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Convergence of $$ \sum_{n=1} ^\infty \dfrac {1}{n(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n} )}$$

Attempt: I believe not a nice attempt: $ n(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n} ) \leq n( 1+1+\cdots+1 )$

$\implies n(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n} ) \leq n^2$

$\implies \dfrac {1}{ n(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n} ) } \geq \dfrac {1}{n^2}$

I guess this result is of not much use. Can somebody please tell me a direction to move ahead in this problem?

Thank you very much for your help .

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    $\begingroup$ Do you know some better approximation for $$\sum_{k=1}^n \frac{1}{k}\,?$$ You've probably come across it, you just need to remember. $\endgroup$ – Daniel Fischer Jan 26 '15 at 1:01
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Hint. Recall that, as $n \to +\infty$, we have $$1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n}\sim \ln n$$ then $$n\left(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n}\right)\sim n\ln n$$ and $$ \sum \frac{1}{n\left(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n}\right)} \sim \sum \frac{1}{n \ln n}, $$ your initial series is then divergent as is the latter series.

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    $\begingroup$ In asymptotic statements, I prefer not to say "for $n$ sufficiently great" but instead "as $n \to \infty$". $\endgroup$ – GEdgar Jan 26 '15 at 1:03
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    $\begingroup$ In fact, we have by virtue of the property that for all integers $k > 1$, $$\int_{x=k}^{k+1} \frac{1}{x} \, dx < \frac{1}{k} < \int_{x=k-1}^{k} \frac{1}{x} \, dx,$$ $$\log n < H_n < 1 + \log n.$$ $\endgroup$ – heropup Jan 26 '15 at 1:44
  • $\begingroup$ @GEdgar Right. For $n$ sufficiently great implies that there's some $N$ for which $1 + \frac{1}{2} + \dots + \frac{1}{n} = \ln n$ for all $n > N$ while "as $n \to \infty$" implies the asymptoticness. $\endgroup$ – MCT Jan 26 '15 at 2:18
  • $\begingroup$ @Olivier Oloa Thank you $\endgroup$ – MathMan Jan 26 '15 at 14:35
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The other way to look at it is, denoting $\displaystyle H_n = \sum\limits_{k=1}^{n} \frac{1}{k}$

The tail of the series: $$\displaystyle \frac{\frac{1}{n+1}}{H_{n+1}}+\frac{\frac{1}{n+2}}{H_{n+2}}+\cdots +\frac{\frac{1}{n+r}}{H_{n+r}} \ge \frac{\sum\limits_{k=1}^{r}\frac{1}{n+k}}{H_{n+r}} = \frac{H_{n+r}-H_{n}}{H_{n+r}} = 1-\frac{H_n}{H_{n+r}}$$

Since, $\displaystyle \lim\limits_{r \to \infty} 1-\frac{H_n}{H_{n+r}} = 1$

The series diverges by Cauchy Criteria.

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Using $GM\ge HM$ $$\left( 1\cdot\frac{1}{2}\cdot \frac{1}{3}\cdots \frac{1}{n}\right)^{\frac{1}{n}}\ge \frac{n}{\frac{1}{1+\frac{1}{2}+ \frac{1}{3}+\cdots \frac{1}{n}}}$$ so that $$\frac{1}{n}\cdot\frac{1}{1+\frac{1}{2}+ \frac{1}{3}+\cdots \frac{1}{n}}\ge (n!)^{\frac{1}{n}}$$ and as $(n!)^{\frac{1}{n}}>1 \forall n\ge2$ we see that $\sum (n!)^{1/n}$ cannot converge and thus $$\sum \frac{1}{n\left(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n}\right)}$$ cannot converge.

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