Suppose I had a sequence $a_n$ that satisfied the following criteria:

$\displaystyle\sum_{n=0}^\infty a_n$ diverges.

$\displaystyle\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=1$ (inconclusive ratio test)

$a_na_{n+1}<0$ (the series is alternating)

Is it provable if for all $a_n$ that satisfies the above that

$$\lim_{r\to1^-}\sum_{n=0}^\infty a_nr^n$$


I can clearly see for any fixed $|r|<1$, the series converges, but I don't know about the behavior as $r\to1$. As far as the examples I've had, this limit exists, for example, if $a_n=(-1)^nn$, the limit is $1/4$.

So can someone show if this limit exists or not?

  • 1
    $\begingroup$ This type of theorem is called a Tauberian theorem. $\endgroup$
    – Neal
    Dec 19, 2016 at 0:06

1 Answer 1


A counterexample would be $$ \begin{align} a_{2n} &= 1+\frac{1}{n+1} \\ a_{2n+1} &= -1 \end{align} $$

We then have $$ \sum_{n=0}^\infty r^n a_n \ge \sum_{n=0}^{\infty} \frac{r^{2n}}{n+1} = -\frac{\log(1-r^2)}{r^2} $$ and the right-hand side of this blows up as $r\to 1^-$.

(Thanks to Wolfram Alpha for evaluating the series in the middle).


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