Euler's constant greater than 0 for all values of n? If Euler's constant is described as the limit as n approaches infinity of the following:
$$t_n = 1 + \frac 12 + \frac 13 \cdots + \frac1n -\ln(n)$$
How can one prove that $t_n$ is greater than $0$ for all values of $n$?
Thanks!

First, put $s_n = 1+\dfrac{1}{2}+\cdots + \dfrac{1}{n}$, then
$1+\dfrac{1}{2}+\cdots + \dfrac{1}{n} - \ln n > 0$ because:
$1 = \displaystyle \int_{1}^2 1dx \geq \displaystyle \int_{1}^2 \dfrac{1}{x}dx$,
$\dfrac{1}{2} = \displaystyle \int_{2}^3 \dfrac{1}{2}dx \geq \displaystyle \int_{2}^3 \dfrac{1}{x}dx$, and 
$.....$
$\dfrac{1}{n-1} = \displaystyle \int_{n-1}^n \dfrac{1}{n-1}dx\geq \displaystyle \int_{n-1}^n \dfrac{1}{x}dx$, and add these integrals we have:
$s_n = s_{n-1} + \dfrac{1}{n} \geq \displaystyle \int_{1}^n \dfrac{1}{x}dx + \dfrac{1}{n}=\ln n + \dfrac{1}{n} > \ln n\Rightarrow t_n = s_n - \ln n > 0$
