Question on convergence of series: Does $\frac{1}{n}\sum_{i=1}^n x_i \to s$ imply $\frac{\max_{i\le n} x_i}{\sum_{i=1}^n x_i} \to 0$? Suppose $x_i \geq 0, i=1, ..., n$ and $\frac{1}{n}\sum_{i=1}^n x_i \to s > 0  \text{ as } n \to \infty$. Is this true?
$$\frac{\max x_i}{\sum_{i=1}^n x_i} \to 0 \text{ as } n \to \infty$$
 A: The fact that $\frac1n\sum_1^n x_i$ converges to $s>0$ does not prevent the sequence $(x_n)$ from being unbounded. For example, you can augment $x_n$ by $\log(n)$ whenever $n$ is a power of $2$ (and do nothing for other $n$) without changing the limit of $\frac1n\sum_1^n x_i$.
So the most general case is where the sequence $(x_n)$ is not bounded. The result you want to show (and maybe this is what you intended) is:
Claim. If $x_i\ge 0$ and $\frac1n\sum_1^n x_i$ converges to a finite limit $s>0$ then
$$
\frac{\max(x_1,\ldots,x_n)}{\sum_1^n x_i}\to0\qquad\text {as $n\to\infty$}.
$$
Proof. It is enough to show that $\frac1n\max(x_1,\ldots,x_n)\to0$. Write $s_n:=x_1+\cdots+x_n$. Since $\frac1ns_n$ converges, it follows that
$$
\frac{x_n}n=\frac {s_n}n-\frac{s_{n-1}}n=\frac {s_n}n-\left(\frac {n-1}n\right)\frac{s_{n-1}}{n-1}
$$
converges to zero. Now let $\epsilon>0$. Since $x_n/n\to0$, there exists $N$ such that $\frac{x_n}n<\epsilon$ whenever $n>N$. Define
$$A:=\max\{x_1,\ldots,x_N\}+1.$$
So if $n>\max(N, \frac A\epsilon)$ we have $$0\le\frac1n\max(x_1,\ldots,x_n)=\max\left(\frac {x_1}n,\ldots,\frac{x_n}n\right)\stackrel{(*)}<\epsilon.
$$
To see (*), argue by cases: if $i\le N$ then $\displaystyle \frac{x_i}n<\frac An<\epsilon$ while if $N<i\le n$ then $\displaystyle\frac{x_i}n\le\frac{x_i}i<\epsilon$.
A: Yes. 
\begin{align}
\frac{\max x_i}{\sum_{i=1}^n x_i} = \frac{\frac{\max x_i}{n}}{\frac{\sum_{i=1}^n x_i}{n}} 
\end{align}
Assuming $\max x_i$ is finite, the numerator tends to $0$ and the denominator tends to $s$, which is non-zero.
