This is a very Interesting question, there are many ways to do it.
Lets see what is the best way to do it.
I have an idea which involves a definite integral, I am working on it, will post it soon.
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Martin Sleziak
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Holy cow
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Did you try searching this site? – Aryabhata Jul 02 '14 at 04:52
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I did search around – Holy cow Jul 02 '14 at 04:57
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I have tried searching for harmonic number inequality site:math.stackexchange.com in Google Images. On of the results is this question, which seems like a possible duplicate. – Martin Sleziak Jul 03 '14 at 07:07
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For each integer $m \ge 2$, we have $\dfrac{1}{m} = \displaystyle\int_{m-1}^{m}\dfrac{1}{m}\,dx \le \int_{m-1}^{m}\dfrac{1}{x}\,dx$.
Sum this up from $m = 2$ to $m = n$ to get $\displaystyle\sum_{m = 2}^{n}\dfrac{1}{m} \le \sum_{m = 2}^{n}\int_{m-1}^{m}\dfrac{1}{x}\,dx = \int_{1}^{n}\dfrac{1}{x}\,dx = \ln n$.
Then, add $1$ to both sides to get $s_n = \displaystyle\sum_{m = 1}^{n}\dfrac{1}{m} \le 1+\ln n$.
EDIT: Here is a good picture to go with this proof.
Source:
JimmyK4542
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As you are going to do it by integration, I will do it by differentiation. Using the Mean Value Theorem (that's the differentiation bit) one can prove that $$\ln x>1-\frac{1}{x}\quad\hbox{for}\quad x>1\ .$$ In particular, $$\ln(n+1)-\ln n=\ln\Bigl(\frac{n+1}{n}\Bigr)>\frac{1}{n+1}\quad\hbox{for}\quad n\ge1\ .$$ This can now be used to provide an easy inductive proof of your inequality.
David
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What I a thinking is $$s_n = \int_0^1 \left(1+x+x^2+....x^{n-1}\right) \hspace{1mm}dx$$
Holy cow
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1To continue, note that we have two inequalities for your integrand $f(x)$: if $0\le x\le1$ then$$f(x)\le1+1+\cdots+1=n$$and$$f(x)\le1+x+\cdots+x^{n-1}+\cdots=\frac{1}{1-x}\ .$$Hence$$S_n\le\int_0^{1-(1/n)}\frac{dx}{1-x}+\int_{1-(1/n)}^1 n,dx\ .$$ – David Jul 02 '14 at 05:27