I am working on an exercise from Zorich's Mathematical Analysis. The problem reads:
If $a_{n}\ge a_{n+1}>0$ for all $n\in\mathbb{N}$ and the series $\displaystyle\sum_{n=1}^{\infty}a_{n}$ converges then $a_{n}=o\left(\frac{1}{n}\right)$ as $n\to\infty$;
I understand that $a_{n}=o\left(\frac{1}{n}\right)$ means $a_{n}=\alpha(n)\cdot\frac{1}{n}$ where $\alpha(n)\to 0$ as $n\to\infty$. So let $\alpha(n)=n\cdot a_{n}$, then $a_{n}=\frac{\alpha(n)}{n}$, but I cannot seem to fully reason out from convergence of $\displaystyle\sum_{n=1}^{\infty}a_{n}$ to $\displaystyle\lim_{n\to\infty}\alpha(n)=0$. My partial result reasoning is as following:
Since $a_{n}\ge a_{n+1}>0$, follow Cauchy's proposition that $\displaystyle\sum_{n=1}^{\infty}a_{n}$ converges if and only if $\displaystyle\sum_{k=1}^{\infty}2^{k}a_{2^{k}}$ converges. Therefore, $$\sum_{k=1}^{\infty}2^{k}a_{2^{k}}=\sum_{k=1}^{\infty}2^{k}\frac{\alpha\left(2^{k}\right)}{2^{k}}=\sum_{k=1}^{\infty}\alpha\left(2^{k}\right)$$ Since it converges, it is necessary that $\displaystyle\lim_{k\to\infty}\alpha\left(2^{k}\right)=0$. But that is only a subsequence of $\alpha(n)$.
(added per Greg Martin's suggestion - thanks) Further from $a_{n+1}\ge a_{n}$ we have \begin{align*} &a_{n}\ge a_{n+1}>0 \Rightarrow\frac{\alpha(n)}{n}\ge\frac{\alpha(n+1)}{n+1}>0\\ \Rightarrow&\frac{\alpha(n+1)}{\alpha(n)}\le\frac{n+1}{n}=1+\frac{1}{n}\\ \Rightarrow&\frac{\alpha(n+2)}{\alpha(n)}=\frac{\alpha(n+2)}{\alpha(n+1)}\cdot\frac{\alpha(n+1)}{\alpha(n)}\le\left(1+\frac{1}{n+1}\right)\left(1+\frac{1}{n}\right)<\left(1+\frac{1}{n}\right)^{2}\\ &\quad\vdots \end{align*} So if $n=2^{k}$, then $2^{k+1}=2\cdot 2^{k}=2n$, thus \begin{align*} \Rightarrow&\frac{\alpha(2^{k}+1)}{\alpha(2^{k})}\le 1+\frac{1}{2^{k}}\\ \Rightarrow&\frac{\alpha(2^{k}+2)}{\alpha(2^{k})}=\frac{\alpha(2^{k}+2)}{\alpha(2^{k}+1)}\cdot\frac{\alpha(2^{k}+1)}{\alpha(2^{k})}<\left(1+\frac{1}{2^{k}}\right)^{2}\\ &\quad\vdots\\ \Rightarrow&\frac{\alpha(2^{k+1}-1)}{\alpha(2^{k})}=\frac{\alpha(2^{k+1}-1)}{\alpha(2^{k+1}-2)}\cdot\frac{\alpha(2^{k+1}-2)}{\alpha(2^{k+1}-3)}\cdots\frac{\alpha(2^{k}-1)}{\alpha(2^{k})}<\left(1+\frac{1}{2^{k}}\right)^{2^{k}-1} \end{align*}
But we know $$\displaystyle\lim_{k\to\infty}\left(1+\frac{1}{2^{k}}\right)^{2^{k}-1}=e$$ Thus $\alpha(2^{k}+d)<e\cdot\alpha(2^{k})$ is bounded for $d<2^{k}$, so $$\lim_{n\to\infty}\alpha(n)\le\lim_{k\to\infty}e\cdot\alpha(2^{k})=e\cdot\lim_{k\to\infty}\alpha(2^{k})=e\cdot 0=0$$
Hope I got it right..
