Harmonic series and monotonicity of $\ln x$ Prove that ln x is strictly monotone increasing and that $\ln x\to \infty$ as $x\to\infty$ by using the property that the harmonic series diverges.
I don't understand how the harmonic series is relevant here. I know the derivative of $\ln x$ is $1/x$, but I don't know how to get a sum to use the harmonic series...
 A: Hint: Use Riemann sums to get an upper and lower bound for $\int_1^n \frac{dx}{x}$. (You really only need a lower bound, but the upper bound is just as simple...)
A: As I commented above, I first found the question very confusing because I read it to mean that we are supposed to use the divergence of the harmonic series to show that $\ln x$ is strictly increasing.  I don't see how to do that.  I also think the question is a bit backwards in some other ways: let me talk through it.
1) What is our definition of $\ln x$?  The one which allows what I think is the intended solution is $\ln x = \int_1^x \frac{dt}{t}$.  Then: if $f(x) = \ln x$, $f'(x) = \frac{1}{x}$, which is positive for all $x > 0$.  It is a standard consequence of the Mean Value Theorem that a function with a positive derivative is strictly increasing.  Or, the other way around, since $\frac{1}{t}$ is continuous and positive, $\int_1^x \frac{dt}{t}$ must be increasing.
2) The manner in which we are supposed to use 1) and $\sum_{n=1}^{\infty} \frac{1}{n} = \infty$ to show $\lim_{x \rightarrow \infty} \ln x = \infty$ has already been well explained in the answers of Unwisdom and Thomas Andrews.  I have nothing to add there.
3) But still this is a somewhat strange question.  If we know 1) and we are willing to use any properties of the logarithm, we can deduce $\lim_{x \rightarrow \infty} \ln x = \infty$ more easily in other ways.  Namely, since $\ln x$ is strictly increasing, to show that it approaches infinity it is enough to show that it is unbounded above.  If we know that $\ln x$ is the inverse function of $e^x$ then:
$\ln e^n = n$, so $\{ \ln n\}$ is unbounded.
Or if we know that $\ln xy = \ln x + \ln y$, then it follows that 
$\ln x^n = n \ln x$, and so we get $\ln e^n = n$, or even $\ln 2^n = n \ln 2$.  Again, the logarithm function is unbounded.
In fact we can get away with even less: let 
$g(x) = \ln(2x) - \ln(x)$.  Then
$g'(x) = \frac{2}{2x} - \frac{1}{x} = 0$,
so (by the Mean Value Theorem) $g(x) = C$ is constant.  Plugging in $x = 1$ we get 
$g(x) = g(1) = \ln 2 - \ln 1 = \ln 2 > 0$ since the logarithm function is strictly increasing.  Then it follows that 
$\ln 2^n = \ln (2^n) - \ln(1) = (\ln(2^n) - \ln(2^{n-1})) + \ldots + (\ln(2) - \ln (1)) = n \ln 2$,
and again we get that the logarithm function is unbounded.
4) Now I want to ask: how do we know that the harmonic series diverges?  The most traditional approach is via the Integral Test -- i.e., it goes the opposite way, and we need to use that $\lim_{x \rightarrow \infty} \ln x = \infty$.  So for this not to be circular we need some other method.  There is another such standard method: 

Cauchy's Condensation Test: Let $\sum_{n=1}^{\infty} a_n$ be a real series with 
  $a_1 \geq a_2 \geq \ldots \geq a_n \geq \ldots \geq 0$.  Then $\sum_{n=1}^{\infty} a_n$ converges if and only if the "condensed series" $\sum_{n=1}^{\infty} 2^n a_{2^n}$ converges.

Applying the Condensation Test to the harmonic series condenses it to $\sum_{n=1}^{\infty} 1$, which diverges.  (This special case of the Condensation Test is actually more famous than the Condensation Test and much earlier -- it is due to the medieval mathematician Nicole Oresme -- but the proof is exactly the same.)
The proof of the Condensation Test is rather short: it is a clever manipulation involving powers of $2$.  My point though is that the argument used in 3) above is also a manipulation involving powers of $2$....a very similar one in fact, which proves a rather special case of the standard properties of logarithm that we use in precalculus and calculus mathematics.  So while it is certainly possible to make the above proof of $\lim_{x \rightarrow \infty} \ln x = \infty$ noncircular, I don't specifically see what is being gained: the geometric argument in question establishes the important asymptotic expansion
$\lim_{n \rightarrow \infty} \frac{ \sum_{k=1}^n \frac{1}{k}}{\ln n} = 1$,
but in my experience one uses this to understand $\sum_{k=1}^n \frac{1}{k}$ in terms of $\ln n$, not the other way around.  For more insight on how to use the Integral Test not just for convergence/divergence but asymptotic analysis of divergent series, see this blurb of Keith Conrad.
A: Consider the function $$f(x)=\frac{1}{\lceil x\rceil}$$ 
where $\lceil \cdot \rceil$ is the ceiling function, taking each real number $x$ to the smallest integer $n$ satisfying $n\geq x$.  
Observe that for any natural number $k$ we have 
$$\int_{1}^{k}f(x)\, \textrm{d}x=\sum_{j=2}^{k}\frac{1}{j}.$$
It is also evident that for all $x$, we have $f(x)\leq \frac{1}{x}$. Thus:
$$
\int_{1}^{k}\frac{1}{x}\, \textrm{d}x\geq \sum_{j=2}^{k}\frac{1}{j}.$$
You should be able to complete it from here.
A: $$f(x)=\ln x \rightarrow f'(x)=\dfrac{1}{x}>0 (for \hspace{2mm} x>0)\rightarrow f \hspace{2mm} is \hspace{2mm} increasing.$$
$$\ln x=\int^{x}_{1}\dfrac{1}{t}dt$$
As
$$\min \{ \frac{1}{t}\hspace{2mm}in\hspace{2mm}t\in [i,i+1]\}=\dfrac{1}{i+1}\rightarrow$$
Let $$n=\lfloor x\rfloor \rightarrow \int^{x}_{1}\dfrac{1}{t}dt>\sum^{n}_{i=1}\int^{i+1}_{i}\dfrac{1}{t}dt>\sum^{n}_{i=1}\dfrac{1}{i+1}[(1+i)-i]=\sum^{n}_{i=1}\dfrac{1}{i+1}$$
$$\ln x>\sum^{n}_{i=1}\dfrac{1}{i+1}\rightarrow \lim_{x\to + \infty}(\ln x)\geq \lim _{x\to +\infty}\sum^{n}_{i=1}\dfrac{1}{i+1}=$$
Using $n=\lfloor x\rfloor$
$$=\lim_{n\to \infty}\sum^{n}_{i=1}\dfrac{1}{i+1}=+\infty$$
