<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] \begin{equation}
\sum_{k=1}^{\infty}\frac{1}{k^4}
\leq \sum_{k=1}^{\infty}\frac{1}{k^2}
= 1 + \sum_{k=1}^{\infty}\frac{1}{\left(k + 1\right)^2}
\leq 1+\int_{1}^{\infty}\frac{1}{x^2} dx
= 1 + 1
= 2
\end{equation}
</s>