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!
 A: 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
$$\sum_{k=1}^n \frac{1}{k}-\log n > 0$$
A: 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:
$s_n = s_{n-1} + \dfrac{1}{n} \geq \displaystyle \int_{1}^n \dfrac{1}{x}dx + \dfrac{1}{n}=\ln n + \dfrac{1}{n} > \ln n\Rightarrow t_n = s_n - \ln n > 0$
A: By Abel's summation $$\sum_{k\leq n}\frac{1}{k}=1+\int_{1}^{n}\frac{\left\lfloor t\right\rfloor }{t^{2}}dt\geq1+\int_{1}^{n}\frac{t-1}{t^{2}}dt=\log\left(n\right)+\frac{1}{n}>\log\left(n\right)
 $$ where $\left\lfloor t\right\rfloor 
 $ is the floor function.
A: Since the function $x \mapsto 1/x: ]0, \infty[ \to \mathbb{R}$ is strictly decreasing, we have $\sum_{k=1}^{n}\frac{1}{k} > \int_{x=1}^{n+1}\frac{1}{x} = \log (n+1)$ for all $n \geq 1$. But, since $\log (n+1) > \log n$ for all $n \geq 1$, we have $\sum_{k=1}^{n}\frac{1}{k} - \log n > 0$ for all $n \geq 1$. 
A: Since $1+x\le e^x$ for $x\in\mathbb{R}$, we have that $\log(1+x)\le x$ for $x\gt-1$. Thus, for $x=-\frac1n$,
$$
\begin{align}
\frac1n-\log\left(\frac{n}{n-1}\right)
&=\log\left(1-\frac1n\right)-\left(-\frac1n\right)\\
&\le0\tag{1}
\end{align}
$$
Inequality $(1)$ means that
$$
\sum_{k=1}^n\frac1k-\log(n)\tag{2}
$$
is a decreasing sequence.
However, using the standard comparison of sum and integral of a decreasing function,
$$
\begin{align}
\sum_{k=1}^n\frac1k
&\ge\int_1^{n+1}\frac1x\,\mathrm{d}x\\
&=\log(n+1)\\[8pt]
&\ge\log(n)\tag{3}
\end{align}
$$

Inequality $(3)$ answers the question. In addition, it shows that $(2)$ is not only a decreasing sequence, but also bounded below by $0$. This means that $(2)$ decreases to a non-negative limit.

This limit is known as $\gamma$, the Euler-Mascheroni Constant, and a fairly simple and efficient method for computing $\gamma$ is derived in this answer.
