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I'm not sure if I should call this "the sum of a sum" but I can't think of another phrase for it.

Anyhow, I'm confused about excercises such as this one

$$\sum_{j=2}^{\infty} \left( \sum_{k=2}^{\infty} k^{-j} \right)$$

where I want to compute the sum. I'm thinking maybe I could rewrite it as

$$\sum_{k=2}^{\infty} \frac{1}{k^{2}} + \sum_{k=2}^{\infty} \frac{1}{k^{3}}+... $$

but I don't know where to go from there or if that is even correct thinking. (I just began my first course in linear analysis.)

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1 Answer 1

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Hint: you can try and rearrange the summands - all of them are positive. In particular, you can try to sum by $ j $ first

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  • $\begingroup$ Let $ S $ denote your sum. First question: is it true that $ S = \sum\limits_{k=2}^\infty \sum\limits_{j=2}^\infty k^{-j} $? Second question: what is the sum $ \sum\limits_{j=2}^\infty k^{-j} $ for $ k \geq 2 $? $\endgroup$
    – Jytug
    Nov 16, 2015 at 12:41
  • $\begingroup$ Thank you for commenting further. I've been trying now to solve $\sum_{j=2}^{\infty} k^{-j}$ but the indice beginning at 2 messes things up for me. If it said from $j=0$ to infinity I would say that for each $k$ it would be $1/(1-k)$... Then I thought maybe I could use that same formula but subtract $1$ and $k$ (since $\sum_{j=0}^{\infty} k^{-j} = 1 + k + k^2 +... = 1/(1-k))$ but that didn't help me. $\endgroup$
    – m.bing
    Nov 16, 2015 at 13:08
  • $\begingroup$ Oh, and to answer the first question, yes, I would say that it is true. $\endgroup$
    – m.bing
    Nov 16, 2015 at 13:11
  • $\begingroup$ It did help! Now I solved it. Thank you and sorry for spamming the comments. :) $\endgroup$
    – m.bing
    Nov 16, 2015 at 13:23
  • $\begingroup$ To be fair, your previous comment was wrong: $\sum\limits_{j=0}^\infty k^{-j} = 1/(1 - (1/k))$ $\endgroup$
    – Jytug
    Nov 16, 2015 at 13:39

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