Is there a simpler way to show that $x_n = {1\over n^3} \sum_{j=1}^n \sum_{i=1}^j i$ is bounded?

Question

Let $n \in \mathbb N$ and: $$ x_n = {1\over n^3} \sum_{j=1}^n \sum_{i=1}^j i $$ Show that the sequence $x_n$ is bounded.

I've solved this by expanding the sums and then finding closed forms of the sums:

$$ \begin{align} x_n &= {1\over n^3} \sum_{j=1}^n \sum_{i=1}^j i = \\ &= {1\over n^3} \sum_{k=1}^n {k(k+1)\over 2} = \\ &= {1\over 2n^3} \left( \sum_{k=1}^n {k^2} + {n(n+1) \over 2}\right) = \\ &= {1\over 2n^3} \left( \frac{n(n+1)(2n+1)}{6} + {n(n+1) \over 2}\right) = \\ &= {1\over 6n^2} (n+1)(n+2) \end{align} $$

Obviously $x_n$ is greater than $0$ (more precisely it is greater than $1 \over 6$). Since the terms of the sequence decrease the upper bound is defined by the lowest $n$. So here $x_n \le 1$ which gives the bounds for the sequence:

$$ 0 < {1\over 6} < x_n \le 1 $$

Personally I don't like the approach I've used since it implies performing clumsy manipulations with the sums (and knowing closed forms of particular sums. Luckily I've already found those sums while solving problems from the section on sums).

Is there a simpler way to show $x_n$ is bounded?

Answer 1

$${1\over n^3} \sum_{j=1}^n \sum_{i=1}^j i \le {1\over n^3} \sum_{j=1}^n \sum_{i=1}^j j= {1\over n^3} \sum_{j=1}^n j^2\le {1\over n^3} \sum_{j=1}^n n^2=1 $$

Hence $0<x_n \le 1$

Answer 2

Note that $$\sum_{j=1}^n\sum_{i=1}^j i=\sum_{j=1}^n\sum_{i=1}^j \sum_{k=1}^i 1=\sum_{1\leq k\leq i\leq j\leq n} 1\ \leq \ n^3\ ,$$ and is about ${1\over6}n^3$, by symmetry.