Prove that $2 < e < 4$ using upper and lower Riemann sums and the definition of $\ln{x}$ 
Prove that $2 < e < 4$ using upper and lower Riemann sums and the
  definition of $\ln{x}$

I think I understand the concept of what I need to do, but I am having some trouble implementing a solution. I guess this would be equivalent to showing that $\ln(2) < 1 < \ln(4)$ since the $\ln$ function is increasing.
What I'm not sure about is how I use the definition of $\ln(x)$ in the Riemann sum. I tried this:
$$\lim_{n\to\infty} \sum_{k=1}^n \frac{1}{n}\int_1^{\frac{k}{n}} \frac{1}{t}dt$$
I wasn't sure how to check the value at each point in order to prove my inequalities. How am I supposed to be doing this?
 A: Let us first show that $\ln 2 \lt 1$. We have $\ln 2=\int_1^2\frac{dt}{t}$.
Divide the interval from $1$ to $2$ into $1$ part. The upper Riemann sum is the width of the interval, times the value of the function $\frac{1}{t}$ at $t=1$. This upper Riemann sum is $1$, and is clearly bigger than the integral.
Now let us show that $\ln 4 \gt 1$. We have $\ln 4=\int_1^4\frac{dt}{t}$.
Divide the interval from $1$ to $4$ into $3$ equal parts, and find the corresponding lower Riemann sum.  The minimum of $\frac{1}{t}$ on the first part is $\frac{1}{2}$. For the other two parts, the minima are $\frac{1}{3}$ and $\frac{1}{4}$. So the lower Riemann sum is $\ge \frac{1}{2}+\frac{1}{3}+\frac{1}{4}$. This is already $\gt 1$, so the integral is $\gt 1$.
We could alternately note that $\ln 4=2\ln 2$. Then we can divide the interval from $1$ to $2$ into $1$ part, and note that our function value is $\frac{1}{2}$ at the right endpoint, and $\frac{1}{t}$ is decreasing. So the lower Riemann sum is $\frac{1}{2}$, and therefore $\ln 2 \gt \frac{1}{2}$, so $2\ln 2 \gt 1$. But this is not quite in the spirit of the game, since we have used a property of $\ln$, namely $\ln 4=2\ln 2$.  This would first have to be established from the definition of $\ln$ as an integral.
Remark: We were a little casual about concluding that the integral from $1$ to $2$ is less than $1$. In principle we only showed that it is $\le 1$. If you are very fussy, you can divide the interval $[1,2]$ into two equal parts. Then the upper sum is $\frac{1}{2}\left(\frac{1}{1}+\frac{1}{3/2}\right)\lt 1$.  
The above calculations will only make full sense if one draws a picture of $y=\frac{1}{t}$, and visually identifies the upper and lower sums mentioned.  They are all either the area of a rectangle, or the sum of the areas of a small number of rectangles.
A: Rewrite it as $\ln 2 -1 < 0 <2\ln 2 -1 $. Then use (from here)
$$
\sum_{n=1}^\infty \frac{(-1)^n}{n(4n^2-1)} = \ln 2 -1 \tag{1}
$$
and
$$
\sum_{n=1}^\infty \frac{1}{n(4n^2-1)} = 2\ln 2 -1 \tag{2}
$$
to get $\sum_{n=1}^\infty \frac{(-1)^n}{n(4n^2-1)} <0<\sum_{n=1}^\infty \frac{1}{n(4n^2-1)}$. It's easy to see that $(1)$ is true, since
$$
\begin{eqnarray}
\sum_{n=1}^\infty \frac{(-1)^n}{n(4n^2-1)}&=&\sum_{n=1}^\infty \underbrace{\frac{2(\pm 1)^n}{2n}}_{ \underbrace{\ln 2}_{\color{red}{\int_1^2 dt/t}}}
- \underbrace{\frac{(\pm 1)^n}{2n+1}}_{1-\pi/4}
- \underbrace{\frac{(\pm 1)^n}{2n-1}}_{\pi/4}=\ln 2 -1.\\
\end{eqnarray}
$$
Here your integral representation shows up ($\color{red}{\text{implicitly}}$).
For $(2)$, it's obvious to see that it converges (proof by comparison) and that it's positive (check). Therefore $(1)$ also converges absolutely.
A: We have that
$$\log x =\int_1^x \frac{dt}{t}$$
Since $y = \dfrac 1 x $ is striclty decreasing it is clear that
$$\frac{1}{x+1}\cdot (x+1-x)< \int_x^{x+1} \frac{dt} t<\frac{1}{x}\cdot (x+1-x)$$
$$\frac{1}{x+1}< \int_x^{x+1} \frac{dt} t<\frac{1}{x}$$
This is
$$\frac{1}{x+1}< \log (x+1)-\log x<\frac{1}{x}$$
$$\frac{1}{{x + 1}} < \log \left( {\frac{{x + 1}}{x}} \right) < \frac{1}{x}$$
Now let $x=1$.
$$\frac{1}{2} < \log 2 < 1$$
This gives
$$\log 2 < 1 < 2\log 2 \Rightarrow \exp \log 2 < \exp 1 < \exp 2\log 2 \Rightarrow 2 < e < 4$$
as desired.

Alernative:
By definition, we're given that
$$\log e = \int_1^e \frac {dt}{t} =1$$
That is, $e$ is this unique number.
Note that since the logarithm is increasing in all its domain, 
$$2<e<4$$ is equivalent to
$$\log 2<1<2 \log 2$$ 
So all we need to prove is that $$\frac 1 2 < \log 2 < 1$$
But this is trivial if you know that for $x >0$
$$\frac{x}{x+1}<\log (x+1)<x$$
This means that, for $x=1$
$$\frac{1}{2}<\log 2<1$$
so we're done.
