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Littlewood's extension of Tauber's theorem states that if $a_{n} = O(1/n)$ and as $x \rightarrow 1^{-}$, we have $\sum a_{n}x^{n} \rightarrow s$, then $\sum a_{n} = s$.

My question is, what if instead I have $a_{n} = O(t^n/n)$ and as $x \rightarrow (t^{-1})^{-}$, I have $\sum a_{n}x^{n} \rightarrow s$, then do we still have $\sum a_{n} = s$?

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    Surely you mean $\sum_n a_n t^{-n} = s$.2012-12-24

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This (with the conclusion corrected as in my comment) is just a change of variables, assuming $t > 0$. Let $b_n = a_n t^{-n}$, so $b_n = O(1/n)$. Then with $y = xt$, as $x \to (1/t)-$ we have $y \to 1-$, and $\sum_n a_n x^n = \sum_n b_n y^n \to s$, so Littlewood says $\sum_n b_n = s$, i.e. $\sum_n a_n t^{-n} = s$.