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Prove that $$\sum_{k=1}^{n} \frac{1}{k}>\ln(n+1)$$ for all $n\geq1$


I am looking for a clear solution to this problem. I've considering trying to prove it by contradiction by starting off assuming that it's not true for all n's, but I'm not sure if this is the best way to go about it. I don't have a lot of experience with summation proofs and would appreciate a clear proof for me to use as an example moving forward.

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    $\begingroup$ How does the sum compare to the integral $\int_1^{n+1} {1\over x}\,dx$? $\endgroup$ – David Mitra Sep 24 '15 at 13:22
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For $x\in(k,k+1]$ we have $\frac{1}{x}<\frac{1}{k}$, so $$ \int_{k}^{k+1}\frac{1}{x}\,dx<\frac{1}{k} $$ Now sum for $k=1,2,\dots,n$

$$\displaystyle\sum_{k=1}^n\frac{1}{k}>\sum_{k=1}^n\int_{k}^{k+1}\frac{1}{x}\,dx=\int_{1}^{n+1}\frac{1}{x}\,dx=\ln(n+1)$$

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What about induction on $n$? The problem boils down to proving: $$ \frac{1}{n}\geq \log(n+1)-\log(n) = \log\left(1+\frac{1}{n}\right) $$ that follows from the concavity of the logarithm function, for instance.

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