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Possible Duplicate:
Abel limit theorem

Let the series $\sum_{k=0}^\infty a_k x^k$ have radius of convergence 1. Assume further that $\sum_{k=0}^\infty a_k = \infty$ . Is it the case that $\lim_{x\to 1^-} \sum_{k=0}^\infty a_k x^k = \infty$ ?

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  • $\begingroup$ Are the coefficients $a_k$ nonnegative? $\endgroup$
    – Julien
    Jan 27, 2013 at 23:03
  • $\begingroup$ I thought I would ask the general case and then if it is false I will look at the special cases. So no. $\endgroup$
    – markus
    Jan 27, 2013 at 23:09
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    $\begingroup$ All I can say for now is that the answer is yes when the $a_k$ are nonegative. $\endgroup$
    – Julien
    Jan 28, 2013 at 0:13

1 Answer 1

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Yes: Let $A_n:=\sum_{k=0}^n a_k$ which we assume tends to $\infty$ as $n \to \infty$. A first idea is to use summation by parts:

$$\sum_{k=0}^Na_kx^k=A_Nx^N-\sum_{k=0}^{N-1}A_k(x^{k+1}-x^k)=A_Nx^N+(1-x)\sum_{k=0}^{N-1}A_kx^k$$

We show that $\sum A_k x^k$ has the same radius of convergence:

$$ |A_k|^{1/k} \leq |k a_{k^\ast}|^{1/k} = k^{1/k} (|a_{k^\ast}|^{1/k^\ast})^{k^\ast/k} $$ where $k^\ast(k) \leq k$ corresponds to $\max_{i=1,\ldots,k} |a_i|$. If $k^\ast$ tends to $\infty$ as $k \to \infty$, then $\limsup |A_k|^{1/k} = \limsup |a_k|^{1/k} = 1$. If $k^\ast$ stays finite, then for $k$ large we see that $|A_k|^{1/k}$ tends to $1$, and either way the radius of convergence for $\sum A_k x^k$ is still $1$. Also note $|A_N x^N| \leq (|A_N|^{1/N}|x|)^N \to 0$ as $N \to \infty$. This allows us to take the limit of the formula $$ \sum_{k=0}^\infty a_k x^k = (1-x)\sum_{k=0}^\infty A_k x^k $$

We now make use of $A_k \to \infty$. First, let $C > 0$ be arbitrary and find $m$ such that $A_k > C$ for all $k>m$. Then we split the sum

$$ \sum_{k=0}^\infty a_k x^k = (1-x) \sum_{k=0}^m A_k x^k + (1-x) \sum_{k=m+1}^\infty A_k x^k $$

Bound this below by

$$ (1-x) C \sum_{k>m} x^k - |1-x|\left| \sum_{k=0}^m A_k x^k \right| $$

As the second term is finite, as $x \to 1$, it vanishes. The first term is precisely $C x^{m+1}$ which tends to $C$ as $x \to 1$.

Thus, $$ \lim_{x \to 1^-} \sum_{k=0}^\infty a_k x^k \geq C $$ and as $C$ is arbitrary, the limit must be $\infty$.

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  • $\begingroup$ Very nice, thank you very much! $\endgroup$
    – markus
    Jan 28, 2013 at 6:33

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