Does $\sum_i \frac{b_i^2}{i}$ diverge assuming the Cesaro mean of $\{b_i\}$ is greater than zero?

Question

Consider a sequence of positive numbers $\{b_i\}$ bounded above. Suppose that the Cesàro mean of these numbers is positive. That is, $$\lim_{n\to\infty}\frac{b_1+\dots+b_n}{n}=b>0.$$ Can it be shown that $$\sum_{i=1}^{\infty} \frac{b_i^2}{i}$$ diverges? Is the converse also true?

Answer 1

Let $B_k = \displaystyle\sum_{i = 1}^{k}b_i^2$. Using summation by parts, we have $S_k := \displaystyle\sum_{i = 1}^{k}\dfrac{b_i^2}{i} = \dfrac{1}{k}B_k+\sum_{i = 1}^{k}\left(\dfrac{1}{i}-\dfrac{1}{i+1}\right)B_i = \dfrac{1}{k}B_k+\sum_{i = 1}^{k}\dfrac{B_i}{i(i+1)}$.

Since $\displaystyle\lim_{n \to \infty}\dfrac{1}{n}\sum_{i = 1}^{n}b_i = b > 0$, there exists an $N$ such that $\dfrac{1}{n}\displaystyle\sum_{i = 1}^{n}b_i \ge \dfrac{b}{2}$ for all $n \ge N$.

Then, by the RMS-AM inequality, $\dfrac{1}{n}\displaystyle\sum_{i = 1}^{n}b_i^2 \ge \left(\dfrac{1}{n}\displaystyle\sum_{i = 1}^{n}b_i\right)^2 \ge \dfrac{b^2}{4}$, i.e. $\dfrac{1}{n}B_n \ge \dfrac{b^2}{4}$ for all $n \ge N$.

Therefore, $S_k = \dfrac{1}{k}B_k+\displaystyle\sum_{i = 1}^{k}\dfrac{B_i}{i(i+1)} \ge \dfrac{b^2}{4} + \displaystyle \sum_{i = 1}^{N - 1} \dfrac{B_i}{i(i+1)} + \sum_{i = N}^{k}\dfrac{\tfrac{b^2}{4}}{i+1}$.

Since $\displaystyle\lim_{k \to \infty}\sum_{i = N}^{k}\dfrac{\tfrac{b^2}{4}}{i+1} = \infty$ (by comparison to the Harmonic series), we have that $\displaystyle\lim_{k \to \infty}S_k = \infty$.

Therefore, if the Cesaro mean of $\{b_n\}_{n = 1}^{\infty}$ is positive, then the series $\displaystyle\sum_{i = 1}^{\infty}\dfrac{b_i^2}{i}$ diverges.

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The converse is not true. Let $b_i = \dfrac{1}{\sqrt{\ln(i+1)}}$. Then, $\displaystyle\sum_{i = 1}^{\infty}\dfrac{b_i^2}{i} = \sum_{i = 1}^{\infty}\dfrac{1}{i \ln(i+1)}$ diverges by the integral test. However, since $\displaystyle\lim_{i \to \infty}b_i = 0$, the Cesaro mean of $\{b_n\}_{n = 1}^{\infty}$ is $0$.

Answer 2

That also follows from combining Titu's lemma (i.e. the Cauchy-Schwarz inequality) together with a condensation strategy. Assume that the Césaro mean of $\{b_n\}_{n\geq 1}$ is $B>0$. Since: $$ \frac{b_1^2}{1}+\frac{b_2^2}{2}+\ldots+\frac{b_m^2}{m}\geq \frac{(b_1+\ldots+ b_m)^2}{1+\ldots + m}\approx \frac{2m}{m+1}\cdot B^2, $$

$$ \frac{b_{m+1}^2}{m+1}+\frac{b_{m+2}^2}{m+2}+\ldots+\frac{b_{m+r}^2}{m+r}\geq \frac{(b_{m+1}+\ldots+ b_{m+r})^2}{(m+1)+\ldots + (m+r)}\approx \frac{2r}{1+2m+r}\cdot B^2, $$ it follows that $\sum_{n\geq 1}\frac{b_n^2}{n}$ is divergent by regrouping its terms in "packets" of size $1,4,16,64,\ldots$ and applying the estimates above to every "packet" (we just need that $m$ and $r$ are roughly the same size).