Question involving approximation, taylor series and proving

Question

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:

$2\left (\frac{1}{7\times 3^{7}}+\frac{1}{9\times 3^{9}}+\frac{1}{11\times 3^{11}}\cdots\right) < \frac{1}{7\times 2^{2} \times 3^{5}}$

Answer 1

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

Answer 2

Since $9\gt 7$, and $11\gt 7$, and $13\gt 7$, and so on, the tail $$\frac{2}{7\cdot 3^7}+\frac{2}{9\cdot 3^9}+\frac{2}{11\cdot 3^{11}}+\frac{2}{13\cdot 3^{13}}+\cdots$$ is less than the sum of the geometric series $$\frac{2}{7\cdot 3^7}\left(1+\frac{1}{3^2}+\frac{1}{3^4}+\frac{1}{3^6}+\cdots\right).$$ But $$1+\frac{1}{3^2}+\frac{1}{3^4}+\frac{1}{3^6}+\cdots=\frac{9}{8}.$$