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I was going over old HW solutions to study for an exam and I noticed something strange.

Picture illustrating power sum I was trying to simplify.

Basically, the number in the denominator was 90 based on the table I looked up the value in, but the solution to this problem said it should be 1440. The problem I'm working on specifies only even n, so I suspect that's part of it somehow.

All I could figure out was that 1440/90 = 16, or 2^4.

I suspect that's what happened there. That the denominator was multiplied by 2^4 for some reason. But why does that happen? And how is it different for odd n?

Ben Grossmann
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q-compute
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1 Answers1

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Hint: you can rewrite the sum as $$ \sum_{k=1}^\infty \frac{1}{(2k)^4} $$ where I've used the substitution $n = 2k$ for $k = 1,2,3,\dots$

Ben Grossmann
  • 225,327
  • So, are you saying it's appropriate to rewrite it in that way in order to simplify, because once you simplify you no longer have an n? Because the n in this problem MUST be even. In the way you wrote it, n could be any integer. But I suppose that works because what you wrote the sum as will always be even? Maybe someone can put this into a more eloquent explanation? – q-compute Dec 14 '16 at 14:24
  • "once you simplify you no longer have an n": I don't understand what this means. Anyway, see my latest edit, maybe that makes more sense. – Ben Grossmann Dec 14 '16 at 14:26
  • I was just saying that pi^4/1440 (what the power sum simplifies to) doesn't have an n in it. But yes, that edit helps a lot. Thanks! – q-compute Dec 14 '16 at 14:30