Let $X\_1,...,X\_n$ be independent uniform random variables in [0,1] and assume $c>1/2$. Is it true that $$\mathbb{P}\left[\sum\_{i=1}^n X\_i \leq n \cdot c\right]$$ is increasing with respect to $n$?


I know that the sum of uniform independent random follows Irwin–Hall distribution, but it seems hard to work with the pdf. Maybe a coupling argument? I want to stress that we want to show the result for all $n$, not only for $n$ sufficiently large.



On the sum of uniform independent random variables
The answer is yes. Let $S\_n:=\sum\_{i=1}^n X\_i$, $x\in\mathbb R$, $n=2,3,\dots$, and 
\begin{equation}
 G\_n(x):=P(S\_n/n\le x)=\frac1{n!}\,\sum\_j(-1)^j\binom nj (nx-j)\_+^n, 
\end{equation}
where $u\_+:=0\vee u$; cf. **[[Irwin--Hall distribution](https://en.wikipedia.org/wiki/Irwin%E2%80%93Hall_distribution)]**. 
The summation here is over all integers $j$, with the usual convention that $\binom nj=0$ if $j\notin\{0,\dots,n\}$. 
Let 
\begin{equation}
 D\_n(x):=G\_{n+1}(x)-G\_n(x). 
\end{equation}
We have to show that $D\_n(x)\ge0$ for $x\ge1/2$. 


Clearly, $D\_n$ is $n-1$ times continuously differentiable, $D\_n=0$ outside $(0,1)$, and $D\_n(1-x)=-D\_n(x)$ (symmetry). 
For small $x>0$, one has $G\_n(x)=\frac1{n!}\,n^nx^n>\frac1{(n+1)!}\,(n+1)^{n+1}x^{n+1}=G\_{n+1}(x)$, whence $D\_n(x)<0$, whence $D\_n>0$ in a left neighborhood of $1$, with $D\_n(1)=0$. Also, $D\_n(1/2)=0$. 
So, it suffices to show that $D\_n$ has no roots in $(1/2,1)$. 


Suppose the contrary. Then, by the symmetry, $D\_n$ has at least $3$ roots in $(0,1)$. Then, by the Rolle theorem, the derivative $G\_n^{(n-1)}$ of $G\_n$ of order $n-1$ has at least $3+n-1=n+2$ roots in $(0,1)$. 
This contradicts the fact (proved below) that, for each integer $j$ such that $n-1\ge j\ge\frac{n-1}2$, $D\_n$ has exactly one root in interval 
$$h\_{n,j}:=[\tfrac jn,\tfrac{j+1}n).$$ 


Indeed, take any $x\in[1/2,1]$. It is not hard to see that, for $j\_x:=j\_{n,x}:=\lfloor nx\rfloor$, 
\begin{equation}
 G\_n^{(n-1)}(x)=n^{n-1}\sum\_{j=0}^{j\_x}(-1)^j\binom nj (nx-j) 
\end{equation}  

\begin{equation}
 =\frac{n^{n-1}}{n-1}(-1)^{j\_x}\binom n{j\_x+1}({j\_x}+1) ((n-1)x-{j\_x}). 
\end{equation}
Similarly, for $k\_x:=j\_{n+1,x}$, 
\begin{equation}
 G\_{n+1}^{(n-1)}(x)=\frac{(n+1)^{n-1}}{2n(n-1)}
(-1)^{k\_x}\binom{n+1}{k\_x+1}(k\_x+1)P(n,k\_x,x), 
\end{equation}
where 
\begin{equation}
 P(n,k,x):=-2 k \left(n^2-1\right) x+k (k n-1)+n \left(n^2-1\right) x^2. 
\end{equation}
Note that $k\_x\in\{j\_x,j\_x+1\}$. 
Now we have to consider the following two cases. 


Case 1: $k\_x=j\_x=j\in[\frac{n-1}2,n-1]$, which is equivalent to $x\in h'\_{n,j}:=[\frac jn,\frac{j+1}{n+1})$. It also follows that in this case $j\ge n/2$. In this case, it is not hard to check that $D\_n(x)$ equals $(-1)^j P\_{n,j,1}(x)$ in sign, where 
\begin{equation}
 P\_{n,j,1}(x):=2 j n^n (n-j)+x \left(-2 (n-1) n^n (n-j)-2 j (n+1)^n \left(n^2-1\right)\right)+j (n+1)^n (j n-1)+(n+1)^n
 \left(n^2-1\right) n x^2, 
\end{equation}
which is convex in $x$. Moreover, $P\_{n,j,1}(\frac jn)$ and $P\_{n,j,1}(\frac{j+1}{n+1})$ each equals $2 n^n - (1 + n)^n<0$ in sign. So, $P\_{n,j,1}<0$ and hence $D\_n$ has no roots in $h''\_{n,j}=[\frac jn,\frac{j+1}{n+1})$. 


Case 2: $k\_x=j+1$ and $j\_x=j\in[\frac{n-1}2,n-1]$, which is equivalent to $x\in h''\_{n,j}:=[\frac{j+1}{n+1},\frac{j+1}n)$. In this case, it is not hard to check that $D\_n(x)$ equals $(-1)^{j+1} P\_{n,j,2}(x)$ in sign, where 
\begin{equation}
 P\_{n,j,2}(x):=(j+1) \left((n+1)^n (j n+n-1)-2 j n^n\right)+2 (j+1) \left((n-1) n^n-(n+1)^n \left(n^2-1\right)\right) x+n
 \left(n^2-1\right) (n+1)^n x^2, 
\end{equation}
which is convex in $x$. Moreover, $P\_{n,j,1}(\frac{j+1}n)$ and $-P\_{n,j,1}(\frac{j+1}{n+1})$ each equals $2 n^n - (1 + n)^n<0$ in sign. So, $P\_{n,j,2}$ has exactly one root in $h''\_{n,j}$ and hence so does $D\_n$. 


Since the interval $h\_{n,j}$ is the disjoint union of $h'\_{n,j}$ and $h''\_{n,j}$, $D\_n$ has exactly one root in $h\_{n,j}=[\tfrac jn,\tfrac{j+1}n)$, and the proof is complete. 


