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On the Remarkable Fact that a Sequence with Convergent Sum when Dotted with the Harmonic Sequence Yields a New Convergent Sum

December 3rd, 2017 No comments

This is kind of incredible: take a sequence with convergent sum

 \left[a_i \right] \cdot \mathbf{1} = \sum_i a_i < \pm \infty

and let

 H = \left[ \begin{array}{c} 1 \\ \frac{1}{2} \\ \frac{1}{3} \\ \vdots \end{array} \right]

 It turns out that

 \left[ a_i \right] \cdot H < \pm \infty

A corollary of this is that if we define

 H^q = \left[ \begin{array}{c} 1 \\ \frac{1}{2^q} \\ \frac{1}{3^q} \\ \vdots \end{array} \right]

 then

 \left[ a_i \right] \cdot H^q < \pm \infty

With this we can prove that the p-series for p>2 converges.  Take the known fact that

 \sum_i \frac{1}{i^2} < \pm \infty \iff \left[ \begin{array}{cccc} 1 & \frac{1}{2^2} & \frac{1}{3^2} & \ldots \end{array} \right] \cdot \mathbf{1}

Then

 \left[ \begin{array}{cccc} 1 & \frac{1}{2^2} & \frac{1}{3^2} & \ldots \end{array} \right] \cdot H = \sum_i \frac{1}{i^{2+1}} < \pm \infty

 Clearly, repeated application of H yields:  

 \left[ \begin{array}{cccc} 1 & \frac{1}{2^2} & \frac{1}{3^2} & \ldots \end{array} \right] \cdot H^q = \sum_i \frac{1}{i^{2+q}} < \pm \infty

Next define

 H^{-q} = \left[\begin{array}{c} 1 \\ 2^q \\ 3^q \\ \vdots \end{array} \right]

 Since 

 \left[ \begin{array}{cccc} 1 & \frac{1}{2^2} & \frac{1}{3^2} & \ldots \end{array} \right] \cdot H^{-1} = \sum_i \frac{1}{i^1}

diverges, it seems clear that H as a function (say, h) is not surjective.

This clears up the question I had about whether a sequence with convergent sum dot H^{-1} was convergent (answer: not generally).

Let me know if you are interested in a proof (which does not rely on the Comparison Test).