So I made an underapproximation and overapproximation for a sum and got that the underapprox is $(e^b-1)(\frac{b/n}{e^{b/n}-1})$ and the overapprox is $(e^b-1)(e^{b/n})(\frac{b/n}{e^{b/n}-1})$. Now I want to rewrite those so that I can take the supremum of the underapprox and infimum of the overapprox. So for the underapprox I want to find something that is less than or equal to $(e^b-1)(\frac{b/n}{e^{b/n}-1})$ and for example we know that exp(x) $\ge$ 1+x but does that help me here? It would become $(b)(\frac{b/n}{e^{b/n}-1})$ I guess because I don't want to use it on the exp term in the denominator.
Ohh, and we also know that exp(x) $\le$ $\frac{1}{1-x}$
As $n\to\infty$ we have
$(e^b-1)\left(\frac{b/n}{e^{b/n}-1}\right)< e^b-1$
and
$(e^b-1)(e^{b/n})\left(\frac{b/n}{e^{b/n}-1}\right)>e^b-1$
because $$\underset{x\to 0}{\text{lim}}\;\frac{x}{e^x-1}=1$$
Actually the sum is $$\int_0^b e^x \, dx=e^b-1$$