Doubt about how to compute $\sum_{n=1}^{2019}{\frac{1+2+3+4...+n}{1^3+2^3+3^3...+2019^3}}$ In the book I am reading, I have encountered the following sum.
$$S = \sum_{n=1}^{2019}{\frac{1+2+3+4...+n}{1^3+2^3+3^3...+2019^3}}$$
From here, I factored the denominator since it does not seem to be dependent on $n,$ and I rewrote the expression as
$$\frac{1}{1^3+2^3+3^3+ \cdots +2019^3} \sum_{n=1}^{2019}{1+2+3+4+ \cdots +n}=\frac{1}{1^3+2^3+3^3 + \cdots +2019^3} \sum_{n=1}^{2019}{\sum_{k=1}^n}k$$
From here, I proceeded to compute the sum since there is  an equation for the sum of the first $n$ terms, and then, I solved the sum of this equation. Also, I applied the equation of the sum of the first $n^3$ terms. Finally, my result was wrong. My question is regarding my initial approach to the problem since everything else is very simple. I have thought about other ways to solve this, and I can not find it. The official answer is
$$\frac{2019}{2010}.$$
Give this problem a try, and if you arrive at the solution, I invite you to help me understand it.

Like you mentioned, the denominator is independent of $n,$ hence we can consider it separately. We have that $D = 1^3 + 2^3 + \cdots + 2019^3 = \sum_{i = 1}^{2019} i^3.$ Observe that $S$ can be written as $$\begin{align*} S = \frac 1 D \sum_{n = 1}^{2019} (1 + 2 + \cdots + n) &= \frac 1 D \sum_{n = 1}^{2019} \sum_{k = 1}^n k \\ \\ &= \frac 1 D \sum_{n = 1}^{2019} \frac{n(n + 1)} 2 \\ \\ &= \frac 1 {2D} \biggl(\sum_{n = 1}^{2019} n^2 + \sum_{n = 1}^{2019} n \biggr) \\ \\ &= \frac 1 {2D} \biggl(\frac {2019 \cdot 2020 \cdot 4039} 6 + \frac {2019 \cdot 2020} 2 \biggr) \end{align*}.$$ One can evaluate $D$ using the sum of cubes formula $\sum_{k = 1}^n k^3 = \frac{n^2 (n + 1)^2} 4$ to find that $S = \frac{2021}{6117570}.$
