<s>[INST] 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:
$2\left (\frac{1}{7\times 3^{7}}+\frac{1}{9\times 3^{9}}+\frac{1}{11\times 3^{11}}\cdots\right) < \frac{1}{7\times 2^{2} \times 3^{5}}$
 [/INST] 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 
$$1+\frac{1}{3^2}+\frac{1}{3^4}+\frac{1}{3^6}+\cdots=\frac{9}{8}.$$ 
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