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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?

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I'm mainly confused about how to use the Riemann sum notation with the definition of the $\ln$ function. How they fit together and how I get them into the appropriate form to evaluate things etc. –  stariz77 Jun 10 '12 at 11:52
    
What is your definition of the log function? –  Juan S Jun 10 '12 at 11:53
    
$$\ln(x) = \int_1^x \frac{1}{t}dt$$ –  stariz77 Jun 10 '12 at 11:54

3 Answers 3

up vote 3 down vote accepted

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.

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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.

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You didn't use upper and lower Riemann sums like the OP wanted. –  Stefan Smith Jun 10 '12 at 21:24
    
@user20520 One can show the OP there are cleverer ways of proving what he wants. You seem to like to criticize answers. Though I think that is healthy, you don't seem to look at the advantages of what I'm showing. Using Riemann sums is almost nonsensical (this is the translation I'm finding for "despropósito", don't get me wrong.) –  Pedro Tamaroff Jun 10 '12 at 22:21
    
I appreciate all the excellent work you post on Math Stack Exchange. But for whatever reason, the OP wanted a proof using Riemman sums. Perhaps if you had written "another way to do it besides the posted answer" or somesuch, I wouldn't have commented. –  Stefan Smith Jun 10 '12 at 23:33

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.

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How would I use the integral representation, rather than the series representation? –  stariz77 Jun 10 '12 at 12:39
    
$\ln(e) = \int_1^e \frac{1}{x}\,dx=1$. So you need to prove $\ln 2 = \int_1^2 \frac{1}{x}\,dx<1$ and $\ln 4 = \int_1^4 \frac{1}{x}\,dx>1$. Estimate $\int_1^2 \frac{1}{x}\,dx$ and $\int_1^4 \frac{1}{x}\,dx$ using upper and lower Riemann sums. –  Stefan Smith Jun 10 '12 at 14:26
    
@stariz77 Does my edit answer your question? –  draks ... Dec 9 '13 at 11:58

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