Question

Find an approximation to $1+\frac{1}{3+}\frac{1}{5+}\frac{1}{7+}\frac{1}{9+}\frac{1}{11+}....$ which differs from the true value by less than $.0001$.

Answer 1

Here, the convergents are $1,\frac{4}{3},\frac{21}{16},\frac{151}{115},.........;$

The fourth convergent $\frac{151}{115}$ differs from the true value by $<\frac{1}{(115{)}^2}$ which is $<\frac{1}{(100{)}^2}=$ $0.0001$