# 12 S2 Q4

Hi! (See title for source). The question asks us to prove that $\ln\left(\frac{2y+1}{2y-1}\right)>\frac1y$ using power expansions of $\frac1y$ (where $y>\frac12$). Is it okay to use the Maclaurin series of $e^x$ in the proof? Something like this:

We have
\begin{align*}
\frac{2y+1}{2y-1} = 1+\frac{\frac1y}{1-\frac{1}{2y}}&=1+\left(\frac1y +\frac{1}{2y^2}+\frac{1}{4y^3}+\frac{1}{8y^4}+\dotsb\right)\\
&\geq 1 + \left( \frac1y+\frac{1}{2y^2}+\frac{1}{6y^3}+\frac{1}{24y^4}+\dotsb \right)\\
&= e^{\frac1y},
\end{align*}
and taking logarithms on both sides, we obtain the result.

### Almost. Your $\geq$ should be

Almost. Your $\geq$ should be a $>$ and you'd be best off with a quick mention that $log$ is a strictly increasing function so taking logarithms of both sides preserves the inequality but you'd be fine without mentioning that anyway, I guess.

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