# Explanation of this inequality

Is there a graphic visualization of $\sum_{k=1}^{n} 1/k \, \, \leq \, \, \,1 \, + \, \int_1^n \! (1/x) \, \mathrm{d} x$ 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.

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This could be of help. –  Fermé somme Jul 3 '14 at 7:06

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.)

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Incidentally, the diagram was generated entirely in plain $\TeX$! –  David Mitra Dec 5 '12 at 15:58

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.

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