An interesting series divergence

Krizalid

Active member
Prove that $\displaystyle\sum_{n=1}^\infty\frac1{n H_n}=\infty$ where $H_n$ is the n-term of the harmonic sum.

chisigma

Well-known member
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

$\chi$ $\sigma$

Krizalid

Active member
(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.

chisigma

Well-known member
(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

$\chi$ $\sigma$

Krizalid

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

chisigma

Well-known member
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

$\chi$ $\sigma$

oasi

New member
(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