Does this series diverge if $a_n\to 0$?

Question

Consider a given sequence $\{a_n\}$ of positive terms and the series $$\frac {a_ 1^2}{a_1}+\frac{a_2^2}{a_1+a_2}+\frac {a_ 3^2}{a_1+a _2+a_3}+\cdots.$$ I can show that this series diverges if the sequence is bounded below. This is not hard. But I'm wondering if it can ever diverge if $\lim_{n\to\infty}a_n=0.$ If it is possible, the convergence to zero would have to be quite slow. So perhaps $a_ n=1/\ln{(n+1)}$ or $a_n= 1/\sqrt{\ln {(n+ 1)}}$ would do the trick. What techniques are available for studying such series? And is there perhaps a better example?

Answer 1

Expanding on Winther's comment, for $x\ge3$, $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\frac{x}{\log(x)} &=\frac{\log(x)-1}{\log(x)^2}\\ &=\frac1{\log(x)}\left(1-\frac1{\log(x)}\right)\\ &\ge\frac1{\log(x)}\left(\frac{\log(3)-1}{\log(3)}\right)\\ \end{align} $$ Thus, $$ \begin{align} \sum_{k=4}^n\frac1{\log(k)} &\le\int_3^n\frac{\mathrm{d}x}{\log(x)}\\ &\le\int_3^n\frac{\mathrm{d}x}{\log(x)}\\ &\le\frac{\log(3)}{\log(3)-1}\left[\frac{x}{\log(x)}\right]_3^n\\ &\le\frac{\log(3)}{\log(3)-1}\frac{n}{\log(n)} \end{align} $$ Therefore, $$ \sum_{n=4}^N\frac{\frac1{\log(n)^2}}{\sum\limits_{k=4}^n\frac1{\log(k)}} \ge\frac{\log(3)-1}{\log(3)}\sum_{n=4}^N\frac1{n\log(n)} $$ which diverges.

Answer 2

Consider $a_n = \dfrac1{\log^s(n+1)}$ for $s>0$. We then have the $n^{th}$ terms of your new sequence to be $$b_n = \dfrac{a_n^2}{a_1+a_2+\cdots+a_n} = \dfrac1{\log^{2s}(n+1) \left(\displaystyle\sum_{k=1}^n \dfrac1{\log^s(k+1)}\right)}$$ Further, we have $$\displaystyle \sum_{k=1}^n \dfrac1{\log^s(k+1)} \leq \dfrac{n}{\log^s(2)}$$ Hence, we have $$b_n \geq \dfrac{\log^s(2)}{n\log^{2s}(n+1)}$$ Now if $s \leq 1/2$, we see that $\displaystyle \sum_{n=1}^{\infty} \dfrac{\log^s(2)}{n\log^{2s}(n+1)}$ diverges. Hence, choosing $a_n= \dfrac1{\log^s(n+1)}$ for $s \in (0,1/2]$ ensure that $a_n \to 0$, whereas $\displaystyle \sum \dfrac{a_n^2}{a_1+a_2+\cdots+a_n}$ diverges.

Answer 3

Let $A_n= \displaystyle \sum_{k=2}^n \frac{1}{(\log k)^{2\alpha} \sum_{m=2}^k 1/(\log m)^\alpha}.$ Since it is easy to deal with the case $0<\alpha\le\frac12$ showing $A_k>\displaystyle\frac{1}{k\log k}$, it is natural to consider $$L= \lim_{n\to\infty}=\frac{\sum_{m=2}^n 1/(\log m)^\alpha}{n/(\log n)^\alpha}, \ \ \ \frac12<\alpha\le1.$$ As long as $\displaystyle \frac{n}{(\log n)^\alpha}$ is both positive and monotonically increasing for $n>2$, the Stolz-Cesàro theorem yields $$\begin{align}L&=\lim_{n\to\infty} \frac{1}{(\log(n+1))^\alpha\left(\frac{n+1}{ (\log(n+1))^\alpha} - \frac{n}{ (\log n)^\alpha} \right)} \\ &= \lim_{n\to\infty} \frac{1}{1+n\left(1- \left(\frac{\log(n+1)}{\log n}\right)^\alpha\right)}=1,\end{align}$$ hence $\sum A_k$ diverges by limit comparison with $\displaystyle \sum \frac{1}{k(\log k)^\alpha}.$ In fact, this argument also shows that $\sum A_k$ converges for $\alpha>1.$ We are then tempted to investigate the behaviour of $B_n= \displaystyle \sum_{k=2}^n\frac{1}{(\log k)^2 (\log\log k)^2 \sum_{m=1}^k 1/( \log k \log\log k) }$. Applying again Stolz-Cesàro shows that $B_n$ diverges, too, and in fact that taking your $a_n$ to be $\displaystyle \frac{1}{\log n(\log\log n)^\beta}$, with any real $1\le\beta<2$, still gives divergence. In particular, the conjecture you state in a comment to robjohn's answer is implied by $\frac{a_n^2}{\sum a_k}\sim \frac{a_n}{n} \iff \frac{a_n}{\sum a_k}\sim\frac1n$, which, assuming monotonical decreasingness of $a_n$ , indeed follows from proving $\lim \frac{n a_n}{\sum a_k}=1$ using Stolz-Cesàro, and the fact that $\frac{a_{n+1}}{a_n}\to1$ faster than $\frac{\log n}{\log(n+1)}\left(\frac{\log\log n}{\log\log(n+1)}\right)^2=: \frac{b_{n+1}}{b_n},$ as a consequence of the convergence of your series with $b_n$ in place of $a_n$.