Prove that limN→∞(1+0N2)(1+1N2)..(1+N−1N2)=√e\displaystyle \lim_{N\to \infty}(1+\frac{0}{N^2})(1+\frac{1}{N^2})..(1+\frac{N-1}{N^2}) = \sqrt e

Given that \dot x(t) = tx\dot x(t) = tx, x(0) = 1x(0) = 1. Then we find the solution x(t)=e^{t^2/2}x(t)=e^{t^2/2}. This implies x(1)=\sqrt ex(1)=\sqrt e.

Then I use Euler’s method to find x(1)x(1). Dividing the unit time interval into NN
pieces, we get the step size h = \frac{1}{N}h = \frac{1}{N}.

x_{n+1}=x_n+f(t_n,x_n)h=x_n+\frac{n}{N}\cdot x_n\cdot \frac{1}{N}=(1+\frac{n}{N^2})\cdot x_nx_{n+1}=x_n+f(t_n,x_n)h=x_n+\frac{n}{N}\cdot x_n\cdot \frac{1}{N}=(1+\frac{n}{N^2})\cdot x_n.

Since x(0)=1x(0)=1, then I get x_N=(1+\frac{0}{N^2})(1+\frac{1}{N^2})..(1+\frac{(N-1)}{N^2})x_N=(1+\frac{0}{N^2})(1+\frac{1}{N^2})..(1+\frac{(N-1)}{N^2}).

So, \displaystyle \lim_{N\to\infty}x_N=\sqrt e\displaystyle \lim_{N\to\infty}x_N=\sqrt e.

Is this limit correct? Are there other ways to show (1+\frac{0}{N^2})(1+\frac{1}{N^2})..(1+\frac{(N-1)}{N^2}) \to \sqrt e(1+\frac{0}{N^2})(1+\frac{1}{N^2})..(1+\frac{(N-1)}{N^2}) \to \sqrt e as N\to \infty.N\to \infty.



2 Answers


Noting for small x>0x>0,