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This time I remembered to include the exponent $k$!

As mentioned in a previous post, this answer to this is key to finding an explicit expression for the sums of the powers of the reciprocals of the figurate numbers, for powers greater then or equal to $2$.

Aside from the beautiful patterns that may result, this approach may provide new relationships among the zeta and other functions. For example, consider triangular numbers ($a=1$) where $k=8$. The expression is:

$$\begin{align*}1/1^8 &+ 1/3^8 + \dots =\\ &= 2^9\left[\binom70\zeta(8) + \binom92\zeta(6) + \binom{11}4\zeta(4) + \binom{13}6\zeta(2) + \binom{15}7\zeta(0)\right] \end{align*}$$

where $\zeta(x)$ is the Riemann zeta function with $\zeta(0)=-1/2$.

It's apparent that for an exponent $k$, I can find a well-defined expressions in the form of $c(i)\zeta(2i)$, where $c(i)$ is a coefficient like $2^9\binom70$.

So, in the limit of $k=\infty$, we have something like $\sum_{i=1}^\infty c(i)\zeta(2i)=1$, giving us a possibly new relationship among the even-valued zeta.

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LaTeX: $\sum_{N=1}^{\infty} \left(\frac{1}{aN - (a-2)}\right)^k, k \ge 2, a \ge 1$ –  user2468 Mar 12 '12 at 17:50
    
Tip: "As mentioned in aprevious post..." please provide hyperlinks whenever you makes such references. –  user2468 Mar 12 '12 at 17:53
    
@BrianM.Scott I removed the \displaystyle in the title since it breaks the layout of the questions. Hope you don't mind. meta.math.stackexchange.com/questions/3135/… –  user17762 Mar 12 '12 at 22:22
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2 Answers 2

Maple says your sum is $S(a,k) = \left( -1 \right) ^{k}\Psi \left( k-1,2/a \right) {a}^{-k} / \Gamma \left( k \right)$. It can also be expressed using the Hurwitz zeta function and its derivative.

Some interesting values include $S(1,k) = \zeta(k)-1$, $S(2,k) = 2^{-k} \zeta(k)$, $S(4,k) = (2^{-k} - 4^{-k}) \zeta(k)$.

EDIT: Perhaps this is something like what you're looking for (for $a > 2$): $$S(a,k) = \frac{1}{2^k} + \frac{1}{a^k} \sum_{j=0}^\infty {j+k-1 \choose j} \zeta(j+k) (-2/a)^j $$

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Thank you, GEdgar. Would you like to see my results as the occur when exploring applications to figurate numbers. – Greg Simay just now edit Thank you, GEdgar. Would you like to see my results as the occur when exploring applications to figurate numbers. –  Greg Simay Mar 14 '12 at 20:22
    
Thank you, Robert Israel and Maple. Would you like to see any results from my exploration of figurate numbers? –  Greg Simay Mar 14 '12 at 20:24
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Actually, since $k \ge 2$ we don't need the derivative. The Hurwitz zeta function is $$ \zeta (z,\nu) = \sum_{i = 0}^{\infty} \frac{1}{(i + \nu)^{z}} $$ for $\mathrm{Re} z \ge 2$ and $\nu$ not a nonpositive integer. So this problem is $$ \frac{1}{a^k}\zeta \Bigl(k,\frac{2}{a}\Bigr) = \sum_{N = 1}^{\infty} \frac{1}{\bigl(a (N - 1) + 2\bigr)^{k}} $$

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Thank you, GEdgar. Would you like to see my results as the occur when exploring applications to figurate numbers. –  Greg Simay Mar 14 '12 at 20:22
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