<s>[INST] Is this inequality true for all k ? $\sum_{n=k}^{n=+\infty} \frac{1}{n^4} \leq (\sum_{n=k}^{n=+\infty} \frac{1}{n^2})^3$ Can it be generalized for other powers ? Wolfram seems to say it is true for k below 20000.
I stumbled upon it randomly when trying to approximate $\sum_{n=1}^{n=+\infty} \frac{1}{n^4}$. 
My reasoning was :
$$\left(\sum_{n=k}^{n=+\infty} \frac{1}{n^2}\right)^2=\sum_{n=k}^{n=+\infty} \frac{1}{n^4} + (\text{double products}) \geq\sum_{n=k}^{n=+\infty} \frac{1}{n^4}$$
So
$$\sum_{n=1}^{n=+\infty} \frac{1}{n^4} \leq \sum_{n=1}^{n=k-1} \frac{1}{n^4}+\left(\sum_{n=k}^{n=+\infty} \frac{1}{n^2}\right)^2 \leq \left(\sum_{n=1}^{n=k-1} \frac{1}{n^4}\right)+\left(\frac{1}{k-\frac{1}{2}}\right)^2$$
where the last inequality comes from An inequality: $1+\frac1{2^2}+\frac1{3^2}+\dotsb+\frac1{n^2}\lt\frac53$.
Then I noticed that, perhaps, I could raise the last term to the power of 3 instead of just 2, making the inequality stronger.
 [/INST] For $k > 1$, 
$$\sum_{n=k}^\infty \dfrac{1}{n^4} < \int_{k-1}^\infty \dfrac{dx}{x^4} = \dfrac{1}{3(k-1)^3}$$
$$\left(\sum_{n=k}^\infty \dfrac{1}{n^2}\right)^3 > \left(\int_{k}^\infty \dfrac{dx}{x^2}\right)^3 = \dfrac{1}{k^3} $$
$\dfrac{1}{k^3} > \dfrac{1}{3(k-1)^3}$ when $3(k-1)^3 > k^3$, which 
is true for $k > 3.2612$.
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