# Why is this series summable?

Let $$\delta, \epsilon \in \mathbb{R}$$, $$\delta >0$$, $$\epsilon >0$$. Let $$\{ a_k\}^\infty$$,$$\{ b_k\}^\infty$$ be sequences of positive integers such that

$$\lim \sup_{k \rightarrow \infty} \frac{a_{k+1}}{ (\prod^k a_j)^{2 +2/\epsilon + \delta}} \frac{1}{b_{k+1}} = \infty$$ and for every suf. large $$k$$ we have $$\sqrt[1+\epsilon]{ \frac{a_{k+1}}{b_{k+1}} }\geq \sqrt[1+\epsilon]{ \frac{a_{k}}{b_{k}} }+1$$. Why is the infinite series $$\sum^\infty \frac{b_k}{a_k}$$ summable? (This claim comes from a theorem in a published paper whose proof we are formalising using the proof assistant Isabelle/HOL. Summability easily follows if we strengthen the assumption substituting $$\lim \sup$$ with $$\lim$$, but is there a way to show summability without strengthening any assumption?) Thank you in advance!

The condition that for all large enough $$k$$ we have $$\sqrt[1+\epsilon]{ \frac{a_{k+1}}{b_{k+1}} }\geq \sqrt[1+\epsilon]{ \frac{a_{k}}{b_{k}} }+1$$ implies that $$\sqrt[1+\epsilon]{ \frac{a_{k}}{b_{k}} }>k-c$$ and hence $$\frac{b_{k}}{a_{k}}<\frac1{(k-c)^{1+\epsilon}}$$ for some real $$c>0$$ and all large enough $$k>c$$, which yields the convergence of series $$\sum^\infty \frac{b_k}{a_k}$$.
(The condition $$\lim \sup_{k \rightarrow \infty} \frac{a_{k+1}}{ (\prod^k a_j)^{2 +2/\epsilon + \delta}} \frac{1}{b_{k+1}} = \infty$$ is not needed here.)