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Prove that the harmonic series $\sum_{i=1}^{\infty} \frac{1}{i}$ is divergent by using induction ?

any ideas ?

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  • $\begingroup$ I edited your question so it was readable, even though it will probably be deleted soon. Why? Because you barged in here without reading any of the rules. 1. Use mathjax to format your questions. 2. Show some effort or work or some context. No one is sitting around anxious to do your homework for you. $\endgroup$ – B. Goddard Apr 19 '17 at 12:29
  • $\begingroup$ Here is a collection (though not complete) of proofs: scipp.ucsc.edu/~haber/archives/physics116A10/harmapa.pdf. $\endgroup$ – farruhota Apr 25 '17 at 13:18
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You can prove the following by induction:

$$\sum_{k=1}^{2^n}\frac1k>\frac n2.$$

Indeed, $$\sum_{k=1}^{2^{n+1}}\frac1k=\sum_{k=1}^{2^n}\frac1k+\sum_{k=2^n+1}^{2^{n+1}}\frac1k>\frac n2+\sum_{k=2^n+1}^{2^{n+1}}\frac1{2^{n+1}}=\frac n2+\frac12.$$

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This is to demonstrate the following inequality $$1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\cdots+\dfrac{1}{2^{n}-1}>\dfrac{n}{2}$$ Let us perform the proof by the method of mathematical induction. It is clear that it is true for $n = 1$ since $1> 1/2$. Suppose that inequality $$1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\cdots+\dfrac{1}{2^{n}-1}>\dfrac{n}{2}$$ is verified with $n = k$, therefore it will be fulfilled: $$S_{k}=1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{2^{k}-1}>\dfrac{k}{2}$$ Let us show that in this case the inequality is also valid with $n = k + 1$, that is: $$S_{k+1}=1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{2^{k+1}-1}>\dfrac{k+1}{2}$$ Now we can put: $$S_{k+1}=1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{2^{k+1}-1}+\left( \dfrac{1}{2^{k}}+\dfrac{1}{2^{k}+1}+\cdots +\dfrac{1}{2^{k+1}-1}\right)$$ doing operations we also have the following inequality $$\left( \dfrac{1}{2^{k}}+\dfrac{1}{2^{k}+1}+\cdots +\dfrac{1}{2^{k+1}-1}\right)>\dfrac{1}{2^{k+1}}+\dfrac{1}{2^{k+1}}+\cdots +\dfrac{1}{2^{k+1}}=\dfrac{1}{2}$$ Thus, the inequality is satisfied with $n = k + 1$ $$S_{k+1}>\dfrac{k}{2}+\dfrac{1}{2}=\dfrac{k+1}{2}$$ According to the principle of mathematical induction we have that the inequality is valid for any natural $n$.

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Since $1/n> \int_n^{n+1} dx/x =\ln(1+1/n) $, $\sum_{n=1}^m 1/n \gt \ln(m+1)$.

This can be written as either a telescoping sum or a proof by induction.

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