Doubt about how to compute $\sum_{n=1}^{2019}{\frac{1+2+3+4...+n}{1^3+2^3+3^3...+2019^3}}$

Question

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.

Answer 1

Let's replace $2019$ by $T$. In the numerator:

$$ \sum_{n=1}^T \sum_{k=1}^n k = \sum_{n=1}^T \frac{n(n+1)}{2} = \frac{(T+1)^3 - (T+1)}{6}$$

In the denominator:

$$ \sum_{k=1}^T k^3 = \frac{(T+1)^4 - 2 (T+1)^3 + (T+1)^2}{4} $$

So:

$$ \frac{ \sum_{n=1}^T \sum_{k=1}^n k }{\sum_{k=1}^T k^3} = \frac{4}{6} \frac{(T+1)^3 - (T+1)}{(T+1)^4 - 2 (T+1)^3 + (T+1)^2} = \frac{2 (T+2)}{3 T (T+1)}$$

For $T=2019$ this is $\dfrac{2021}{3030 \cdot 2019}$. The official answer is wrong.

Or maybe the question was supposed to be $$ \frac{1}{2}\sum_{n=1}^{2019} \frac{1 + 2 + \ldots + n}{1^3 + 2^3 + \ldots + n^3} $$ which would work out to $\dfrac{2019}{2020}$.

Answer 2

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 $S = \frac{2021}{6117570}.$