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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?

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    $\begingroup$ I have not checked the details but your suggestion of $a_n = \frac{1}{\log(n)}$ seems to do the trick. We have $s_n \equiv \sum^n a_k \sim \int_1^n \frac{{\rm d}x}{\log(x)} \sim \frac{n}{\log(n)}$ so $\frac{a_n^2}{s_n} \sim \frac{1}{n\log(n)}$ whose sum diverges. $\endgroup$ – Winther Jan 10 '16 at 11:50
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    $\begingroup$ We can take $a_1,\dots,a_{N_0} = 1$ with $N_0 \in \mathbb{N}$ such that the partial sum (of the mentioned series) of the terms till the $N_0$-th term is at least $1$. Then choose $N_1 > N_0$ and $a_{N_0+1},\dots,a_{N_1} = \tfrac{1}{2}$ such that the partial sum till the $N_1$-th term is at least $2$, etcetera. $\endgroup$ – Ovidius Jan 10 '16 at 12:16
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    $\begingroup$ I think you should try the continuous problem : $\int_{0}^{+ \infty} \frac{f'(x)^2}{f(x)} \mathrm{d}x$ where $f'(x) \to 0$. $\endgroup$ – M.LTA Jan 10 '16 at 12:33
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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.

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  • $\begingroup$ Great answer. Your final estimation is interesting. It lends support to a thought that I had - that the series might diverge if (and only if?) $\sum_{n=1}^{\infty}{\frac{a_n}{n}}$ diverges. I'll give this some thought! $\endgroup$ – Auslander Jan 10 '16 at 22:46
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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.

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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 $a_n$.

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  • $\begingroup$ Oh right. That's a more detailed response than I was expecting! Thanks for your input. So you're saying that the conjecture holds provided the additional assumptions about the convergence of $a_{n+1}/a_n\to 1$ but may not hold more generally? Or are you saying that this convergence follows from your argument, rather than an assumption? $\endgroup$ – Auslander Jan 11 '16 at 12:21
  • $\begingroup$ In order to prove it showing the asymptotic equivalence I mention in the answer, the only assumption we need is that $a_n $ is monotonically decreasing. We do know that $a_{n+1}/a_n\to1: \sum a_n =+\infty \implies \lim a_{n+1}/a_n \ge 1,$ and on the other hand, if it were larger than $1$, $\sum a_n$ would diverge considerably faster than $\sum (\log n)^{-\alpha}$, which it doesn't. And in fact, I then state, the ratio approaches $1$ faster than that $b_{n+1}/b_n$, because if it didn't, taking $a_n:=b_n$ would give divergence; and it doesn't. This information, finally, allows us to show... $\endgroup$ – Vincenzo Oliva Jan 11 '16 at 12:55
  • $\begingroup$ @David: ... again applying Stolz-Cesàro, that $\lim\frac{n a_n}{\sum a_k} =1$, implying your conjecture. $\endgroup$ – Vincenzo Oliva Jan 11 '16 at 12:55
  • $\begingroup$ Thanks for clarifying! The theorem is new to me so I have some homework to do. $\endgroup$ – Auslander Jan 12 '16 at 4:55
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    $\begingroup$ @David: Errata (errata corrige) corrige: if you look at the Wikipedia page, you'll see that only the denominator of the studied ratio needs to be monotonocally increasing. In our case, this means we can drop the assumption on the monotonicity of $n a_n$! :D Good luck with L'Hospital's rule for sequences, you're welcome! $\endgroup$ – Vincenzo Oliva Jan 12 '16 at 7:21

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