show that for $n=1,2,...,$ the number $1+1/2+1/3+...+1/n-\ln(n)$ is positive

Question

show that for $n=1,2,...,$ the number $1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}-\ln(n)$ is positive, that it decreases as $n$ increases, and hence that the sequence of these numbers converges to a limit between $0$ and $1$ (Euler's constant).

I'm trying to prove this by induction on $n$ and I made the base step, I could not with the inductive step because to do so suppose that for $n=1,2,\dots,$ it is true that $1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n}-\ln(n)$ is positive and let's see that $1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}+\frac{1}{n+1}-\ln(n+1)$ is positive, We see that \begin{align} &1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n}+\frac{1}{n+1})-\ln(n+1)\\ =&1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n}-\ln(n)+\frac{1}{n+1}-\ln(n+1)+\ln(n)\\ >&\frac{1}{n+1}-\ln(n+1)+\ln(n) \end{align} But I do not know how to prove that $\frac{1}{n+1}-\ln(n+1)+\ln(n)>0$ what do you say? Can you do what I did?

Answer 1

There is a different way (more long) that is not using integrals. First observe that $$n=\prod_{k=1}^{n-1}\frac{k+1}k$$

Hence

$$\sum_{k=1}^n\frac1k-\ln(n)=\frac1n+\sum_{k=1}^{n-1}\left(\frac1k-\ln\left(\frac{k+1}k\right)\right)$$

Then if we define in $[1,\infty)$ the function

$$f(x):=\frac1x-\ln\left(\frac{x+1}x\right)$$

we can see that $f'(x)=-\frac1{x^2(x+1)}$ is negative, hence $f$ is strictly decreasing. From here is easy to check that $f$ is positive because $f(1)>0$, $\lim_{x\to\infty}f(x)=0$ and $f$ is continuous. Then

$$\frac1k-\ln\left(\frac{k+1}k\right)>0,\quad\forall k\in\Bbb N_{>0}$$

Answer 2

Let :

$$S=1+\frac12 + \frac13+ \frac14 \dots \frac1n $$

$$S=\frac1n (\frac n1+\frac n2+\frac n3+.....\frac nn)$$

$$S=\frac 1n(\frac{1}{\frac1n}+\frac{1}{\frac 2n}+....\frac{1}{\frac nn}) > \int_{\frac 1n}^{1} \frac{1}{x} dx$$ (Using sum as integration)

See here :

$$S >\ln n$$

P.S. - Sorry for bad graph scales.

Answer 3

Note that the sequence in question, let's call it $\alpha_n$, can be written as $$\alpha_n = \Big(\sum^n_{k=1} 1/k\Big) - ln(n).$$

Note that \begin{align} ln(n) &:= \int^n_1 \frac {1} {t} dt \\\ &= \int_1^2 \frac {1}{t} dt + \int_2^3 \frac {1}{t} dt + \dots + \int^n_{n-1} \frac {1}{t} dt \\\ & \le (2-1) \cdot \frac {1}{1} + (3-2) \cdot \frac {1}{2} + \dots + (n-(n-1)) \cdot \frac {1}{n-1} \\\ &= 1 + \frac {1}{2} + \dots + \frac {1}{n-1}. \end{align}

Therefore, for a fixed $n$, $\alpha_n \ge \frac {1} {n} > 0$, so the sequence is positive.

Answer 4

The basic idea is that $\frac1{x}$ is decreasing, so $\frac1{n-1} \gt \int_{n-1}^n \frac{dx}{x} \gt \frac1{n}$.

But that integral is $\ln(n)-\ln(n-1)$, so $\frac1{n-1} \gt \ln(n)-\ln(n-1) \gt \frac1{n}$.

Summing from $2$ to $n$, and changing the variable to $k$, $\sum_{k=2}^n\frac1{k-1} \gt \sum_{k=2}^n(\ln(k)-\ln(k-1)) \gt \sum_{k=2}^n\frac1{k} $, or $\sum_{k=1}^{n-1}\frac1{k} \gt \ln(n) \gt \sum_{k=2}^n\frac1{k} = \sum_{k=1}^{n-1}\frac1{k}-1+\frac1{n} $.

I'll use the standard notation $H_n =\sum_{k=1}^n \frac1{k} $.

Therefore $0 \lt H_{n-1}-\ln(n) \lt 1-\frac1{n} $ or $\frac1{n} \lt H_{n}-\ln(n) \lt 1 $.

To show that the difference decreases,

$\begin{array}\\ (H_n-\ln(n))-(H_{n+1}-\ln(n+1)) &=(H_n-H_{n+1})+(\ln(n+1)-\ln(n))\\ &=\frac{-1}{n+1}+\int_n^{n+1} \frac{dx}{x}\\ &\gt 0 \qquad \text{ as shown above }\\ \end{array} $