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Are these inequalities useless for getting better estimates? If not what is needed? My motivation for asking this question is to get a glimpse to the mind of masters that can tell if a line of reasoning looks like to be useful or should be abandoned and alternative path should be taken ( just by intuitionistic reasoning that would seem reasonable without having to be correct).

Background: While working on this questions, found out some inequities. The motivation was to find some inequalities and squeeze them for better approximation, what I got didn't seem to get me any closer to anything useful.

My question is: could some inequity masters to show that these results can be used to get somewhere closer to answer or , loosely explain loosely why these inequalities are not on the right track anywhere closer to a better approximation?

$\frac {1}{2}+\frac {1}{3} > \frac {1}{4}+\frac {1}{4} = \frac {1}{2}$

$\frac {1}{4}+\frac {1}{5} + \frac {1}{6}+\frac {1}{7} > \frac {1}{8}+\frac {1}{8} + \frac {1}{8}+\frac {1}{8} = \frac {1}{2}$

In general : $ \frac{1}{2} < \sum _{2^n}^{2^{n+1}-1} \frac{1}{k} $

or more generally : $ \frac{1}{\alpha} < \sum _{\alpha^n}^{\alpha^{n+1}-1} \frac{1}{k} $

also

$$ \sum _{\alpha^n}^{\alpha^{n+1}-1} \frac{1}{k} - \frac{1}{\alpha} < \sum _{{(\alpha+1)}^n}^{{(\alpha+1)}^{n+1}-1} \frac{1}{k} - \frac{1}{\alpha+1}$$

Some trivial inequalities I tried to use:

$2^6<100<2^7$,

$3^4<100<3^5$

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jimjim
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1 Answers1

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I'm not sure what you mean by "useless for getting better estimates", but your inequalities are quite useful as they are.

First, they enable you to prove that the harmonic sum $H_n =\sum_{k=1}^n \frac1{k} $ diverges as $n \to \infty$, and gives a reasonable estimate for the growth of this sum.

Second, your estimate forms the basis for the Cauchy condensation test for whether or not a series converges. Look it up.

All in all, if you discovered this by yourself, I am impressed.

marty cohen
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  • I had seen the original idea for the proof of divergence (the first 2 inequalities), I have no original thought here just like a juxtaposition monkey randomly trying different different things to see if can sandwich the harmonic series between two other estimateable series and then squeeze them. By no means I am trying to find a better estimate than $\ln n$ for estimation of harmonic series – jimjim Aug 20 '15 at 00:51
  • @Arjang Why not? For instance, $\log(2n+1)$ is closer to $H_n$ than $\log(n)$ is. Asymptotically, $\log(2)-\gamma$ < $\gamma$. – Jaume Oliver Lafont Feb 11 '16 at 14:30