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got stuck on this proof given out of a practice test for my exam. the course is taught in rudin's principles of mathematics.

it says: let $a_n$ be a positive non-increasing sequence, prove: $\sum a_n$ converges if and only if $\sum n a_{n^2}$ converges. Note: the square is on the subscript.

i've been attempting it the same way Rudin proved Theorem 3.27, which was similar but with $2^k a_{2^k}$. however, i can't line up the patterns.

i've been doing two cases, one with $n > k^2$ and one with $n < k^2$. then attempting to show in either case, one sequence of partial sums bounds the other, so that either both are bounded are both are unbounded.

however, i can't quite get the bounding patterns. This may be the wrong method, which would explain why i can't get them. any help is appreciated.

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mac5
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  • one direct way would be to say that $\sum_n a_n$ converges iff there exists some $\epsilon > 0$ such that $a_n$ is $o(\frac{1}{n^{1+\epsilon}})$ or $o(\frac{1}{n (\ln n)^{1+\epsilon}})$ or $o(\frac{1}{n (\ln n)(\ln \ln n)^{1+\epsilon} })$ or $o(\frac{1}{n (\ln n)(\ln \ln n)(\ln \ln \ln n)^{1+\epsilon} })$, etc. (the Cauchy condensation test you are using with $\sum_k 2^k a_{2^k}$) – reuns Mar 28 '16 at 05:49
  • It might help to write $\sum_{n=1}^\infty a_n = \sum_{n=0}^\infty \sum_{k=n^2+1}^{(n+1)^2} a_{k} = [a_1] + [a_2+a_3+a_4] + [a_5+a_6+ \ldots + a_9] + \ldots$. Now use that $a_n$ is decreasing, so that for example $a_5,a_6,\ldots,a_9 \geq a_9$, to bound the sum from below. – Winther Mar 28 '16 at 05:57
  • and the more even direct way, using decreasing-ness if by grouping $a_{n^2}, \ldots, a_{(n+1)^2-1}$ and bounding by $((n+1)^2-1-n^2))a_{(n+1)^2-1}$ – reuns Mar 28 '16 at 05:58
  • Winther, im not seeing exactly how that would help here. isnt that exactly the same strategy for proving this for 2^k * a(2^k) – mac5 Mar 28 '16 at 06:21
  • user1952009, your first response is not something we've covered yet. and i'm not understanding your second response – mac5 Mar 28 '16 at 06:42
  • It's relevant since $a_5 + a_6 + \ldots + a_9 \geq 5a_9 > 3a_{9} = 3a_{3^2}$ and $a_{10} + \ldots + a_{16} \geq 7a_{16} > 4a_{16} = 4a_{4^2}$ so $\sum a_n > 1a_{1^2} + 2a_{2^2} + 3a_{3^2} + 4 a_{4^2} + \ldots$ – Winther Mar 28 '16 at 06:47

1 Answers1

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On the one hand, $$ [a_1]+[a_2+a_3+a_4]+[a_5+\cdots + a_9] + \cdots \geq a_1 + 3 a_4 + 5a_9 + \cdots. $$

On the other hand, $$ [a_1]+[a_2+a_3+a_4]+[a_5+\cdots + a_9] + \cdots \leq a_1 + 3 a_1 + 5a_4 + \cdots. $$

Bounding the odd sequences crudely yields $$ \sum n a_{n^2}\leq \sum a_n \leq a_1+\sum 3n a_{n^2}, $$ from which the result follows.

pre-kidney
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  • this shows boundedness of partial sums which implies convergence of non-negative sequences, right? – mac5 Mar 28 '16 at 16:07
  • Yes. Since the sequence $a_n$ is positive non-increasing, $\sum a_n$ converges if and only if it is finite. (In other words, the only way for such a sum not to converge is if it diverges to positive infinity.) – pre-kidney Mar 29 '16 at 00:09