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?

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