$\sum_{n\ge1}\frac{a_n}{(s_n)^\alpha} \text{converges} \iff \alpha > 1$ , where $\sum_{n\ge1} a_n$ is divergent

Question

Let $\forall n \quad a_n >0, \quad s_n=\sum_{k=1}^n a_k$

If $\sum_{n\ge1} a_n$ diverges, than show that

$$\sum_{n\ge1}\frac{a_n}{(s_n)^\alpha} \text{converges} \iff \alpha > 1$$

Answer 1

Let $\alpha>1$. We have $$\frac{a_n}{(s_n)^\alpha}\leq \frac{a_n}{s_n(s_{n-1})^{\alpha-1}}=\frac{s_n-s_{n-1}}{s_n(s_{n-1})^{\alpha-1}}$$ moreover since the function $x\mapsto 1-x^\frac{1}{\alpha}$ is convex on $(0,1]$ then $1-x^\frac{1}{\alpha}\leq 1-x$ so if $x=\left(\frac{s_{n-1}}{s_n}\right)^\alpha$ we have $$1-\frac{s_{n-1}}{s_n}\leq 1-\left(\frac{s_{n-1}}{s_n}\right)^\alpha\iff \frac{s_n-s_{n-1}}{s_n(s_{n-1})^{\alpha-1}}\leq \frac{1}{s_{n-1}^{\alpha-1}}-\frac{1}{s_{n}^{\alpha-1}}$$ so $$\sum_{k=n+1}^{n+p}\frac{a_k}{(s_k)^\alpha}\leq \sum_{k=n+1}^{n+p}\frac{s_k-s_{k-1}}{s_k(s_{k-1})^{\alpha-1}}\leq \sum_{k=n+1}^{n+p} \frac{1}{s_{k-1}^{\alpha-1}}-\frac{1}{s_{k}^{\alpha-1}}\leq \frac{1}{s_{n}^{\alpha-1}} $$ hence the series is convergent by Cauchy criterion.

We have $$\displaystyle\sum_{k=n+1}^{n+p}\frac{a_{k}}{s_k}\geq \frac{\sum\limits_{k=n+1}^{n+p}{a_{k}}}{s_{n+p}}=\frac{s_{n+p}-s_n}{s_{n+p}}\to 1,\quad p\to\infty$$ hence by Cauchy criterion the series is divergent if $\alpha=1$

Finally, if $\alpha\leq 1$ and since for $n$ large enough $\frac{a_n}{s_n^\alpha}\geq \frac{a_n}{s_n}$, the series is also divergent.

Answer 2

One can proceed as one does with Riemann sums : we show that the series is dominated by a “domino” series. So ,putting $f(t)=t^{1-\alpha}$ and $w_n=f(s_n)$, we have

$$ w_{n-1}-w_n = f(s_{n-1})-f(s_{n-1}+a_n) = f’(t), t\in [s_{n-1},s_{n-1}+a_n] \tag{1} $$

When $\alpha \gt 1$, we deduce $w_{n-1}-w_n \geq (\alpha-1)\frac{a_n}{s_n^{\alpha}}$, which shows the convergence we need. When $\alpha \lt 1$, we deduce $w_n-w_{n-1} \leq (1-\alpha)\frac{a_n}{s_n^{\alpha}}$, which shows the divergence we need.

Finally, when $\alpha=1$, it will suffice to show that for any $i$, there is a $j \gt i$ such that

$$ \sum_{k=i}^j \frac{a_k}{s_k} \geq 1 \tag{2} $$

Since the series $\sum_{}{a_k}$ diverges, there is a $j$ such that

$$ \sum_{k=i}^j a_k \geq s_i \tag{3} $$

so that

$$ \sum_{k=i}^j \frac{a_k}{s_k} \geq \sum_{k=i}^j \frac{a_k}{s_i} \geq 1 \tag{4} $$

as wished.