Inequality with Force of interest I have to show the following inequalities and don't know how to proceed:
If $t \geq 0$, then show
$$ d \leq d_m \leq \delta \leq r_m \leq r \text{ for } m \geq 1$$
where $\delta$ is the force of interest, $r$ the annual effective interest rate, $r_m$ the nominal interest rate, convertible m times per year, $d$ the annual effective discount rate and $d_m$ the equivalent nominal rate of interest in advance credited $m$ times per year.
So I know that $r = (1+ \frac{r_m}{m})^m -1 \geq 1 + m \cdot \frac{r_m}{m} -1 = r_m$ follows by Bernoulli. Any suggestions on how to proceed now? I know that $$\delta = \lim_{m \to \infty} r_m = \lim_{m \to \infty} d_m$$
Furthermore, $d = 1-(1-\frac{d_m}{m})^m$ and $d_m = m(1-(1-d)^{\frac{1}{m}})$. How can I show know the inequality $d \leq d_m$? 
 A: To prove that $d\leq d_m\leq \delta$, fix $0<d<1$ and observe that 
$$
d_m= \frac{1-(1-d)^{\frac1m}}{\frac1m}, \ \ \ \  d_1=d.
$$
Now, let $g(x)=(1-d)^x$. We have $g(0)=1$ and 
$$
d_m=\frac{g(0)-g(\frac1m)}{\frac1m}= - \frac{g(\frac1m)-g(0)}{\frac1m}.
$$
Since $g(x)$ is a decreasing convex function in $x$,  we see that $d_m$ is increasing as $m$ increases. Moreover, 
$$
\lim_{m\rightarrow\infty} d_m = -\lim_{x\rightarrow 0+} \frac{g(x)-g(0)}{x} = -\ln (1-d)=\delta.$$
Therefore, we have $d_m\leq \delta$. 
A similar approach works for $r_m$ as well. 
For the other question about $r_m-d_n < \frac{r^2}{\min(m,n)}$, write $k=\min(m,n)$ and
$$
r_m-d_n\leq r_k-d_k = k((1+r)^{\frac1k}-1 - 1 + (1-d)^{\frac1k})=k((1+r)^{\frac1k}+(1+r)^{-\frac1k}-2).
$$
Then,
$$
\begin{align}
k((1+r)^{\frac1k}+(1+r)^{-\frac1k}-2)&= \frac k{(1+r)^{\frac1k}}\left( (1+r)^{\frac1k} - 1 \right)^2 \\
&=\frac {k r^2}{(1+r)^{\frac1k} \left( 1+(1+r)^{\frac1k} + (1+r)^{\frac2k} + \cdots + (1+r)^{\frac{k-1}k} \right)^2 }\\
& < \frac{ kr^2} {k^2} = \frac {r^2}k = \frac{ r^2}{\min(m,n)}.
\end{align}
$$
