Let $x$ be a real number. To prove...

Question

Let $x$ be a real number. Define the sequence $(x_n)_{n\ge1}$ recursively by $x_1=1$ and $x_{n+1}=x^n+nx_n$ for $n\ge1$. Prove that, $$\prod_{n=1}^\infty \bigg(1-\dfrac{x^n}{x_{n+1}}\bigg)=e^{-x}$$

Could not approach the sum. Please help!

Answer 1

Hint:

We are to prove that, $$\prod_{n=1}^\infty \bigg(1-\dfrac{x^n}{x_{n+1}}\bigg)=e^{-x}$$

Let's have a look to how the LHS actually looks like. A little manipulation in $x_{n+1}=x^n+nx_n$ would yield,

$$\displaystyle\prod_{n=1}^\infty \bigg(1-\dfrac{x^n}{x_{n+1}}\bigg)=\prod_{n=1}^\infty n\dfrac{x_n}{x_{n+1}}=(1.2.3...)\dfrac{x_1}{x_2}\dfrac{x_2}{x_3}\dfrac{x_3}{x_4}...=(1.2.3...)\dfrac{1}{x_2}\dfrac{x_2}{x_3}\dfrac{x_3}{x_4}...$$

Let $a_n=\dfrac{n!}{x_{n+1}}.$ Note if $a_n^{-1}\to e^{x}$ the rest of the result becomes immediate.

For showing $a_n^{-1}\to e^{x}$ keep obtaining the previous iterations untill $x_1$ by using $x_{n+1}=x^n+nx_n$ as,

$$\dfrac{x_{n+1}}{n!}=\dfrac{x^n+nx_n}{n!}=\dfrac{x^n}{n!}+\dfrac{x_n}{(n-1)!}=\dfrac{x^n}{n!}+\dfrac{x^{n-1}+(n-1)x_{n-1}}{(n-1)!}$$ etc.