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?
 A: \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}
A: Without resorting to known p-series, do the following: You've already done part of the work in comparing subsets of terms via a Cauchy-condensation-type move. The next step is to realize that $4 = 2^2, 8 = 2^3,$ and so on. You then have
$$ \sum_{j \geq 1} \frac{1}{(2^3)^{j-1}} = \sum_{j \geq 1} \left( \frac{1}{8} \right)^{j-1} = \frac{8}{7}$$
(you should check this by using the geometric series formula). Thus by comparison we have
$$ \sum_{j \geq 1} \frac{1}{j^4} \leq \frac{8}{7}.$$
Notice that this estimate is off by $-0.060533$ from the actual value of the sum of $\frac{\pi^4}{90}.$
A: 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,
$$\sum_{k=1}^\infty \frac{1}{k^4} \leq \sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6}$$
