Prove that $\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{\lfloor \ln n \rfloor}}{n}$ is divergent.

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- May 22nd 2009, 09:19 PMNonCommAlgInfinite series (1)
Prove that $\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{\lfloor \ln n \rfloor}}{n}$ is divergent.

- May 24th 2009, 06:45 AMDeadstar
Does it diverge? According to maple it converges to 1.412025371...

- May 24th 2009, 11:02 AMNonCommAlg
here's another reason why i don't trust maple! (Nod) the series is divergent, as i said, and to prove it, write the series in the form $\displaystyle \sum_{n=1}^{\infty} (-1)^{n-1}a_n, \ a_n \geq 0,$ first and then see that $\displaystyle \lim_{n\to\infty} a_n \neq 0.$

- May 26th 2009, 09:43 AMchisigma
The original series is...

$\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{\lfloor \ln n \rfloor}}{n}$ (1)

... and it can be transformed in the form...

$\displaystyle \sum_{n=1}^{\infty} (-1)^{n-1} a_{n}$ (2)

... where $\displaystyle \forall n$ is $\displaystyle a_{n}>0$ only changing the order of its terms. But (1) is not absolutely convergent and in this case changing the order of its terms can transform the series from convergent to divergent or vice-versa, so that (Thinking)...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - May 26th 2009, 10:13 AMOpalg
... but if we keep the terms in the same order then we won't alter the convergence/divergence. Roughly speaking, the terms of the series have the same sign throughout any interval of the form $\displaystyle e^k<n<e^{k+1}$. The sum of the absolute values of the terms in that interval is $\displaystyle \sum_{e^k<n<e^{k+1}}\frac1n \approx \int_{e^k}^{e^{k+1}}\frac1x\,dx = \ln(e^{k+1}) - \ln(e^k) = 1.$ So the series alternates between a batch of positive terms adding up to approximately1, followed by a batch of negative terms adding up to approximately –1. This oscillation means that it cannot converge.

- May 26th 2009, 12:31 PMchisigma
The series...

$\displaystyle \sum_{n=1}^{\infty} (-1)^{n-1}= 1-1+1-1+... $ (1)

... properly speaking is neither convergent nor divergent, it is*indeterminated*[for someone it converges to $\displaystyle \frac{1}{2}$ but is not the case to discuss about that here...]. If we 'alterate' in some way [also in 'microscopic way'...] the (1), it can...

a) 'explode' [diverge]

b) 'implode' [converge]

c) remain 'stable' [indeterminated]

In the case...

$\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{\lfloor \ln n \rfloor}}{n}$ (2)

... what of the hypothesis a),b), c) is 'true'?...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - May 27th 2009, 02:01 AMchisigma
Considering the identity...

$\displaystyle (-1)^{\lfloor \ln n \rfloor} = (-1)^{ \ln n - \{\ln n\}} = e^{- i\cdot \pi (\ln n - \{\ln n\})} = n^{-i\cdot \pi} \cdot e^{i\cdot \pi \cdot \{\ln n\}} $ (1)

... where the symbol $\displaystyle \{*\}$ means 'fractional part of', the proposed series can be written as...

$\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{\lfloor \ln n \rfloor}}{n} = \sum_{n=1}^{\infty} \frac{e^{i\cdot \pi \cdot \{\ln n\}}}{n^{1+i\cdot \pi}}$ (2)

Now if we consider the sequence $\displaystyle a_{n}= \frac{1}{n^{1+i\cdot \pi}}$ , it generates the series...

$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{1+i\cdot \pi}} = \zeta(1+i\cdot \pi)$ (3)

... that is convergent since the function $\displaystyle \zeta(*)$ is analytic in the whole complex plane with the only exception of the point $\displaystyle s=1$. So if the sequence $\displaystyle a_{n}$ produces a convergent series, may be not hazarded suppose that the sequence $\displaystyle \alpha_{n} = a_{n}\cdot e^{i\cdot \pi \cdot \{\ln n\}}$ produces also a convergent series. It is obvious that this conclusion must be supported in rigorous fashion... at first however seems to be not an absurd 'a priori'...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - May 27th 2009, 12:33 PMOpalg
The zeta function is analytic in $\displaystyle \mathbb{C}\setminus\{1\}$, but it is

*not*equal to the sum of the series $\displaystyle \textstyle\sum n^{-s}$ on the whole of that domain. In fact, the series only converges in the region $\displaystyle \text{Re}\,s>1$. To extend it beyond there, you have to use analytic continuation. The equation (3) is not correct because the left side does not converge. - Apr 6th 2010, 05:54 PMchiph588@
$\displaystyle \zeta(1+i\cdot \pi)=\sum_{n=1}^{\infty} \frac{1}{n^{1+i\cdot \pi}}$ does diverge, but $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{1+i\cdot \pi}}\neq\infty $. This sum oscillates forever, it's kind of analogous to $\displaystyle \sum_{n=0}^\infty (-1)^n $ in that sense.

This can be shown by a use of partial summation, if you are familiar. - Jul 4th 2010, 03:24 PMchiph588@
Let's consider $\displaystyle \log_2(x) $ instead.

It's easy to see the sum is the same as $\displaystyle \sum_{n=1}^\infty (-1)^na_n $, where $\displaystyle a_n=\frac{1}{2^n}+\frac{1}{2^n+1}+\cdots+\frac{1}{ 2^{n+1}-1} $.

The conclusion then follows, but isn't this technically changing the order of summation since we're not evaluating the sum in the same order? - Jul 4th 2010, 10:23 PMCaptainBlack
- Jul 4th 2010, 10:26 PMCaptainBlack
- Jul 5th 2010, 10:37 PMchisigma
The following is from 'T. J. A. Bromwich An introduction to the theory of infinite series [London: Macmillan, 1908]' ...

http://digilander.libero.it/luposabatini/Bromwich2.JPG

The [british] term*oscillatory*is much better than the [italic]*indeterminate*to indicate a*bounded*sequence that has no finite limit, so that I'll be glad to use it in the future (Itwasntme)...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Jul 6th 2010, 12:07 AMOpalg
- Jul 6th 2010, 08:29 AMchiph588@