MHB An interesting series divergence

AI Thread Summary
The discussion centers on proving the divergence of the series ∑(1/(n H_n)), where H_n is the n-th harmonic number. Participants analyze the relationship between H_n and ln(n), noting that H_n approaches ln(n) as n increases, specifically that H_n < 2 ln(n) for large n. They argue that this relationship allows for the comparison of the series to known divergent series, despite initial claims that certain inequalities do not provide useful information. The conversation highlights the importance of establishing bounds for H_n to demonstrate the divergence of the series effectively. Overall, the participants reach a consensus on the analytical approach needed to prove the divergence.
Krizalid1
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Prove that $\displaystyle\sum_{n=1}^\infty\frac1{n H_n}=\infty$ where $H_n$ is the n-term of the harmonic sum.
 
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Krizalid said:
Prove that $\displaystyle\sum_{n=1}^\infty \frac1{n H_n}=\infty$ where $H_n$ is the n-term of the harmonic sum.

Because is $\displaystyle \lim_{n \rightarrow \infty} H_{n}-\ln n=\gamma>0$ and $H_{n}>\ln n$ then for n 'large enough' is...

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

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

Kind regards

$\chi$ $\sigma$
 
(1) is false. The fact $H_n>\ln n$ for large $n$ implies that $\displaystyle\frac{1}{{\ln n}} > \frac{1}{{{H_n}}} \Rightarrow \frac{1}{{n{H_n}}} < \frac{1}{{n\ln n}},$ however this doesn't provide information.
 
Krizalid said:
(1) is false. The fact $H_n>\ln n$ for large $n$ implies that $\displaystyle\frac{1}{{\ln n}} > \frac{1}{{{H_n}}} \Rightarrow \frac{1}{{n{H_n}}} < \frac{1}{{n\ln n}},$ however this doesn't provide information.

... 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

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

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 $\displaystyle \lim_{n \rightarrow \infty} \frac{H_{n}}{2\ \ln n}=\frac{1}{2}$...

Kind regards

$\chi$ $\sigma$
 
Krizalid said:
(1) is false. The fact $H_n>\ln n$ for large $n$ implies that $\displaystyle\frac{1}{{\ln n}} > \frac{1}{{{H_n}}} \Rightarrow \frac{1}{{n{H_n}}} < \frac{1}{{n\ln n}},$ however this doesn't provide information.
i'm agree with krizalid
 

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