I have to find an $a_n$ such that $\sum_{n=1}^{\infty}a_nx^n $ converges absolutely on $[-1,1]$. I have choosen $a_n = \frac {1}{n^2}$ and did the root test and see that $\lim_{n \to \infty} (|a_n|) =1$. Now I know that if the result of the root test was < 1 then it would converge $\forall |x| < 1 $. How can i show it it holds even on -1 and 1?
2026-03-25 21:23:02.1774473782
Absolute convergence of $\sum_{n=1}^{\infty} a_nx^n$
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$|a_nx^{n}| \leq \frac 1 {n^{2}}$ so you can apply Comparison Test.