"Distance" between summands in Taylor Series of Cosine

The $n$-th curve crosses through the line $y=c$ at $x$ coordinate $$ \frac{x^{2n}}{\left(2n\right)!}=c $$ or, equivalently, $x=(k!c)^{1/k}$ where $k=2n$. Therefore, the distance between two adjacent curves along this line is $$ \left(\left(k+2\right)!c\right)^{1/(k+2)}-\left(k!c\right)^{1/k} $$ which converges to $2/e\approx0.73576$ as $k \rightarrow \infty$. To see this, let $F_{k}\equiv(\sqrt{2\pi c^{2}k})^{1/k}$ and use Stirling's approximation to get the asymptotic $$ \frac{k+2}{e}F_{k+2}-\frac{k}{e}F_{k} $$ and apply the identity $\log(ab)=\log a+\log b$ and L'Hopital's rule to get $$ \begin{multline} \lim_{k}F_{k} =\lim_{k}\exp\left(\frac{\ln(2\pi c^{2}k)}{2k}\right) =\exp\left(\lim_{k}\frac{\ln(2\pi c^{2}k)}{2k}\right)\\ =\exp\left(\lim_{k}\left\{ \frac{\ln(2\pi c^{2})}{2k}+\frac{\ln k}{2k}\right\} \right) =1. \end{multline} $$ Note that the quantity $2/e$ is

• independent of $c$ and
• "relatively" close to $\pi/4\approx0.78540$.

I don't know what you mean by distance between graphs, because they pairwise intersect and therefore have $0$ distance, but I will assume you mean the distance between the $y=1$ intercepts.

We have $x^{2n}/(2n)!=1$ exactly if $x=(2n)!^{1/(2n)}$, so let us study the sequence $n!^{1/n}$, which by your observation should grow about $\pi/8$ when $n$ increases by one. You can easily check that the growth is not exactly constant, but we can prove that $n!^{1/n}\sim n/e$, so $\lim_{n\to\infty}n!^{1/n}/n=1/e$. Furthermore, we can prove that $\lim_{n\to\infty}(n+1)!^{1/(n+1)}-n!^{1/n}=1/e$, which does not in fact follow from the previous limit. Now since $\pi/8\approx0.3927$ and $1/e\approx0.3679$ I guess you can say you were kinda close for someone eyeballing it.