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series convergence with a floor

Lisa91

New member
Jan 15, 2013
29
I have one series [tex] \sum_{n=13}^{\infty}(-1)^{\left\lfloor\frac{n}{13}\right\rfloor} \frac{ \ln(n) }{n \ln(\ln(n)) } [/tex]. How to investigate its convergence? I wanted to group the terms of this series but I don't know whether it's a good idea as we have 13 terms with minus and then 13 with plus and so on. What do you think?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,779
Hi Lisa91!

Actually, that does seem like a good idea.

If you can proof that the sequence without the (-1) is declining towards zero, then summing 13 consecutive terms and repeating that will also be declining towards zero.
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
I have one series [tex] \sum_{n=13}^{\infty}(-1)^{\left\lfloor\frac{n}{13}\right\rfloor} \frac{ \ln(n) }{n \ln(\ln(n)) } [/tex]. How to investigate its convergence?
Hint: Prove that (i) $T_k=\left |\sum_{i=13k}^{13(k+1)}a_i\right|$ is strictly decreasing and has limit $0$. (ii) The sequence $(S_n)_{n\geq 13}$ of the partial sums of the given series is a Cauchy sequence.
 
Last edited:

chisigma

Well-known member
Feb 13, 2012
1,704
I have one series [tex] \sum_{n=13}^{\infty}(-1)^{\left\lfloor\frac{n}{13}\right\rfloor} \frac{ \ln(n) }{n \ln(\ln(n)) } [/tex]. How to investigate its convergence? I wanted to group the terms of this series but I don't know whether it's a good idea as we have 13 terms with minus and then 13 with plus and so on. What do you think?
The series is the sum of 13 series and can be written as...

$\displaystyle \sum_{k=0}^{12} \sum_{j=0}^{\infty} (-1)^{j}\ \frac{\ln (13\ j + k)}{(13\ j + k)\ \ln \{\ln (13\ j + k)\}}$ (1)

Each of the inner series is 'alternating' and converges for the Leibnitz's criterion, so that the whole series converges...

Kind regards

$\chi$ $\sigma$
 

Lisa91

New member
Jan 15, 2013
29
I tried to do it this way but I don't know how to prove that it decreases and that the limit is zero. I tried estimating it RHS and LHS using [tex] \frac{t}{t+1}< \ln(t+1) < t [/tex] but in one case I've got -1...

[tex] (-1)^{j}\ \frac{\ln (13\ j + 1)}{(13\ j + 1)\ \ln \{\ln (13\ j + 1)\}} [/tex].
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
The series is the sum of 13 series and can be written as...

$\displaystyle \sum_{k=0}^{12} \sum_{j=0}^{\infty} (-1)^{j}\ \frac{\ln (13\ j + k)}{(13\ j + k)\ \ln \{\ln (13\ j + k)\}}$ (1)

Each of the inner series is 'alternating' and converges for the Leibnitz's criterion, so that the whole series converges...
A priori, you don't know if the series is convergent, so ¿what associativity and/or conmutativity properties are you using?
 

chisigma

Well-known member
Feb 13, 2012
1,704
I tried to do it this way but I don't know how to prove that it decreases and that the limit is zero. I tried estimating it RHS and LHS using [tex] \frac{t}{t+1}< \ln(t+1) < t [/tex] but in one case I've got -1...

[tex] (-1)^{j}\ \frac{\ln (13\ j + 1)}{(13\ j + 1)\ \ln \{\ln (13\ j + 1)\}} [/tex].
We pratically can consider the term $\ln \{\ln (13\ j + k)\}$ as a constant term [it is very slowly changing with j...] and can analyse the term...

$\displaystyle \frac{\ln (13\ j + k)}{13\ j + k}$ (1)

Any doubt about the fact that, at least for j 'large enough', it is decreasing with j and that its limit is 0?...

Kind regards

$\chi$ $\sigma$
 

Lisa91

New member
Jan 15, 2013
29
We pratically can consider the term $\ln \{\ln (13\ j + k)\}$ as a constant term [it is very slowly changing with j...] and can analyse the term...

$\displaystyle \frac{\ln (13\ j + k)}{13\ j + k}$ (1)

Any doubt about the fact that, at least for j 'large enough', it is decreasing with j and that its limit is 0?...

Kind regards

$\chi$ $\sigma$
Thank you! I don't have any doubts about the fact that the limit of this guy is zero [tex] \frac{\ln (13\ j + k)}{13\ j + k} [/tex].

Do you think the explanation 'we may consider it as a constant term' is good for the exam? I think I feel what you mean. Indeed, it increases very slowly but still I am looking for something more formal.
 

chisigma

Well-known member
Feb 13, 2012
1,704
Thank you! I don't have any doubts about the fact that the limit of this guy is zero [tex] \frac{\ln (13\ j + k)}{13\ j + k} [/tex].

Do you think the explanation 'we may consider it as a constant term' is good for the exam? I think I feel what you mean. Indeed, it increases very slowly but still I am looking for something more formal.
All right!... all what You have to do is to consider that, for j 'large enough', is $\displaystyle |\ln \{\ln (13\ j + k)\}| >1$ so that is...


$\displaystyle |\frac{\ln (13\ j + k)}{(13\ j + k)\ \ln \{\ln (13\ j + k)\}}|< |\frac {\ln (13\ j + k)}{(13\ j + k)}|$ (1)

Kind regards

$\chi$ $\sigma$
 

Lisa91

New member
Jan 15, 2013
29
[tex] \ln \{\ln (13j + k) >1 [/tex]
[tex] \ln (13j + k)>e [/tex]
[tex] 13j + k >e^{e} [/tex]

so if we take j=1 and k=0,1,... it's true and we can also take k=9 and j=0,1...

Is it ok?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,779
Sorry, I have lost you in your argument.

But did you consider that for [TEX]n \ge 3[/TEX]:

[TEX]0 < {1 \over n \ln(\ln n)} < {\ln n \over n \ln(\ln n)} < {n \over n \ln(\ln n)} = {1 \over \ln(\ln n)} \to 0[/TEX]

The LHS and RHS both approach zero when [TEX]n \to \infty[/TEX].
Moreover they do so monotonously.
 

Lisa91

New member
Jan 15, 2013
29
Sorry, I have lost you in your argument.

But did you consider that for [TEX]n \ge 3[/TEX]:

[TEX]0 < {1 \over n \ln(\ln n)} < {\ln n \over n \ln(\ln n)} < {n \over n \ln(\ln n)} = {1 \over \ln(\ln n)} \to 0[/TEX]

The LHS and RHS both approach zero when [TEX]n \to \infty[/TEX].
Moreover they do so monotonously.

Thank you so much! It's so beautiful!!