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$$\sum _{n=1}^{\infty }\frac{1}{n\left(1 +\frac{1}{2}+...+\frac{1}{n}\right)}$$

I am trying to use limit comparison test to test this series,

Let $a_n = \frac{1}{n\left(1 +\frac{1}{2}+\ldots+\frac{1}{n}\right)}$ and $b_n = \frac{1}{n}$

$\lim_\limits{{n\to \infty }}\left(\frac{a_n}{b_n}\right)$

$=\lim _\limits{{n\to \infty }}\left(\frac{\frac{1}{n\left(1+\frac{1}{2}+\ldots+\frac{1}{n}\right)}}{\frac{1}{n}}\right)$

$=\lim _\limits{{n\to \infty }}\left(\frac{1}{\left(1+\frac{1}{2}+\ldots+\frac{1}{n}\right)}\right)$

This limit will always larger than $0$, therefore

$\sum _{n=1}^{\infty }\:\frac{1}{n}$ diverges, $\sum _{n=1}^{\infty }\frac{1}{n\left(1 +\frac{1}{2}+\ldots+\frac{1}{n}\right)}$ diverges.

Is this working valid or there is a better solution for this question?

M.Mass
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Steven Lau Zhou Sheng
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    What do you mean when you write "This limit will always larger than $0$"? The limit is $0$. – Ranc Jun 22 '17 at 17:52
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    I would compare it with $\sum\bigl(1/(n\log n)\bigr)$. – Lubin Jun 22 '17 at 17:52
  • For the reason why $\frac{1 + \frac{1}{2} + \cdots + \frac{1}{n}}{\ln n} \to 1$ as $n \to \infty$: adapt the proof of the integral test to get a pair of bounds on the sum, then apply the squeeze theorem. – Daniel Schepler Jun 22 '17 at 17:56
  • $$\sum_{n=1}^\infty \frac{1}{nH_n}\sim1+ \sum_{n=2}^\infty \frac{1}{n\log(n)}$$ – Mark Viola Jun 22 '17 at 18:13
  • You can prove a more general result: if a series of positive terms $\sum a_n$ diverges then so does $\sum \frac{a_n}{ a_1 + a_2 + \ldots + a_n}$. This is the special case where $a_n = \frac{1}{n}$. – Winther Jun 22 '17 at 21:02

3 Answers3

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Since $\sum_{k=1}^n \frac1{k} = \ln(n) + O(1) $, $\dfrac1{n\sum_{k=1}^n \frac1{k}} =\dfrac1{n(\ln(n)+O(1))} $ the sum of which diverges by, for example, condensation.

marty cohen
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I will try to work up some kind of drawing. As in a comment, the most elementary aspects of proof of the Integral Test give this: writing $$ H_n = 1 + \frac {1}{2} + \frac {1}{3} + \cdots + \frac {1}{n}, $$ we get the very ordinary $$ H_n > \log ( n+1) $$ and $$ H_n < 1 + \log n. $$ Here the logarithm is base $e \approx 2.71828.$

I drew in color, but my scanner does poorly with graph paper in color enter image description here

Will Jagy
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As the general term is decreasing, we can apply Cauchy's condensation criterion. The convergence of the given series (or rather of $\sum \frac 1{n\ln n}$) is thus equivalent to that of $$\sum 2^k\cdot \frac1{2^k\ln 2^k} =\sum\frac1{k\ln 2}$$ so divergent.

Hagen von Eitzen
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