Expected value of square[X/sigmaX] = 1/n^2(1+1/pi)? Please see the below link for the complete description. I already have an answer shown in the link, based on many Excel simulations ($n=4$ to $100$, $x_i$ generated by RAND() function of Excel). I want to know if the answer $\frac{1}{n^2}\left(1+\frac{1}{\pi}\right)$ has a theoretical basis. If no, how to derive as a function of $n$? Thanks.
https://drive.google.com/file/d/1j7yvuysxiBI7uMrOiaHzS8kxzUfhy3VL/view?usp=sharing

(Added by David Roberts) The file says:
$$
E\left[\left(\frac{x_1}{\sum_{i=1}^n x_i}\right)^2\right] = 
\frac{1}{n^2}\left(1+\frac{1}{\pi} \right)
$$
where $x_i \sim U(0,1)$.
 A: Your conjecture is obviously false for $n=1$, and I think it is false for any $n$. Using the Irwin-Hall formula, one can write an explicit but very complicated expression for the expectation in question, and that expression will not involve $\pi$. 
Also, the conjecture is false for all large enough $n$. More specifically, letting 
$$
m_n:=Er_n^2,\quad r_n:=\frac{nx_1}{\sum_{i=1}^n x_i}=\frac{x_1}{\bar x_n},\quad \bar x_n:=\frac1n\,\sum_{i=1}^n x_i, 
$$
we have 
$$
\lim\inf_n n^2 E\Big(\frac{x_1}{\sum_{i=1}^n x_i}\Big)^2=\lim\inf_n m_n\ge\frac43>1+\frac1\pi. \tag{1}
$$
Indeed, by the strong law of large numbers, $\bar x_n\to Ex_1=\frac12$ almost surely (a.s.) and hence $r_n\to2x_1$ a.s. Also, $Ex_1^2=\frac13$. So, by the Fatou lemma,
$$\lim\inf_n m_n=\lim\inf_n Er_n^2\ge E(2x_1)^2=4Ex_1^2=4/3,
$$
so that the first inequality in (1) follows. 

Working a bit harder, we can show that actually $m_n\to\frac43$, that is, 
$$
E\Big(\frac{x_1}{\sum_{i=1}^n x_i}\Big)^2\sim\frac4{3n^2}. \tag{2}
$$ 
Indeed, by Hoeffding's inequality for sums of independent bounded random variables, we have $P(\bar x_n<1/4)\le e^{-n/8}$. Also, obviously $0\le r_n\le n$. So,
$$Er_n^21_{\bar x_n<1/4}\le n^2P(\bar x_n<1/4)\le n^2e^{-n/8}\to0. 
$$
On the other hand, $r_n^21_{\bar x_n\ge1/4}\le1/(1/4)^2<\infty$ and $r_n^21_{\bar x_n\ge1/4}\to4x_1^2$ a.s., again by the strong law of large numbers. So, by dominated convergence, 
$$Er_n^21_{\bar x_n\ge1/4}\to4Ex_1^2=\frac43. 
$$
Thus,
$$m_n=Er_n^2=Er_n^21_{\bar x_n<1/4}+Er_n^21_{\bar x_n\ge1/4}\to0+\frac43=\frac43,
$$
as claimed. That is, we have (2).  
