Recursive Si integral $I_{n+1}=\int^{I_{n}}_{0}\frac{\sin x}{x}dx$ I have seen this problem somewhere on the internet but I could not prove it.
Let $$I_{0}=\int^{\infty}_{0}\frac{\sin x}{x}dx$$ and then define
$$I_{n+1}=\int^{I_{n}}_{0}\frac{\sin x}{x}dx.$$
Show that 
$$\lim_{n\rightarrow\infty}\sqrt{n}\ I_{n}=3.$$
 A: It is easy to see that $I_n\to0$ as $n\to\infty$. So by Stolze's Theorem, one has
\begin{eqnarray}
\lim_{n\rightarrow\infty}n I^2_{n}&=&\lim_{n\rightarrow\infty}\frac{n}{I^{-2}_{n}}\\
&=&\lim_{n\rightarrow\infty}\frac{1}{I^{-2}_{n+1}-I^{-2}_{n}}\\
&=&\lim_{n\rightarrow\infty}-\frac{I^{2}_{n+1}I^{2}_{n}}{I^{2}_{n+1}-I^{2}_{n}}\\
&=&\lim_{x\to0}-\frac{x^2\left(\int_0^x\frac{\sin t}{t}dt\right)^2}{\left(\int_0^x\frac{\sin t}{t}dt\right)^2-x^2}\\
&=&\lim_{x\to0}-\frac{x^2\left(\int_0^x(1-\frac16t^2+O(t)^5)dt\right)^2}{\left(\int_0^x(1-\frac16t^2+O(t)^5)dt\right)^2-x^2}\\&=&9.
\end{eqnarray}
A: Denote the integral by $S(I_n)$
$$
I_{n+1}+I_n=S_{n+1}+S_n\\
I_{n+1}-I_n=S_{n+1}-S_n \\
\rightarrow
2 I_{n}= 2 S_{n}
$$
writing (Taylor expansion)
$$
S_{n}=I_{n-1}-\frac1{18}I_{n-1}^3+O(I_{n-1}^5)
$$
in the limit of big $n$ this leads to a "continuum approximation" (justification follows)
$$
18\,\partial_nI_n\sim-I^3_n+O(I_{n}^5) \quad (\star)
$$
or (the constant of Integration can be neglected in the limit $n\rightarrow\infty$)

$$
I_n \sim\frac{3}{\sqrt{n}} \quad (\star \star)
$$


Justification of $(\star)$:
We assume that $I_n$ is in the form of $(\star \star)$:
$$
I^{(k)}_n\ll I^{(m)}_n\quad (1)
$$
for $m<k$, so we can approximate the finite difference by a continious derivative.
$$
I_n-I_{n-1}\ll 1/\sqrt{n}\quad (2)
$$
so we can replace $I_{n-1}$ by $I_n$.
$$
I^{k}_n\ll I^{m}_n\quad (3)
$$
for $m<k$, so we can neglect higher order Terms in the Taylor approximation to $S(I_n)$.
Combining $(1)$ & $(2)$ & $(3)$ shows that our approximation $(\star)$ is consistent. 
