Finding $\lim_{n\to \infty} \sum_{i=1}^n\sum_{j=1}^i \frac{j}{n^3}.$ I have some problems with task.
I have an idea how to solve, but I am not sure, can you check, please?
$\lim_{n\to \infty} \sum_{i=1}^n\sum_{j=1}^i \frac{j}{n^3};$
$$\lim_{n\to \infty} \sum_{i=1}^n\sum_{j=1}^i \frac{j}{n^3} = \lim_{n\to \infty} \frac{1}{n^3}\sum_{i=1}^n \Biggl(1+2+3+4+...+(i-1)+i\Biggr)=\\= \lim_{n\to \infty} \frac{1}{n^3}\Biggl(1+2+3+4+...+(i-1)+\sum_{i=1}^ni\Biggr)= \frac{i(i+1)n(n+1)}{4n^3}=0 $$
is it correct?
Thank you for your help!
 A: Here is a treatment where it is visualized what to do:
$$
\lim_{n\to \infty} \sum_{i=1}^n\sum_{j=1}^i \frac{j}{n^3}\\
= \lim_{n\to \infty} \frac{1}{n^3}\sum_{i=1}^n (\sum_{j=1}^i j)\\
= \lim_{n\to \infty} \frac{1}{n^3}\sum_{i=1}^n \frac{i(i+1)}{2}\\
= \lim_{n\to \infty} \frac{1}{2 n^3}\sum_{i=1}^n i(i+1)\\
= \lim_{n\to \infty} \frac{1}{2 n^3} \frac{1}{3} n (n + 1) (n + 2)\\
= \frac{1}{6}
$$
A: hint
if we use the identities
$$1+2+3+...+i=\frac 12(i^2+i)$$
and
$$1^2+2^2+3^2+...+n^2=$$
$$\frac n6(n+1)(2n+1)$$
the limit becomes
$$\lim_{n\to+\infty}\frac{ 1}{2n^3}\Bigl(\frac n6(n+1)(2n+1)+\frac 12n(n+1)\Bigr)=\frac{1}{6}$$
A: Here is a combinatorial proof. Observe that since $j=\sum_{k=1}^j$ we can express the desired sum as $$\displaystyle\sum_{i=1}^n\sum_{j=1}^i\sum_{k=1}^j 1=\sum_{1\leq i\leq j\leq k\leq n}1.$$
This sum counts how many ways we can pick 3 integers from $1$ to $n$ in increasing order. These may be enumerated by choosing $3$ bars from $n+3$ stars and bars, so that the total number of ways is $$\displaystyle\binom{n+3}{3}=\frac{1}{6}n(n+1)(n+2)=\frac16 n^3+O(n^2).$$
