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?

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.
