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>




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(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 $x\_i \sim U(0,1)$.



Expected value of square[X/sigmaX] = 1/n^2(1+1/pi)?
Your conjecture is obviously false for $n=1$, and I think it is false for any $n$. Using the [Irwin-Hall formula](https://en.wikipedia.org/wiki/Irwin%E2%80%93Hall_distribution), 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](https://en.wikipedia.org/wiki/Law_of_large_numbers#Strong_law), $\bar x\_n\to Ex\_1=\frac12$ almost surely (a.s.) and hence $r\_n\to2x\_1$ a.s. Also, $Ex\_1^2=\frac13$. So, by the [Fatou lemma](https://en.wikipedia.org/wiki/Fatou%27s_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. 




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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](https://en.wikipedia.org/wiki/Hoeffding%27s_inequality#General_case_of_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). 


