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 $(R_{k,k} + R_{k+1,k-1})/2$...