(n+1)^2} {a^n}} {1 + \frac {n^2} {a^n}} \right| = |a|$$
which means the radius of convergence is $\frac 1 {|a|}$.