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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:

  1. for $0<a<1$, the power series converges for every $x$ - following Cauchy–Hadamard theorem - the radius of convergence is $\infty$.

  2. for $a>1$ - I'll try again to use Cauchy–Hadamard - but at this point I got a little confused. would appreciate a hint!

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  • $\begingroup$ what exactly confused you? $\endgroup$ – Holo Jul 13 '18 at 15:12
  • $\begingroup$ for the second section? or for $a$? $\endgroup$ – Jneven Jul 13 '18 at 15:13
  • $\begingroup$ The second section, if $a>1$ the $\limsup$ is just the limit, so what is $\lim( (a^{n^2})^{1/n})$? $\endgroup$ – Holo Jul 13 '18 at 15:14
  • $\begingroup$ that would be $$\infty$$ $\endgroup$ – Jneven Jul 13 '18 at 15:20
  • $\begingroup$ Exactly, so the radius is? And when $a=1$ the limit is just $1$, so the radius is? $\endgroup$ – Holo Jul 13 '18 at 15:23
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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 $0$.

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  • $\begingroup$ this question appeared in a test of "Calculus 1". I said that for every $a>0$ the series converges for $-1<x<1$. I got no points as they claimed that the radius of convergence is $0<x<2$. why is that? $\endgroup$ – Jneven Jul 13 '18 at 15:16
  • $\begingroup$ Asserting that $0<x<2$ is the radius of convergence makes no sense. What happens (when $a=1$) is that the region of convergence is $(0,2)$. $\endgroup$ – José Carlos Santos Jul 14 '18 at 8:15
  • $\begingroup$ that's my thoughts exactly. they wrote the correct answer is : $|x-1|<$ or $0<x<2$, so i'm considering to submit an appeal $\endgroup$ – Jneven Jul 14 '18 at 9:08
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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 $x$.

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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.

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