# Comparing the tails of two related convergent series

Let $$b_1,b_2,\dots$$ be positive real numbers such that $$s_1<\infty\quad\text{and}\quad z_1<\infty,$$ where $$s_k:=\sum_{j=k}^\infty b_j\quad\text{and}\quad z_k:=\sum_{j=k}^\infty\frac{b_j}{\sqrt{s_j}}$$ for natural $$k$$. Does it then necessarily follow that $$\limsup_{k\to\infty}\frac{z_k}{\sqrt{s_k}}<\infty\,\text{?}$$ (One may note here that $$z_k\ge\sqrt{s_k}$$ for all $$k$$, whence $$\liminf_{k\to\infty}\frac{z_k}{\sqrt{s_k}}\ge1$$.)

(This question is a "discretized" version of a problem considered in this answer.)

In fact $$z_k$$ (or better any partial sum for it) is a lower Riemann sum for the function $${1\over \sqrt{x}}$$ wrto the infinitesimal decreasing sequence $$s_k>s_{k+1}>\dots s_{m}$$, so $$z_k< \int_0^{s_k}{1\over \sqrt{x}}dx=2\sqrt{s_k}$$, and the constant $$2$$ is sharp.
Let us prove the following, stronger claim: $$z_k<2\sqrt{s_k}\tag{1}$$ for all natural $$k$$. Also, the condition $$z_1<\infty$$ will not be needed or used.
Take indeed any natural $$k$$. Letting $$h_i:=\frac1{\sqrt{s_{i+1}}}-\frac1{\sqrt{s_i}},$$ we have $$\frac1{\sqrt{s_j}}=\frac1{\sqrt{s_k}}+\sum_{i=k}^{j-1}h_i$$ for all natural $$j\ge k$$. So, \begin{align}z_k&=\sum_{j=k}^\infty\frac{s_j-s_{j+1}}{\sqrt{s_j}} \\ &=\sum_{j=k}^\infty(s_j-s_{j+1})\Big(\frac1{\sqrt{s_k}}+\sum_{i=k}^{j-1}h_i\Big) \\ &=s_k\,\frac1{\sqrt{s_k}}+\sum_{i=k}^\infty h_i\sum_{j=i+1}^\infty(s_j-s_{j+1}) \\ &=\sqrt{s_k}+\sum_{i=k}^\infty\Big(\frac1{\sqrt{s_{i+1}}}-\frac1{\sqrt{s_i}}\Big)s_{i+1} \\ &<\sqrt{s_k}+\sum_{i=k}^\infty(\sqrt{s_i}-\sqrt{s_{i+1}})=2\sqrt{s_k}. \end{align} So, we have (1), as claimed.
The constant factor $$2$$ in (1) cannot be improved. Indeed, take any $$a\in(0,1)$$ and for all natural $$j$$ let $$b_j=a^j$$, so that $$s_j=b_j/(1-a)$$ and $$z_j=(1+\sqrt a)\sqrt{s_j}$$. It remains to let $$a$$ be arbitrarily close to $$1$$.