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Given two sequences $(a_k),(b_k)$ with $a_k\geq0,b_k>0$ such that the power series $\sum_{k=0}^\infty a_k b_kr^{k}$ and $\sum_{k=0}^\infty a_kr^k$ converge for each $r>0$. My question now is: Does there exist a constant $c$ (depending only on $(b_k))$ such that \begin{align*} \sum_{k=0}^\infty a_kb_kr^{k}\geq c\sum_{k=0}^\infty a_kr^{k} \end{align*} for all large $r>0$? This is obvious for if $(b_k)$ is bounded away from zero, but what if $\liminf b_k=0?$ In my case, $b_k=1/k!$, and I have no idea how to approach this problem... Any help is highly appreciated! Thanks in advance.

sranthrop
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2 Answers2

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No, take $a_k = 1/k!,b_k=1/2^k.$ Then $\sum a_kr^k = e^r, \sum a_kb_kr^k = e^{r/2}.$

zhw.
  • 105,693
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On the contrary. Let $f(r) = \sum_{k} a_k b_k r^k$ and $g(r) = \sum_k a_k r^k$, and suppose all $a_k > 0$. If $b_k \to 0$, then for any $\epsilon > 0$ take $N$ so that $b_k < \epsilon$ for $k \ge N$. Now for $r$ sufficiently large (say $r > R$), $$\sum_{k=0}^{N-1} a_k b_k r^k < r^{N-1} \sum_{k=0}^{N-1} a_k b_k < a_N b_N r^N < \sum_{k=N}^\infty a_k b_k r^k $$ Thus for $r > R$ we have $$f(r) < 2 \sum_{k=N}^\infty a_k b_k r^k < 2 \epsilon \sum_{k=N}^\infty a_k r^k < 2 \epsilon g(r)$$

Robert Israel
  • 448,999