Proof Regarding Effective Annual Rate Inequality

Question

I tried to use the $$\lim_{m \to \infty}(1 + \frac{i}{m})^{m} = e^i$$ as a starting point but couldn't work it out quite right.

I am using the definition that the EAR is $$(1 + \frac{i}{m})^{m} - 1$$

Answer 1

We want to show that $\left(1+\frac im\right)^m$ is an increasing function of $m$. This is intuitively clear to finance folk (as compounding more often seems as though it should generate more money) but it turns out to be surprisingly hard to prove. It is possible to do that via calculus but in this situation we know that $m\in \mathbb N$ and that allows us to find an elementary approach (noting that "elementary" is not a synonym for "easy"). I'll show both methods. In both we suppose that $i>0$.

Method I: Calculus

We define $f(x)=\left(1+\frac ix\right)^x$ and let $g(x)=\ln\,f(x)$. We want to prove that $f'(x)>0$. Since $\ln$ is increasing, it suffices to show that $g'(x)>0$. We note that $$g'(x)=\ln\left(1+\frac ix\right) -\frac i{i+x}$$

Letting $G(x)=\left(1+\frac ix\right)e^{-\frac i{i+x}}$ we see that $g'(x)=\ln\,G(x)$. Since the Taylor expansion for $e^{-x}$ is alternating (for $x>0$) we have $$ G(x)> \left(1+\frac ix\right)\left(1-\frac i{i+x}\right)=1$$ It follows that $$g'(x)>\ln\,1=0$$ and we are done.

Method II: Elementary

Specifically, then, we want to show that $$\left(1+\frac im\right)^m<\left(1+\frac i{m+1}\right)^{m+1}$$ We remark that, for $x>y>0$ we have $$\frac {x^{m+1}-y^{m+1}}{x-y}=\sum_{j=0}^mx^jy^{m-j}<(m+1)x^m$$

Cross-multiplying and rearranging yields $$x^m\left((m+1)y-mx\right)<y^{m+1}$$ Now let $x=1+\frac im$ and $y=1+\frac i{m+1}$. We see that for these values $\left((m+1)y-mx\right)=1$ and we are done.