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I want to show that $\sum_{n=1}^{\infty} \sum_{k=n}^{\infty}\frac{1}{k^{3}} < \infty$.

Therefore I want to show that $\sum_{n=1}^{\infty} \sum_{k=n}^{\infty}\frac{1}{k^{3}}$ converges. From Wolfram-Alpha I can conclude this by the comparision test. My problem is then to use the comparision test. Which positive series should I compare $\sum_{n=1}^{\infty} \sum_{k=n}^{\infty}\frac{1}{k^{3}}$ with?

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I cannot see a good comparison, but you can do it as follows: We have \begin{align*} \sum_{n=1}^\infty \sum_{k=n}^\infty \frac 1{k^3} &= \sum_{k=1}^\infty \sum_{n=1}^{k} \frac 1{k^3}\\ &= \sum_{k=1}^\infty \frac 1{k^2} < \infty \end{align*} as $\sum_{n=1}^k \frac 1{k^3} = \frac 1{k^2}$ and $\sum_{k=1}^\infty \frac 1{k^2}$ converges.

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If you want to compare with something, $\sum {1\over k^3}$ cries out to be compared with $\int {1\over x^3}\,{\rm d}x$. For $n>1$ you have $$\sum_{k=n}^\infty {1\over k^3}<\int_{n-1}^\infty {1\over x^3}\,{\rm d}x={1\over 2(n-1)^2}$$ For $n=1$ we then can use $$\sum_{k=1}^\infty {1\over k^3}=1+\sum_{k=2}^\infty {1\over k^3}<1+{1\over 2}$$ Another integral comparison now gives $$\sum_{n=1}^\infty \sum_{k=n}^\infty {1\over k^3}<{3\over 2}+{1\over 2}+\sum_{n=3}^\infty {1\over 2(n-1)^2}<{3\over 2}+{1\over 2}+\int_2^\infty {1\over 2(x-1)^2}\,{\rm d}x={5\over 2}$$ Of course, martini's method is superior, as it gives the exact answer $\sum_{k=1}^\infty {1\over k^2}$, which is well known to be equal to ${\pi^2 \over 6}=1.6449\dots$

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