# Sum of cubes of all inverted integers

$$\sum \limits_{n \in \mathbb{N}} \frac{1}{n^{3}}$$

Is there a way to sum this sequence analytically?

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You should read a little about Riemann's Zeta, Apèry's Constant, and Euler's way of computing all values for even exponents, and why he failed to compute odd values $\geq 3$. – Pedro Tamaroff Feb 20 '12 at 21:45
A technique due to Euler that works for the even powers is the following. Let $f(x)=\sin\sqrt{x}/\sqrt{x}$. This can be expanded in a Taylor series, and can also be expressed as the infinite product $(1-x/\pi^2)(1-x/4\pi^2)\ldots$. Equating the linear terms gives the result for an exponent of 2, and equating the quadratic term gives the result for 4. From what Peter says, it sounds like there's nothing similar that works for the odd powers. – Ben Crowell Feb 20 '12 at 21:47
This is sorta the second half of this question. There are other MSE questions floating around related to this one, like its irrationality and relation to powers of $\pi$. – anon Feb 20 '12 at 22:59

No one has ever found a closed form for this number, and it hasn't been for lack of trying. All the "obvious" possibilities (e.g., $p\pi^3/q$ for reasonable sized integers $p$ and $q$) have been ruled out.