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 we have 
$$\log (k+1)-\log k=\int_k^{k+1}\frac{1}{u}\,du\le \frac{1}{k} \tag 1$$
Next, we can sum both sides of $(1)$ to reveal that 
$$\sum_{k=1}^n \left(\log (k+1)-\log k\right)=\log (n+1)\le \sum_{k=1}^n \frac{1}{k}\tag 2$$
Thus, we see that $(2)$ implies that
$$\sum_{k=1}^n \frac{1}{k}-\log (n+1)\ge 0$$
and therefore
$$\sum_{k=1}^n \frac{1}{k}-\log n > 0$$
