- Thread starter
- #1

- Thread starter Lisa91
- Start date

- Thread starter
- #1

- Admin
- #2

- Mar 5, 2012

- 9,006

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.

- Jan 29, 2012

- 661

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.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?

Last edited:

- Feb 13, 2012

- 1,704

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

$\chi$ $\sigma$

- Thread starter
- #5

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

- 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...

- Feb 13, 2012

- 1,704

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...

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

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

$\chi$ $\sigma$

- Thread starter
- #8

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].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

$\chi$ $\sigma$

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.

- Feb 13, 2012

- 1,704

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...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.

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

Kind regards

$\chi$ $\sigma$

- Thread starter
- #10

- Admin
- #11

- Mar 5, 2012

- 9,006

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.

- Thread starter
- #12

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!!