When is this sum of perfect powers bounded For any positive integers $n,d$, let
$$
A_d(n)=\frac{\sum_{k=1}^n k^{2d}}{n(n+1)(2n+1)}
$$
It is easy to see (and well-known) that for fixed $d$, $A_d(.)$ is
a polynomial of degree $2d-2$. Writing $A_d(x)$ then makes sense
for any $x\in{\mathbb R}$, not just the positive integers.
Is it known for which real numbers $x$ the sequence $(A_d(x))_{d\geq 1}$ is
bounded ?
UPDATE : this question was also asked on MO 
 A: In a comment above Markus Scheuer gave links to materials which readily contain an answer. I will use highly informative lecture notes by Omran Kouba pointed out by him.
It follows from the formulæ on page 6 of these notes that what we are looking at are actually the Bernoulli polynomials:
$$
S_{2d}(x):=x(x+1)(2x+1)A_d(x)=\frac1{2d+1}B_{2d+1}(x+1)
$$
which by Corollary 5.2 (page 18) of the same notes gives that $(-1)^{d+1}\frac{(2\pi)^{2d+1}}{2(2d)!}S_{2d}(x)$ converges to $\sin(2\pi x)$ uniformly on every compact subset of $\mathbb C$.
It thus follows that for large $d$, $A_d(x)$ behaves like $(-1)^{d+1}\frac{2(2d)!}{(2\pi)^{2d+1}x(x+1)(2x+1)}\sin(2\pi x)$, i. e. grows quite rapidly - by Stirling's formula, $\log(A_d(x))$ grows like $2d\log\frac d{\pi e}+$ const.
Well in fact as Ewan Delanoy points out in the comment below this only shows unboundedness of $A_d(x)$ when $x$ is not a half-integer. To capture this case too, first note that
$$
A_d(0)=\lim_{x\to0}\frac{B_{2d+1}(x+1)}{(2d+1)x(x+1)(2x+1)}
=\lim_{x\to0}\frac{B_{2d+1}'(x+1)}{(2d+1)(6x^2+6x+1)}=B_{2d}(0)
$$
since $B_n'(t)=nB_{n-1}(t)$ and $B_{2d}(1)=B_{2d}(0)$, and similarly
$$
A_d(-\frac12)=-2B_{2d}(\frac12)=2(1-\frac1{2^{2d-1}})B_{2d}(0).
$$
For other half-integers use $B_n(x+1)=B_n(x)+nx^{n-1}$ (all the needed equalities are in the mentioned notes) to obtain
$$
A_d(x)=\frac{(x-1)(2x-1)A_d(x-1)+x^{2d-1}}{(x+1)(2x+1)}
$$
for $x\ne-1,-1/2$ and
$$
A_d(x-1)=\frac{(x+1)(2x+1)A_d(x)-x^{2d-1}}{(x-1)(2x-1)}
$$
for $x\ne1,1/2$. This gives $A_d(-1)=A_d(0)$,
$$
A_d(1/2)=A_d(-3/2)=\frac1{3\cdot2^{2d-1}},
$$
$$
A_d(1)=A_d(-2)=\frac16
$$
and shows (by induction) that at all other half-integers and integers $A_d$ is unbounded.
Thus $A_d(x)$ goes to infinity with $d$ everywhere except that it goes to zero at $x=1/2,-3/2$ and is constantly $1/6$ at $x=1,-2$.
PS all this could be slightly simplified using $A_d(-1-x)=A_d(x)$...
