Question

Show that $\frac{1}{2+}\frac{2}{3+}\frac{3}{4+}\cdots =\frac{3-e}{e-2}$; hence show that $e$ lies between $2\frac{2}{3}$ and $2\frac{8}{11}$.

Answer 1

$\frac{1}{1+}\frac{2}{2+}\frac{3}{3+}\frac{4}{4+}\frac{5}{5+}.......$

$=\frac{1}{1+}\frac{1}{1+}\frac{1}{2+}\frac{2.4}{3.4+}\frac{3.5}{5+}.......$

$=\frac{1}{1+}\frac{1}{1+}\frac{1}{2+}\frac{2}{3+}\frac{3}{4+}.......$

Thus, $e-1=1+=\frac{1}{1+}\frac{1}{2+}\frac{2}{3+}.......$

If x denotes the value of the given expression we have,

$e-1=1+\frac{1}{1+x}$

$e-2=\frac{1}{1+x}$

$3-e=\frac{x}{1+x}$

$x=\frac{3-e}{e-2}$

Now, $x<\frac{1}{2}$ and $>\frac{1}{2+}\frac{2}{3}$ or $\frac{3}{8}$

$\therefore \frac{3-e}{e-2}<\frac{1}{2}$ and $>\frac{3}{8}$

$\therefore 2\frac{2}{3}<e<2\frac{8}{11}$