What is the value of $a_1a_2\cdots a_{2019}$?

Question

Let $a_1=\frac 34$ and for any $n\geq2$ $4a_n=4a_{n-1}+\frac {2n+1}{1^3+2^3+\cdots n^3 }$. What is the value of $a_1a_2\cdots a_{2019}$?

I tried $1^3+2^3+\cdots +n^3=\frac {n^2(n+1)^2}{4}$ and

I wroted

$4(a_n-a_{n-1})=\frac {2n+1}{1^3+2^3+\cdots n^3 }$. Then

$$4(a_2-a_1)=\frac {2×2+1}{1^3+2^3 }\\ 4(a_3-a_2)=\frac {2×3+1}{1^3+2^3+3^3 }\\ \cdots$$

When we add the equations side by side, the left side becomes simpler. But the right side gets complicated. I tried to find closed form for $a_n$ but it seems very complicated.

Answer 1

Observe $$a_n - a_{n-1} = \frac{2n+1}{n^2(n+1)^2} = \frac{(n+1)^2 - n^2}{n^2 (n+1)^2} = \frac{1}{n^2} - \frac{1}{(n+1)^2}. \tag{1}$$ So $$a_m - a_1 = \sum_{n=2}^m a_n - a_{n-1} = \frac{1}{2^2} - \frac{1}{(m+1)^2},$$ hence $$a_m = 1 - \frac{1}{(m+1)^2} = \frac{m(m+2)}{(m+1)^2}. \tag{2}$$ Therefore, $$\prod_{m=1}^{2019} a_m = \prod_{m=1}^{2019} \frac{m}{m+1} \frac{m+2}{m+1} = \frac{2021}{4040}.$$