Explanation of the inequality $\sum_{k=1}^{n} 1/k \le 1 + \int_1^n(1/x) \, \mathrm{d} x$ Is there a graphic visualization of $\sum_{k=1}^{n} 1/k \, \, \leq \, \, \,1 \, + \, \int_1^n \! \frac1x \, \mathrm{d} x$ as intuitive as the integral test ?
I can't see why the inequality is true.
I know I could plot the Harmonic partial sum function nevertheless...
Thanks.
 A: The inequality to prove is $\sum\limits_{k=2}^n1/k\leqslant\int\limits_1^n\mathrm dx/x$. For each $k\geqslant2$, $1/k$ is the area of the rectangle $[k-1,k]\times[0,1/k]$, which is entirely below the curve $x\mapsto1/x$, hence $1/k$ is less than the area below the curve from $x=k-1$ to $x=k$. Considering these $n-1$ disjoint rectangles from $k=2$ to $k=n$ and adding their areas yields the inequality.
A: Yes, it's the usual diagram used to illustrate the integral test.
Take $f(x)=1/x$ in the following diagram.

Then $a_1=1$, $a_2=1/2$, $\ldots$. Note that in the diagram, the infinite sum is the sum of the areas of the drawn rectangles, while the integral is the area under the graph of $f$ over the interval $[1,\infty)$.
Note that this integral is greater than  $\sum\limits_{n=2}^\infty a_n$; so
$$\sum\limits_{n=1}^\infty a_n = a_1+ \sum\limits_{n=2}^\infty a_n\le a_1+\int_1^\infty f(x)\,dx.$$ 
For the "finite version", as you have, use the same diagram; but "cut it off" at the appropriate point. You'll be able to see why your inequality holds.
(I may post a nicer diagram later; but I had this one on hand.)
