# Let $a_n=\frac{1}{n^{\sqrt n}}$. Determine the exact interval of convergence of the power series $\sum a_nx^n$.

Let $a_n=\frac{1}{n^{\sqrt n}}$. Determine the exact interval of convergence of the power series $\sum a_nx^n$.

This is what we thought:
$\limsup|a_n|^{\frac1n}=1$. Therefore $R=1$, so we only have to check at $x=-1,x=1$.
If $x=1$, can we compare it with $\sum\frac1{n^p}$, where $p>1$? If $x=-1$, we can use the alternating series, and I would conclude it converges, as $\lim \frac{1}{n^{\sqrt n}}=0$ and it is decreasing.

Is this all correct ? According to our solution manual the exact interval must be $(-1,1]$. But we don't understand why.

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This is correct. Note that your comparison works also when $x=-1$ and yields absolute convergence in this case. Your manual makes a mistake no matter what. If it converges at $1$ it must converge abosolutely at $-1$, since $\sum|a_n(-1)^n|=\sum |a_n|=\sum a_n$. –  1015 Feb 21 '13 at 19:42
ok, thanks, you convinced me –  Kasper Feb 21 '13 at 19:46

Nice work. You are correct. It seems you've discovered an error in your solution manual. The correct answer should be $[-1, 1]$.
As you point out, the series converges at BOTH $-1$ and at $1$. If it does indeed converge at $1$, and you've shown it does, and the solution manual even claims it does, so we know it converges absolutely when $x = -1$, as @julien points out, given $$\sum|a_n(-1)^n|=\sum |a_n|=\sum a_n.$$