Given $a>0 $, let $\sum_{n=1}^{\infty}a^{n^{2}}(x-1)^{n}$ check for which values of $x$ as $x\in\mathbb{R}$, the power series converges.
I divided my answer for two diffrent cases:
for $0<a<1$, the power series converges for every $x$ - following Cauchy–Hadamard theorem - the radius of convergence is $\infty$.
for $a>1$ - I'll try again to use Cauchy–Hadamard - but at this point I got a little confused. would appreciate a hint!
On
Note that $$\limsup\sqrt[n]{a^{n^2}} =\limsup{a^{n}}=\begin{cases}0&|a|<0,\\1&a=1,\\\infty&a>1\end{cases}$$ Hence we have radius of convergence $\infty$, $1$, $0$, respectively.
On
For a simpler (if less general and powerful) approach, we know that if a sum $\sum_{n=1}^{\infty}{c_n}$ converges, then $\lim_{n\to\infty}{c_n}=0$, by considering successive differences of partial sums.
Hence, fixing $a>1$ and supposing that your sequence converges, we must also have $\lim_{n\to\infty}{a^{n^2}(x-1)^n} = 0$. This is equivalent to the condition $$\lim_{n\to\infty}{(a^n\lvert x-1 \rvert)^n} = 0$$ by, say, continuity of taking absolute values.
Suppose that $x\ne 1$. Then for sufficiently large $n$, we can write $\lvert{x-1}\rvert > \alpha/a^n$ for some constant $\alpha > 1$. So we have the inequality $$(a^n\lvert x-1\rvert)^n > \alpha^n.$$ Taking limits gives us $\lim_{n\to\infty}{\alpha^n} = 0$, which contradicts the fact $\alpha>1$.
Suppose on the other hand that $x=1$. Then $x-1=0$, so your original sum is trivially $0$, and converges.
In summary, for $a>0$, $f(x)$ converges for $x=1$ and for no other choice of $x$.
Note that$$\sqrt[n]{\bigl|a^{n^2}(x-1)^n\bigr|}=a^n|x-1|.$$If $0<a<1$, then $\lim_{n\to\infty}a^n|x-1|=0$ and therefore the series converges, for every $x$. If $a=1$, the series converges if $|x-1|<1$ and diverges otherwise. Therefore, the radius of convergence is $1$. Finally, if $a>1$, $\lim_{n\to\infty}a^n|x-1|=+\infty$ and therefore the series diverges (unless $x=1$) and so the radius of convergence is $0$.