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 $\sum a_{n} = s$?