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#### Krizalid

##### Active member

- Feb 9, 2012

- 118

- Thread starter Krizalid
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- #1

- Feb 9, 2012

- 118

- Feb 13, 2012

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Because is $\displaystyle \lim_{n \rightarrow \infty} H_{n}-\ln n=\gamma>0$ and $H_{n}>\ln n$ then for n 'large enough' is...Prove that $\displaystyle\sum_{n=1}^\infty \frac1{n H_n}=\infty$ where $H_n$ is the n-term of the harmonic sum.

$\displaystyle \sum_{k>n} \frac{1}{k\ H_{k}}> \sum_{k>n} \frac{1}{2\ k\ \ln k}$ (1)

... and the second series in (1) diverges...

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$\chi$ $\sigma$

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- #3

- Feb 9, 2012

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- Feb 13, 2012

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... of course... but is also for n 'large enough' $\displaystyle H_{n}< 2\ \ln n \implies \frac{1}{n\ H_{n}}>\frac{1}{2\ n\ \ln n}$ and that provides very good information...

Kind regards

$\chi$ $\sigma$

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- #5

- Feb 9, 2012

- 118

Okay that works but it's not clear why exactly $H_n<2\ln n.$ Can you prove it analytically?

- Feb 13, 2012

- 1,704

Immediate consequence of what is written in post #2 is that $\displaystyle \lim_{n \rightarrow \infty} \frac{H_{n}}{\ln n}=1$ so that is $\displaystyle \lim_{n \rightarrow \infty} \frac{H_{n}}{2\ \ln n}=\frac{1}{2}$...Okay that works but it's not clear why exactly $H_n<2\ln n.$ Can you prove it analytically?

Kind regards

$\chi$ $\sigma$

i'm agree with krizalid