Conditions for convergence of $\sum\limits_{n=1}^\infty{a^nf(n)}$

Question

assume $a>0$, and for all $n$ we have $0 \leq f(n) \leq 1$.

Is there a necessary and sufficient condition on the series $f(n)$ for which

$\sum\limits_{n=1}^\infty{a^nf(n)}<\infty$ ?

Thanks!

Answer 1

The most straightforward sufficient condition we'd usually think of is to have that $a < 1$ - then as $0 \leq f(n) \leq 1$ for all $n$, it follows that we have that $$ \sum_{n=1}^\infty a^n f(n) \leq \sum_{n=1}^\infty a^n < \infty \text{ as } a < 1,$$ so by the comparison test the series converges. However, this condition isn't necessary; for example, if we take $f(n) = a^{-2n}$, then the series converges for all $a > 1$.

Generally, when it comes to do with results regarding convergence of infinite summations, they are rarely both necessary and sufficient (or if they are, they require some strict condition on the terms being summed, such as monotonicity); for example, if we were considering a series $\sum_{n=1}^\infty g(n)$, then $g(n) \to \infty$ as $n \to \infty$ is a necessary but not sufficient condition for convergence of the series.

While there are some tests, such as the ratio and root tests, for convergence of series, they leave the possibility of giving inconclusive results, even for these restricted types of series (this boils down to the limit $L$ in the ratio and root tests having some dependence on $a$ which cannot avoid the $L = 1$ indeterminate case).

In some sense, one can prove for general series there is no "universal" test for convergence. While I'm not too familiar with the proof, I wouldn't be surprised if this held for this class of series also, so for an answer to your question, I'd say that the answer is probably no.

Answer 2

Almost necessary and sufficient condition for the convergence of this series is provided by the root test: Let $r=\limsup_{n\to\infty} \lvert\,f(n)\rvert^{1/n}$. Then $$ \sum_{n=0}^\infty a^nf(n), $$ converges if $r\lvert a\lvert<1$ and diverges if $r\lvert a\lvert>1$.

It is almost necessary and sufficient condition as it does not tell us anything about the case $r\lvert a\lvert=1$.