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.
 A: 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 $k > 3.2612$.
A: Let's see what we can say about
comparing
$s_1(k)
=\sum_{n=k}^{\infty} \frac1{n^a}
$
with
$s_2(k)
=\left(\sum_{n=k}^{\infty} \frac1{n^b}\right)^c
$
for large enough $k$,
where
$a > 1$ and $b > 1$
so the sums converge.
Using the integral approximation,
$s_1(k)
=\sum_{n=k}^{\infty} \frac1{n^a}
\approx \int_k^{\infty} \frac{dx}{x^a}
=-\frac1{(a-1)x^{a-1}}\big|_k^{\infty}
=\frac1{(a-1)k^{a-1}}
$.
Therefore
$s_2(k)
=\left(\sum_{n=k}^{\infty} \frac1{n^b}\right)^c
\approx \left(\frac1{(b-1)k^{b-1}}\right)^c
=\frac1{(b-1)^c k^{c(b-1)}}
$,
so
$\dfrac{s_1(k)}{s_2(k)}
\approx \dfrac{\frac1{(a-1)k^{a-1}}}{\frac1{(b-1)^c k^{c(b-1)}}}
= \dfrac{{(b-1)^c k^{c(b-1)}}}{{(a-1)k^{a-1}}}
= \dfrac{{(b-1)^c }}{{(a-1)}}k^{c(b-1)-(a-1)}
$.
Therefore
if
$c(b-1) > a-1$,
then
$s_1(k)
> s_2(k)
$
for all large enough $k$;
if
$c(b-1) < a-1$,
then
$s_1(k)
< s_2(k)
$
for all large enough $k$;
If
$c(b-1) = a-1$,
then
$\dfrac{s_1(k)}{s_2(k)}
\approx \dfrac{{(b-1)^c }}{{(a-1)}}
$,
so the result
depends on this ratio.
For your case of
$a=4$ and $b=2$,
the key difference is
$c(b-1)-(a-1)
=c-3
$.
If
$c > 3$,
then
$s_1(k) > s_2(k)$
for large enough $k$;
if
$c < 3$,
then
$s_1(k) < s_2(k)$
for large enough $k$.
If $c=3$,
which is your case,
the ratio  is
$\dfrac{(b-1)^c }{(a-1)}
=\dfrac{1}{3}
< 1
$,
so
$s_1(k) 
\approx \frac13 s_2(k)
< s_2(k)
$
for large enough $k$,
which confirm's
Robert Israel's result
(good thing too,
because any result of mine
that differs from
a result of his
is probably wrong).
