Harmonic series and monotonicity of $\ln x$

Question

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...

Answer 1

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.

Answer 2

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...)

Answer 3

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.

Answer 4

$$f(x)=\ln x \rightarrow f'(x)=\dfrac{1}{x}>0 (for \hspace{2mm} x>0)\rightarrow f \hspace{2mm} is \hspace{2mm} increasing.$$ $$\ln x=\int^{x}_{1}\dfrac{1}{t}dt$$ 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 $n=\lfloor x\rfloor$ $$=\lim_{n\to \infty}\sum^{n}_{i=1}\dfrac{1}{i+1}=+\infty$$