Today I Learned

April 10th

Today I learned a fun example of a telescoping product, from the Purple Comet. We evaluate\[P:=\prod_{k=2}^N\left(\frac1{k^3-1}+\frac12\right).\]Just looking at this, it is not obvious why it should telescope, but I promise it does.

It turns out that the $\frac12$ is quite crucial to the structure here. For fixed $k,$ we note that\[\frac1{k^3-1}+\frac12=\frac12\cdot\frac{k^3+1}{k^3-1}.\]These factors of $\frac12$ we now ignore so that we want to evaluate\[2^{N-1}P=\prod_{k=2}^N\frac{k^3+1}{k^3-1}.\]As for the remaining fraction, it turns out to telescope in competing ways. We write\[\frac{k^3+1}{k^3-1}=\frac{k+1}{k-1}\cdot\frac{k^2-k+1}{k^2+k+1}.\]This first part telescopes, cancelling after skipping term, as\[\prod_{k=2}^N\frac{k+1}{k-1}=\frac31\cdot\frac42\cdot\frac53\cdot\frac64\cdot\ldots\cdot\frac N{N-2}\cdot\frac{N+1}{N-1}=\frac{N(N+1)}2.\]As for the other side, we note the complex roots of $k^2-k+1$ (the primitive sixth roots of unity) are shifted one to the right of the complex roots of $k^2+k+1$ (the primitive cubic roots of unity). Algebraically, this means that\[k^2+k+1=(k+1)^2-(k+1)+1.\]Thus, we do have telescoping, cancelling on the next term, as\[\prod_{k=2}^N\frac{k^2-k+1}{(k+1)^2-(k+1)+1}=\frac{2^2-2+1}1\cdot\frac1{N^2+N+1}.\]Putting this all together, we see that\[2^{N-1}P=\left(\prod_{k=2}^N\frac{k+1}{k-1}\right)\left(\prod_{k=2}^N\frac{k^2-k+1}{k^2+k+1}\right)=\frac{N(N+1)}2\cdot\frac3{N^2+N+1}.\]So we have the following.

This follows from rearranging the final expression in the above discussion. $\blacksquare$

As an aside, we note that we can evaluate the infinite product $2^{N-1}P$ as\[\prod_{k=2}^\infty\frac{k^3+1}{k^3-1}=\lim_{N\to\infty}\frac{3N(N+1)}{2\left(N^2+N+1\right)}.\]As $N\to\infty,$ the quadratics are approximately equal, so the product approaches $\boxed{3/2}.$

What amuses me most about this product is that it has managed to somewhat elegantly combine two kinds of telescoping. This is typically not too difficult when one has telescoping series, for we can just directly sum the desired telescoping behavior to get something perhaps contrived. For example,\[\sum_{k=1}^\infty\left(\frac1k-\frac1{k+2}+\frac1{k^3+1}-\frac1{(k+3)^3+1}\right)\]will telescope cleanly. A similar process can be done for products, but it's impressive that this product comes out so cleanly.