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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!

Shaun
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GIFT
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    The last two equalities do not make sense. Why is $i$ remaining if this is only an index? – Dietrich Burde Sep 04 '19 at 19:05
  • @DietrichBurde because the first sum has index i, but the second sum has the last element also i, as I understand the first sum works only for the last element of the second sum and the previous elements are only numbers and they make no contribution – GIFT Sep 04 '19 at 19:24
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    That's not how summation notation works. One should have: $$\sum_{i=1}^n \sum_{j=1}^i a_{i,j}=\sum_{j=1}^i a_{1,j}+\sum_{j=1}^2 a_{2,j}+\cdots +\sum_{j=1}^n a_{n,j}.$$ Alternatively, you can view the double sum as being over all $i,j$ such that $n\geq i\geq j\geq 1$. – Semiclassical Sep 04 '19 at 19:29
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    Hint: the limit is $\int_0^1\int_0^x y,dy,dx$. – metamorphy Sep 04 '19 at 19:39
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    @metamorphy One can push that back a step further to $\int_0^1 \int_0^x \int_0^y 1,dz,dy,dz$ in which case the problem is purely geometrical. – Semiclassical Sep 04 '19 at 19:52

3 Answers3

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Here is a treatment where it is visualized what to do:

$$ \lim_{n\to \infty} \sum_{i=1}^n\sum_{j=1}^i \frac{j}{n^3}\\ = \lim_{n\to \infty} \frac{1}{n^3}\sum_{i=1}^n (\sum_{j=1}^i j)\\ = \lim_{n\to \infty} \frac{1}{n^3}\sum_{i=1}^n \frac{i(i+1)}{2}\\ = \lim_{n\to \infty} \frac{1}{2 n^3}\sum_{i=1}^n i(i+1)\\ = \lim_{n\to \infty} \frac{1}{2 n^3} \frac{1}{3} n (n + 1) (n + 2)\\ = \frac{1}{6} $$

Andreas
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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

$$\lim_{n\to+\infty}\frac{ 1}{2n^3}\Bigl(\frac n6(n+1)(2n+1)+\frac 12n(n+1)\Bigr)=\frac{1}{6}$$

hamam_Abdallah
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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 $$\displaystyle\binom{n+3}{3}=\frac{1}{6}n(n+1)(n+2)=\frac16 n^3+O(n^2).$$

Semiclassical
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