Zero Limit of Sum of Squares of Terms with Bounded Range

[DaTario]

Homework Statement

Let $\sum_{i=1}^N a_{i,N} /N = 1$ with $0 < a_{i,N} < M > 1$ for all $i$ and $N$.
Show that $\lim_{N \to \infty} \sum_{i=1}^N (a_i /N)^2 = 0$.

Relevant Equations

We know that each $a_{i,N}/N$ is positive and less than one implying that their square is even smaller.

I don't know how to show that this limit is zero.
It seems that $\sum_{i=1}^N a_{i,N} /N = 1$ and the fact that $0 < a_{i,N} < M > 1$ implies that some $a_{i,N}$ are less than one.
Another conclusion I guess is correct to draw is that $\lim_{N \to \infty} \sum_{i=1}^N a_{i,N}^2 /N < 1$.

 

[statdad]

Try to use the fact that the $a_{i,N}$ are bounded by $M$ so that $\left(\dfrac{a_{i,N}}{N}\right)^2$ is less than some fraction times $\dfrac{a_{i,N}}{N}$.

 

[DaTario]

Thank you, staddad.