# 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 $2d-2$. Then we can speak of $A_d(x)$ 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 ?

This question was also asked on StackExchange.

• I don't know, but it seems the closed interval [-1,1] should be the answer. – The Masked Avenger Mar 8 '15 at 19:58
• @TheMaskedAvenger I played with it a bit, and for most rationals (including those in [-1,1]) it seems to grow quite rapidly. – მამუკა ჯიბლაძე Mar 9 '15 at 14:37
• Played with it some more - actually it seems that $\frac{x(x+1)(2x+1)A_d(x)}{A_d(-1/4)}$ converges quite nicely to $\frac3{32}\sin(2\pi x)$; in particular, $A_d(x)/A_d(-1/4)$ stays bounded for every $x$ (while $A_d(-1/4)$ grows very rapidly) – მამუკა ჯიბლაძე Mar 11 '15 at 8:43

Let $S_d(n)=n(n+1)(2n+1)A_d(n)=\sum_{k=1}^nk^{2d}$, then numerical evidence suggests $$\lim_{d\to\infty}\frac{4^{2d+1}}{E_{2d}}S_d(x)=\sin(2\pi x),$$ where $E$ are the Euler numbers. Using Faulhaber's formula this reduces to $$2\lim_{d\to\infty}4^{2d-k}\frac{B_{2(d-k)}}{(2(d-k))!}/\frac{E_{2d}}{(2d)!}=(-1)^{k+1}\pi^{2k+1}$$ where $B$ are the Bernoulii numbers.
It turns out that $A_d(x)$ is unbounded at all $x$ except for $A_d(1/2)=A_d(-3/2)=\frac1{3\cdot2^{2d-1}}$ and $A_d(1)=A_d(-2)=1/6$.
• @AntonioVargas Yes I thought about using asymptotics but could not establish a rigorous argument since I do not know what is behind these $\sim$s... – მამუკა ჯიბლაძე Mar 12 '15 at 0:43