Comparison test. 
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?
 A: 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.
A: 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 ${\pi^2 \over 6}=1.6449\dots$
