Showing $\frac{1}{n} < \int_{n-1}^n \frac{dt}{t}$, $n≥2$ These notes on harmonic sum present the following inequality:
$$\frac{1}{n} < \int_{n-1}^n \frac{dt}{t}$$ for $n≥2$.
How can this be shown? By induction and by evaluation the integral?
 A: See if this makes it any clearer. The red line is the function $f(x) = \frac{1}{x}$ and the red shaded area is the $$\int_{n-1}^{n}{\frac{1}{x}dx}$$ whereas the blue rectangle is a rectangle of $\text{width} = 1$ and $\text{height} =\frac{1}{n}$ and thus it's area = $1 \over n$. Can you see why the proposition is true and thus formalize it mathematically?

A: Just look at a graph of $y=1/t$ between $t=n-1$ and $t=n$, versus the constant function $y=1/n$ between the same two $t$-values. The latter graph is a rectangle, with its area contained entirely under the first graph.
A: By the mean value theorem, there is an $m \in (n-1, n)$ such that
$$\int_{n-1}^n \frac{1}{t} dt = \log(n)-\log(n-1) = \frac{1}{m} (n-(n-1)) = \frac{1}{m}.$$
Because $1/t$ is decreasing, 
$$\frac{1}{n} < \frac{1}{m} = \int_{n-1}^n \frac{1}{t} dt.$$
Note: It is not necessary to find the antiderivative. Also note, this answer is less general than noting that $1/n<f(x)$ for $x \in (n-1, n)$. It depends on $f(x)$ being decreasing.
A: For $t\in[n-1,n]$, we have $\frac1n\le\frac1t$. Therefore,
$$
\frac1n=\int_{n-1}^n\frac1n\,\mathrm{d}t\le\int_{n-1}^n\frac1t\,\mathrm{d}t
$$
A: Integral with $n\in\mathbb{N}$. 
$\displaystyle \int\limits_n^{n+1}\frac{dt}{t} =\ln(1+\frac{1}{n})=(\frac{1}{n}-\frac{1}{2n^2})+\sum\limits_{k=1}^\infty (\frac{1}{(2k+1)n^{2k+1}}-\frac{1}{(2k+2)n^{2k+2}})$
It is $\displaystyle \frac{1}{n+1}\leq \frac{1}{n}-\frac{1}{2n^2}$ and therefore $\displaystyle \frac{1}{n+1}<\int\limits_n^{n+1}\frac{dt}{t}$ because of the positivity of the rest of the series.

Note:
The logarithmic quality of the integral comes from $\frac{d(xy)}{xy}=\frac{dx}{x}+\frac{dy}{y}$ if $x$ and $y$ are choosed parameterized and differentiable which is the case of $t$ in the integral.
A: This is a direct consequence of the Mean Value theorem for definite integrals:
$$\int_n^{n+1}\frac1t\,\mathrm d\mkern 1mu t= \frac1c(n+1-n)\quad\text{for some }c\in [n,n+1].$$
