Krdytkyu

up vote 4 down vote favorite Suppose the sequence ${a_n}_{n=1}^{infty}$ satisfies $$midsumlimits^{n}_{k=1}{a_{k}}midleq Csqrt{n} spacespacespacespacespacespacespacespacespacespacespacespacespacespacespace n=1, 2, 3, cdots$$ where $C$ is a fixed (but arbitrary) number. Prove that the series $$sumlimits^{infty}_{n=1}{frac{a_n}{n}}$$ converges. My attempt: Suppose $b_n:= frac{1}{n}$ ; Abel's lemma on summation by parts gives $$sumlimits^{k}_{n=1}{frac{a_n}{n}}=sumlimits^{k-1}_{n=1}{[sumlimits_{i=1}^{n}{a_i}cdot(b_n-b_{n+1})] + sumlimits_{i=1}^{k}{a_i}cdot b_k}$$ $$<sumlimits^{k-1}_{n=1}{midsumlimits_{i=1}^{n}{a_i}midcdot(b_n-b_{n+1}) +mid sumlimits_{i=1}^{k}{a_i}mid cdot b_k}$$ $$lesumlimits_{n=1}^{k}[Csqrt{n}cdot{(frac{1}{n}-frac{1}{n+1}})]+Csqrt{k}cdot frac{1}{k+1}$$ $$=sumlimits_{n=1}^{k}{frac{Csqr