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Is it possible to have a sequence $a_n \geq 0$ which is decreasing and such that $\sum a_n < \infty$ but $\sum n a_n^2 = \infty$? I have seen Find a sequence $a_n$ so that $\sum |a_n|$ converges but $\sum n |a_n|^2$ diverges where the condition holds but the $a_n$ given is not decreasing.

nullUser
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4 Answers4

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The answer is no, there are two steps to do:

1) If $a_n$ is decreasing and $\sum_{n=1}^{\infty} a_n < \infty$, show that $a_n \leq 1/n$ for all sufficiently large $n$.

2) Conclude that if $a_n$ is decreasing and $\sum_{n=1}^{\infty} a_n < \infty$, then $\sum_{n=1}^{\infty} na_n^2 < \infty$.

Michael
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    How does (2) follow for $a_n=\frac{1}{n}$, which satisfies (1)? – vadim123 Apr 01 '18 at 02:52
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    But $\sum_{n=1}^{\infty} 1/n = \infty$, so that example does not apply. – Michael Apr 01 '18 at 02:53
  • Also (1) seems false, take a geometric sum and pad it out with zeroes between the terms – bitesizebo Apr 01 '18 at 02:54
  • @Z.Xie : How would a zero-padded sequence be decreasing? – Michael Apr 01 '18 at 02:55
  • @Michael Ahh, didn't see that – bitesizebo Apr 01 '18 at 02:56
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    @vadim123 assuming 1., note $na_n^2 = (na_n)a_n \le a_n$ for large $n.$ – zhw. Apr 01 '18 at 03:14
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    @Michael, I was wrong, and admit defeat. I have sent $5 to my favorite charity as penance, and removed my downvote. – vadim123 Apr 01 '18 at 03:16
  • @vadim123 : Thanks! Not many people will admit a mistake. And we all make them. I will delete my 1-line solution I left as a comment on your other page, as it has nothing to do with the question there. – Michael Apr 01 '18 at 03:17
  • Okay, but about proving part (1) though.... I don't see why (1) holds. – nullUser Apr 01 '18 at 03:17
  • @Michael I am not the downvoter, I just don't understand how to prove part (1). – nullUser Apr 01 '18 at 03:20
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    proving part (1) is kind of tricky and requires cauchy condensation style reasoning. It depends essentially on the sequence being decreasing – Rolf Hoyer Apr 01 '18 at 03:21
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    @vadim123 . Three yrs later. If $a_n\ge 0$ and $(a_n)n$ is decreasing but $na_n$ doesn't converge to $0$ then for some $r>0$ the set $S={n:na_n\ge r}$ is infinite. But then $\sum_n a_n$ does not meet the Cauchy Criterion: For any $n$ we have $\sup{n\le n'}\sum_{j=n}^{n'} a_j\ge$ $\sup_{n\le n'\in S}\sum_{j=n}^{n'} a_j \ge$ $\sup_{n\le n'\in S}(n'-n)a_{n'}\ge$ $ \sup_{n\le n'\in S}(n'-n)r/n'$ $=r.$ – DanielWainfleet Aug 25 '21 at 18:19
  • @DanielWainfleet : +1. Everyone seems to have a different way of proving 1. Here is how I was thinking about it: Suppose $a_n>1/n$ for infinitely many $n$. Then we can find a sequence of positive integers ${n[k]}{k=1}^{\infty}$ such that (i) $a{n[k]} > 1/n[k]$ for all $k$; (ii) $n[k+1] >2n[k]$ for all $k$. So then for all $k$: $$\sum_{i=n[k]+1}^{n[k+1]} a_i \geq \frac{n[k+1]-n[k]}{n[k+1]}\geq 1/2$$ – Michael Aug 25 '21 at 19:20
  • @Michael. Here is a 17th-centyry-style proof: If $\sum_{n\in\Bbb N}a_j=L\in\Bbb R$ then $\sum_{j>m}a_j$ is infinitely close to $0$ when $m$ is infinitely large. So if $m$ is infinitely large and if $n\ge 2m$ then $\frac {1}{1-m/n}\le 2$ so $na_n=$ $\frac {1}{1-m/n}(n-m)a_n\le$ $2(n-m)a_n\le$ $2\sum_{j=m+1}^na_j\le$ $ 2\sum_{j>m}a_j$ is infinitely close to $0.$ – DanielWainfleet Aug 27 '21 at 11:41
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The part 1 of @Michael's answer lies on the following fact: $na_{n}\rightarrow 0$ for decreasing/nonincreasing $(a_{n})$, $a_{n}\geq 0$ and that $\displaystyle\sum a_{n}<\infty$.

Consider $na_{2n}\leq a_{2n}+a_{2n-1}+\cdots+ a_{n+1}$ and also consider $na_{2n+1}$. Finally, consider $(2n+1)a_{2n+1}=2(na_{2n+1})+a_{2n+1}$. Note that we have $a_{n}\rightarrow 0$. So now consider the even subsequence and odd subsequence of $(ka_{k})$.

user284331
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Inspired by the zhw and RRL comments on Marty's answer, here is an example of a nonnegative and nonincreasing sequence $\{a_n\}_{n=1}^{\infty}$ such that $\sum_{n=1}^{\infty} a_n < \infty$ but $a_i > \frac{1}{i \log(i)}$ for infinitely many $i$.

We design the sequence to be constant over a sequence of successive "frames" of time. For frame $k \in \{1, 2, 3, ...\}$ define: $$a_n = \frac{5}{(2^{k^2})\log(2^{k^2})} , \quad \mbox{ for } n \in \{2^{(k-1)^2}+1, ..., 2^{k^2}\}$$ Define $a_1=a_2$. For $k \in \{1, 2, 3, ...\}$ define $T_k = 2^{k^2}-2^{(k-1)^2}$, which is the size of frame $k$.

By construction, this sequence satisfies $a_n = \frac{5}{n\log(n)} > \frac{1}{n\log(n)}$ for indices $n=2^{k^2}$ (for all $k \in \{1, 2, 3, ...\}$). Also: \begin{align} \sum_{n=2}^{\infty} a_n &= \sum_{k=1}^{\infty}\frac{5T_k}{2^{k^2}\log(2^{k^2})}\\ &= \frac{5}{\log(2)}\sum_{k=1}^{\infty} \frac{(T_k/2^{k^2})}{k^2} < \infty \end{align} where the last inequality uses the fact that $\lim_{k\rightarrow\infty} \frac{T_k}{2^{k^2}} = 1$.

The example can of course be modified to make $\{a_n\}$ strictly decreasing if desired. For example, just define $\{b_n\}$ by $b_n = a_n + 1/n^2$.

Michael
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  • I missed the fact that you posted this (+1) after we tried to inform the user below that the answer was incorrect. – RRL Jun 16 '21 at 17:32
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No.

If $\sum a_n$ converges and $a_n$ is decreasing, then $a_n < \dfrac1{n \log n}$ for all large enough $n$ since the sum of the latter diverges.

Then $na_n^2 \lt \dfrac{n}{n^2 \log^2 n} = \dfrac{1}{n \log^2 n} $ and the sum of this converges.

marty cohen
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    I don't follow why $a_n < 1/(n \log n)$, could you elaborate? – nullUser Apr 01 '18 at 02:44
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    Certainly the inequality $a_n < 1/(n\log(n))$ does not necessarily hold for all $n$. – Michael Apr 01 '18 at 02:48
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    I assume you meant $a_n < 1/(n\ln n)$ for large enough $n.$ I believe this is false. Do you have a reference or a proof? (The classic result here is $na_n\to 0$ and I don't think that can be improved.) – zhw. Apr 01 '18 at 03:45
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    I'd be interested in a proof that $a_n < 1/(n \log n)$ for large $n$. All we can conclude with the given information is $\liminf n \log n ,a_n = 0$ and there are cases where $\limsup n \log n ,a_n > 0$. – RRL Apr 01 '18 at 04:25
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    If $a_n=\frac{1}{1.1^n}$, then $a_2=\frac{1}{1.21}>\frac{1}{2\ln 2}\approx \frac{1}{1.38}$. So, to be safe with the claim, it should state "for sufficiently large $n$". – farruhota Apr 01 '18 at 07:23
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    @marty Inspired by the zhw and RRL comments, I give another answer that shows your conjecture is false. – Michael Apr 01 '18 at 13:19