Does the series converge?

We prove that $\sum\limits_{k=1}^{\infty} \frac {b_1+\ldots + b_k} {a_1+\dots a_k}\leq 8 \sum\limits_{k=1}^{\infty} \frac { b_k} {a_k}<+\infty$

Let $\{c_n\}_{n\geq 1}$ be a sequence of positive numbers to be chosen later. Using Cauchy-Shwartz inequality

$\frac {b_1+\ldots+ b_k} {a_1+\ldots + a_k}\leq \frac {b_1+\ldots +b_k} {(c_1+\ldots +c_k)^2} \sum_{i=1}^{k} \frac {c_i^2} {a_i}.$

$(*):=\sum\limits_{k=1}^{\infty} \frac {b_1+\ldots + b_k} {a_1+\dots + a_k}\leq\frac{b_1}{a_1} + \sum_{i\geq 2} \frac {c_i^2} {a_i}\sum_{k\geq i} \frac {b_1+\ldots +b_k} {(c_1+\ldots +c_k)^2}.$

Now, we chose $c_n$ , let $D_0=0,\, D_1=0,\, D_n=b_1+(b_1+b_2)+\ldots+(b_1+\ldots+b_{n-1})$ and $c_n=\sqrt{D_{n+1}D_n}-\sqrt{D_nD_{n-1}}$ , so that

$b_1+\ldots +b_n= D_{n+1}-D_{n} \text{ and } (c_1+\ldots +c_n)^2= D_{n+1}D_{n}.$

This gives $\sum_{k\geq i} \frac {b_1+\ldots +b_k} {(c_1+\ldots +c_k)^2}=\sum_{k\geq i} \frac {D_{k+1}-D_{k} } {D_{k+1}D_{k}}=\sum_{k\geq i} \frac {1} {D_{k}}-\frac 1 {D_{k+1}}\leq \frac 1 {D_i}.$ Thus $(*)\leq \frac{b_1}{a_1} + \sum_{i> 1} \frac {c_i^2/D_i} {a_i},$ and it is enough to prove that $\frac {c_i^2} {D_i b_i}\leq 8$ , for this we first note that $b_1+\ldots +b_k \leq \sqrt{2 b_k D_{k+1}}$ , indeed

$(b_1+\ldots +b_k)^2\leq 2(b_1^2+\ldots+b_k^2)+2 \sum_{r<s\leq k} b_rb_s \leq 2[b_1b_1+b_2(b_1+b_2)+\ldots +b_k(b_1+\ldots +b_k)] \leq \\\nonumber 2b_k\left[b_1+(b_1+b_2)+\ldots+(b_1+\ldots+b_{k})\right]=2b_kD_{k+1}.$

Consequently, $\frac {c_i^2} {D_i b_i}=\frac {(\sqrt{D_{i+1}}-\sqrt{D_{i-1}})^2} {b_i}=\frac {(D_{i+1}-D_{i-1})^2} {b_i(\sqrt{D_{i+1}}+\sqrt{D_{i-1}})^2}\leq \frac {(2(b_1+\ldots +b_i))^2} {b_iD_{i+1}}\leq 8.$