show that for $n=1,2,...,$ the number $1+1/2+1/3+...+1/n-\ln(n)$ is positive 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?
 A: 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.
A: 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.
A: 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
$$\frac1k-\ln\left(\frac{k+1}k\right)>0,\quad\forall k\in\Bbb N_{>0}$$
A: 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}
$
