# Does the converse to Kronecker's lemma hold?

Odds are that this question has been answered already and even that the argument is not too complicated, but here it goes:

Assume that $(a_{k})_{k\in\mathbb{N}}$ is a sequence of real numbers and $(b_{k})_{k\in\mathbb{N}}$ is a sequence of positive numbers with $\lim_{n}b_{n}=\infty$.

Can we infer from the hypothesis $$\sum_{k}a_{k}$$ is not convergent that $$\limsup_{n}|\frac{1}{b_{n}}\sum_{k=1}^{n}a_{k}b_{k}|>0?$$ In other words, does the converse to Kronecker's lemma hold?

For the purpose of my question, you can assume that $(a_{n})_{n}$ is positive and decreasing and that $b_{n}=n$.

Put $\displaystyle a_k=\frac{1}{k\log k}$ for $k\geq 2$, and say $a_1=10$;note that the series $a_k$ is divergent, and that $a_k$ is decreasing. Put $b_k=k$. We have $$\sum_{2}^n\frac{1}{\log k}\leq \frac{1}{\log 2}+\int_2^x\frac{dt}{\log t}$$ Integrating by parts $$\int_2^x\frac{dt}{\log t} =\frac{x}{\log x}-\frac{2}{\log 2}+\int_2^x \frac{dt}{(\log t)^2}$$
Now there exists $A$ such that for $t\geq A$, we have $\log t\geq 2$, and hence for $x\geq A$ $$\displaystyle \int_A^x\frac{dt}{(\log t)^2}\leq \frac{1}{2} \int_A^x\frac{dt}{\log t}$$ With some computations, this show that there exists a constant $c_1$ such that $$\int_2^x \frac{dt}{\log t}\leq 2\frac {x}{\log x}+c_1$$ Now we get with a constant $c_2$ $$\frac{1}{b_n}\sum_{k=1}^n b_ka_k\leq \frac{2}{\log n}+\frac{c_2}{n}$$ and $\displaystyle \frac{1}{b_n}\sum_{k=1}^n b_ka_k \to 0$.
• Yes, thanks. Actually for every $a_{n}\to 0$ $$\frac{1}{n}\sum_{k=1}^{n}a_{k}\to 0$$ by Cesaro convergence. Thus it suffices to find (as you did) $a_{n}\to 0$ such that $$\sum_{k}\frac{a_{k}}{k}$$ is divergent. – David Aug 20 '16 at 0:36