Proving the Uniqueness of Positive Solution of $x(x+1)(x+2)\cdots(x+2020)-1=0$

[anemone]

Show that the equation $x(x+1)(x+2)\cdots(x+2020)-1=0$ has exactly one positive solution $x_0$ and prove that this solution $x_0$ satisfies $\dfrac{1}{2020!+10}<x_0<\dfrac{1}{2020!+6}$.

 

[jvicens]

To show that the equation $x(x+1)(x+2)\cdots(x+2020)-1=0$ has exactly one positive solution $x_0$, we can use the Intermediate Value Theorem. This theorem states that if $f$ is a continuous function on an interval $[a,b]$ and $f(a)$ and $f(b)$ have opposite signs, then there exists at least one value $c \in (a,b)$ such that $f(c)=0$.

In this case, we can let $f(x)=x(x+1)(x+2)\cdots(x+2020)-1$ and consider the interval $[1,2020]$. We can see that $f(1)=-1$ and $f(2020)=2020!-1>0$, so by the Intermediate Value Theorem, there exists at least one value $x_0 \in (1,2020)$ such that $f(x_0)=0$. This means that the equation has at least one positive solution.

To prove that this solution $x_0$ satisfies $\dfrac{1}{2020!+10}<x_0<\dfrac{1}{2020!+6}$, we can use mathematical induction.

First, we will show that $\dfrac{1}{2020!+10}<1$. This can be proven by noting that $2020!+10>2020!$ and $\dfrac{1}{2020!}$ is a very small number, so $\dfrac{1}{2020!+10}<\dfrac{1}{2020!}<1$.

Next, we will show that if $\dfrac{1}{2020!+10}<x_0<\dfrac{1}{2020!+6}$, then $\dfrac{1}{2020!+10}<x_0+1<\dfrac{1}{2020!+6}$. This can be done by multiplying both sides of the inequality by $x_0+1$ and using the fact that $x_0(x_0+1)(x_0+2)\cdots(x_0+2020)=1$.

Therefore, by mathematical induction, we can conclude that $\dfrac{1}{2020!+10}<x_0<\dfrac{1}{