Series convergence with a floor

[Lisa91]

I have one series $$\sum_{n=13}^{\infty}(-1)^{\left\lfloor\frac{n}{13}\right\rfloor} \frac{ \ln(n) }{n \ln(\ln(n)) }$$. 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?

 

[I like Serena]

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]

I have one series $$\sum_{n=13}^{\infty}(-1)^{\left\lfloor\frac{n}{13}\right\rfloor} \frac{ \ln(n) }{n \ln(\ln(n)) }$$. 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.

 

[chisigma]

I have one series $$\sum_{n=13}^{\infty}(-1)^{\left\lfloor\frac{n}{13}\right\rfloor} \frac{ \ln(n) }{n \ln(\ln(n)) }$$. 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]

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 $$\frac{t}{t+1}< \ln(t+1) < t$$ but in one case I've got -1...

$$(-1)^{j}\ \frac{\ln (13\ j + 1)}{(13\ j + 1)\ \ln \{\ln (13\ j + 1)\}}$$.

 

[Fernando Revilla]

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]

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 $$\frac{t}{t+1}< \ln(t+1) < t$$ but in one case I've got -1...

$$(-1)^{j}\ \frac{\ln (13\ j + 1)}{(13\ j + 1)\ \ln \{\ln (13\ j + 1)\}}$$.

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]

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

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]

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

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]

$$\ln \{\ln (13j + k) >1$$
$$\ln (13j + k)>e$$
$$13j + k >e^{e}$$

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?

 

[I like Serena]

Sorry, I have lost you in your argument.

But did you consider that for $$n \ge 3$$:

$$0 < {1 \over n \ln(\ln n)} < {\ln n \over n \ln(\ln n)} < {n \over n \ln(\ln n)} = {1 \over \ln(\ln n)} \to 0$$

The LHS and RHS both approach zero when $$n \to \infty$$.
Moreover they do so monotonously.

 

[Lisa91]

Sorry, I have lost you in your argument.

But did you consider that for $$n \ge 3$$:

$$0 < {1 \over n \ln(\ln n)} < {\ln n \over n \ln(\ln n)} < {n \over n \ln(\ln n)} = {1 \over \ln(\ln n)} \to 0$$

The LHS and RHS both approach zero when $$n \to \infty$$.
Moreover they do so monotonously.

Thank you so much! It's so beautiful!