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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.)

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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.

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    $\begingroup$ It is this simple! $\endgroup$ – Iosif Pinelis Dec 12 '19 at 13:46
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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 $1$.

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