# On the continued fraction of $e$

The following question comes up during analysis of Padé approximants to $e^x$ (see my related question in MathOverflow for more background). Recall that the continued fraction expansion of $e$ is $$e = [2,1,2,1,1,4,1,1,6,1,1,8,\ldots].$$ Consider \begin{align*} R_{k,k} &= [2,1,2,1,1,4,\ldots,1,1,2(k-1),1,1] \text{ and } \\ R_{k+1,k-1} &= [2,1,2,1,1,4,\ldots,1,1,2(k-1),1,1,k]. \end{align*} Empirically, we always have $$|R_{k+1,k-1} - e| < |R_{k,k} - e|.$$ Can anyone explain why?

Edit: Now I can explain why. We have $$\frac{R_{k,k} + R_{k+1,k-1}}{2} = [2,1,2,1,1,4,\ldots,1,1,2k,1,1,2(k-1),\ldots,1,1,4,1,1,3].$$ When $k$ is even (odd), this will differ from the continued fraction in an even (odd) position, by being too small. This means that the value is larger (smaller) than $e$. Since $R_{k,k}$ is an upper (lower) bound on $e$, and $R_{k+1,k-1}$ is a lower (upper) bound on $e$, we get what we want.

Now it remains to prove this continued fraction expansion of $(R_{k,k} + R_{k+1,k-1})/2$...

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I hope you don't mind that I edited in a link to your related question on MathOverflow. – Rahul Feb 18 '12 at 10:20
I presume you're very much aware of the explicit expressions for the Padé approximants of $\exp\,z$? – J. M. Feb 18 '12 at 10:36
@J.M.: That's how I proved the rest of my MathOverflow question. But my estimates are not good enough to decide which of $|R_{k+1,k-1}-e|,|R_{k,k}-e|$ is bigger. – Yuval Filmus Feb 18 '12 at 10:47
It seems that it is possible to obtain $R_{k+1,k-1}$ from $R_{k,k}$ by applying a Bauer-Muir transformation w.r.t. a convenient sequence $\left( w_{k}\right) _{k\geq 0}$, $w_{3k}=1/k$, $w_{3k\pm 1}=0$. – Américo Tavares Feb 18 '12 at 23:19
@Américo: Can you expand on this? Doesn't the Bauer-Muir transformation conserve the value of the continued fraction? – Yuval Filmus Feb 18 '12 at 23:52