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My argument: $$1=(\frac{1}{2})^2+(\frac{1}{3})^2+\cdots+(\frac{1}{2})^3+(\frac{1}{3})^3+\cdots=\sum_{k=2}^\infty (\frac{1}{k})^2+\sum_{k=2}^\infty (\frac{1}{k})^3+\cdots .$$

Explanation) First, for any natural number $n\geq2$, the following holds: $$\sum_{k=1}^\infty (\frac{1}{n})^k=(\frac{1}{n})+(\frac{1}{n})^2+(\frac{1}{n})^3+\cdots= \frac{\frac{1}{n}}{1-\frac{1}{n}}=\frac{1}{n-1}.$$ (the infinite geometric series.)

So, we obtain $$1=(\frac{1}{2})+(\frac{1}{2})^2+\cdots=\sum_{k=1}^\infty (\frac{1}{2})^k,$$ $$\frac{1}{2}=(\frac{1}{3})+(\frac{1}{3})^2+\cdots=\sum_{k=1}^\infty (\frac{1}{3})^k,$$ $$\frac{1}{3}=(\frac{1}{4})+(\frac{1}{4})^2+\cdots=\sum_{k=1}^\infty (\frac{1}{4})^k,$$ and so on.

From above equalities, $$1=(\frac{1}{2})+(\frac{1}{2})^2+\cdots$$ $$=((\frac{1}{3})+(\frac{1}{3})^2+\cdots)+(\frac{1}{2})^2+(\frac{1}{2})^3+\cdots$$ $$=((\frac{1}{4})+(\frac{1}{4})^2+\cdots)+(\frac{1}{3})^2+(\frac{1}{3})^3+\cdots+(\frac{1}{2})^2+(\frac{1}{2})^3+\cdots$$ $$=\cdots .$$ If $p\ge2$, then p-series absolutely converges. Hence we can change the order of terms in series as follows: $$1=(\frac{1}{2})^2+(\frac{1}{3})^2+(\frac{1}{4})^2+\cdots$$ $$+(\frac{1}{2})^3+(\frac{1}{3})^3+(\frac{1}{4})^3+\cdots$$ $$+(\frac{1}{2})^4+(\frac{1}{3})^4+(\frac{1}{4})^4+\cdots$$ $$=\sum_{k=2}^\infty (\frac{1}{k})^2+\sum_{k=2}^\infty (\frac{1}{k})^3+\sum_{k=2}^\infty (\frac{1}{k})^4+\cdots.$$ Thus, "1" becomes the sum of p-series(exactly from $k=2$ to infinity).

Is this explanation correct?

mathvoca
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    Please use MathJax to format your posts. This is hard to read. Here's a tutorial – saulspatz Nov 07 '19 at 05:07
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    Also please represent your sums using the $\sum$ symbol with limits. It is often easy to hide errors in the ellipses because it is not always clear what terms are included in the sum. – Ross Millikan Nov 07 '19 at 05:14

1 Answers1

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So what you are saying is

$1 =\sum_{m=2}^{\infty} \sum_{k=2}^{\infty} \dfrac1{k^m} $.

Let's check.

$\begin{array}\\ \sum_{m=2}^{\infty} \sum_{k=2}^{\infty} \dfrac1{k^m} &=\sum_{k=2}^{\infty} \sum_{m=2}^{\infty} \dfrac1{k^m} \qquad\text{(reverse order of summation)}\\ &=\sum_{k=2}^{\infty}\dfrac{1/k^2}{1-1/k} \qquad\text{(just a geometric series)}\\ &=\sum_{k=2}^{\infty}\dfrac{1}{k^2-k} \qquad\text{(multiply num and dec by }k^2)\\ &=\sum_{k=2}^{\infty}\dfrac{1}{k(k-1)} \qquad\text{(rewrite)}\\ &=\sum_{k=2}^{\infty}(\dfrac1{k-1}-\dfrac1{k}) \qquad\text{(now we can telescope)}\\ &=1\\ \end{array} $

Yup.

marty cohen
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