<s>[INST] Show that the series $\sum^\infty_{k=1}\frac{1}{k^4}$ is bounded. Show that the series $\sum^\infty_{k=1}\frac{1}{k^4}$ is bounded.
What I've tried: Since
$\frac{1}{2^4}+\frac{1}{3^4}\leq \frac{1}{2^4}+\frac{1}{2^4}=\frac{1}{2^3}$
$\frac{1}{4^4}+\frac{1}{5^4}+\frac{1}{6^4}+\frac{1}{7^4}\leq \frac{1}{4^4}+\frac{1}{4^4}+\frac{1}{4^4}+\frac{1}{4^4}=\frac{1}{4^3}$
and so on. So the series in question is less than $1+\frac{1}{2^3}+\frac{1}{4^3}+\frac{1}{8^3}...$. If I can show the latter is bounded I'm done. But I can't so maybe there is and easier solution?
 [/INST] First, be careful with your index of summation - your series is not defined at $k = 0$. Anyways, note that $\frac{1}{k^4} \leq \frac{1}{k^2}$ for each positive integer $k$. Thus,
$$\sum_{k=1}^\infty \frac{1}{k^4} \leq \sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6}$$
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