Question involving approximation, taylor series and proving Question:
Consider the approximation $$\ln(2)\approx 2\left ( \frac{1}{3}+\frac{1}{3\times 3^{3}}+\frac{1}{5\times 3^{5}} \right )$$
Prove that the error in this approximation is less than $$\frac{1}{7\times 2^{2} \times 3^{5}}$$
Attempt:
It looks like the expression comes from the taylor series expansion so:
$\ln(1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+... \text{ for }\ -1< x< 1$
$\ln(1-x)=-x-\frac{x^{2}}{2}-\frac{x^{3}}{3}-\frac{x^{4}}{4}+...$
$\therefore \ln\left ( \frac{1+x}{1-x} \right )=2\left ( x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+\frac{x^{7}}{7} \right )$
$\text{Now let}\ x=\frac{1}{3}$
$\therefore \ln(2)=2\left ( \frac{1}{3}+\frac{1}{3\times 3^{3}}+\frac{1}{5\times 3^{5}}+\frac{1}{7\times 3^{7}}\right )$
So we have to prove that:
$2\left (\frac{1}{7\times 3^{7}}+\frac{1}{9\times 3^{9}}+\frac{1}{11\times 3^{11}}\cdots\right) < \frac{1}{7\times 2^{2} \times 3^{5}}$

We can say that $2\left (\frac{1}{7\times 3^{7}}+\frac{1}{9\times 3^{9}}+\frac{1}{11\times 3^{11}}\cdots\right)<2 \left (\frac{1}{7\times 3^{7}}+\frac{1}{7\times 3^{9}}+\frac{1}{7\times 3^{11}}\cdots\right)= 2 \left(\frac{1}{7\times 3^7}\right)
\div \left(1-\frac{1}{3^2}\right)=\frac{1}{7\times 2^2\times3^5}$
