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Let $(a_n)$ be a sequence of real numbers such that $$f(x) = \sum_{n=1}^{\infty} a_n x^{2^n}$$ converges for $|x| < 1$ and $f(x)$ converges to $a$ as $x \to 1^{-}$. Then I have to prove that $\sum a_n$ exists and is equal to $a$. (This is a particular instance of so called High Indices Theorem, which was originally proved by Hardy and Littlewood.) Here, we are given a hint that, assuming $\sum a_n = a \ (A)$, we have $\sum a_n = a$ if and only if $$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} k a_k = 0.$$ This is not hard to prove, by comparing the difference between the partial sum $s_N$ and $p_N = \sum a_n e^{-n/N}$.

The result follows once we show that $a_n \to 0$ as $n \to \infty$. But I failed to make any useful estimation except for a trivial bound $a_n \ll_{\varepsilon}(1+\varepsilon)^{2^n}$ obtained by considering the radius of convergence of $f$, and this is obviously of no use. I'm a bit depressed, because my professor told me it would follow from the hint above.

Can somebody give any idea or additional hint to me?

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This is an interesting question. However, I am skeptical that it is actually called the "High Indice Theorem", for instance because so far as I know "indice" is not an English word. Presumably you mean High Indices Theorem? – Pete L. Clark Apr 12 '11 at 15:15
Where have you used the fact that $f(x) \to a$ as $x \to 1^{-}$? Without using that fact, I don't think you can prove what you are trying to. Also, a quick search on google books reveals that it is High Indices Theorem, as Pete mentioned. – Aryabhata Apr 12 '11 at 16:40
You can find a proof of this theorem in Tauberian Theory by Jacob Korevaar, pp. 50-53. The "ramp method" is used to prove other Tauberian theorems. I doubt there is a much simpler proof in your case (exponents $2^n$). – Plop Apr 12 '11 at 19:54
Thanks for correcting my error. I'm not living in English-speaking world, so I often make errors. Moreover, Google always gave me correct search result whenever I tried to search it on google, so I even did not know that I was wrong! Anyway, this problem is from the book 'Multiplicative Number Theory I. Classical Theory'. It contains many fascinating problems, but very occasionally it tacitly assumes some additional conditions on a problem. So I suspect that this is also the case. Well, I have intentionally shunned consulting a reference so far, but now it seems unnecessary as Plop said. – Sangchul Lee Apr 12 '11 at 23:32
Can you please take a look at this question?… – Kermani Mar 12 at 8:10

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