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So this is a question from a test I'm trying to solve: Given the series $a_{p}=\sum_{n=2}^\infty\frac{1}{n^{p}}$ (I know this series converges), I need to prove that the series ${\sum_{p=2}^\infty}a_{p}$ also converges.

At first I thought to prove this with induction, since for every $p\ge2$ each series converges, therefore their partial limit sequences also converge, and from the properties of limits the sum of a finite number of series will also tend to some limit.

But I realized that since I have an infinite amount of series, I probably can't use induction. Any help would be appreciated.

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We have that

$$ a_p=\zeta(p) -1 \sim \frac{1}{2^p} \text{ as } p \to \infty, $$

(refer to Growth of Riemann zeta function $\zeta(k)$)

therefore

$${\sum_{p=2}^\infty}a_{p}$$

converges by limit comparison test.

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For an alternative way note that every term in the double series is positive, so we may interchange the order of summation by Tonelli's theorem. We can evaluate the geometric series in $p$ to find $$ \sum \limits_{p=2}^\infty \sum \limits_{n=2}^\infty \frac{1}{n^p} = \sum \limits_{n=2}^\infty \sum \limits_{p=2}^\infty \frac{1}{n^p} = \sum \limits_{n=2}^\infty \left(\frac{1}{1-\frac{1}{n}} - 1 - \frac{1}{n}\right) = \sum \limits_{n=2}^\infty \left(\frac{1}{n-1} - \frac{1}{n}\right) \, . $$ Now we either employ the estimate $\frac{1}{n-1} - \frac{1}{n} = \frac{1}{(n-1)n} < \frac{1}{(n-1)^2}$ to obtain an upper bound or, even better, we look through a telescope to spot the exact value of the series.

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$\begin{array}\\ a_{p} &=\sum_{n=2}^\infty\frac{1}{n^{p}}\\ &=\frac1{2^p}\sum_{n=2}^\infty\frac{1}{(n/2)^{p}}\\ &=\frac1{2^p}(1+\sum_{n=3}^\infty\frac{1}{(n/2)^{p}})\\ &\le\frac1{2^p}(1+\sum_{n=3}^\infty\frac{1}{(n/2)^{2}}) \qquad\text{since }(n/2)^2 \le (n/2)^p\\ &\le\frac1{2^p}(1+4\sum_{n=3}^\infty\frac{1}{n^2})\\ &\lt\frac1{2^p}(1+4\sum_{n=3}^\infty\frac{1}{n(n-1)}) \qquad\text{standard trick}\\ &=\frac1{2^p}(1+4\sum_{n=3}^\infty(\frac{1}{n-1}-\frac1{n}))\\ &=\frac1{2^p}(1+4\frac1{2})\\ &=\frac52\frac1{2^p}\\ \text{so}\\ \sum_{p=2}^{\infty} a_p &< \sum_{p=2}^{\infty}\frac52\frac1{2^p}\\ &=\frac52\\ \end{array} $

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