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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.

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    [This](http://math.stackexchange.com/questions/831594/proving-inequalities-about-integral-approximation/831617#831617) could be of help.2014-07-03

2 Answers 2

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Yes, it's the usual diagram used to illustrate the integral test.

Take $f(x)=1/x$ in the following diagram.

enter image description here

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 $\TeX$!2012-12-05
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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.