0
$\begingroup$

I think I can explain the question better first visually like the next table \begin{array}{|c|c|} \hline \sum_{n=1}^{\infty}\frac{1}{n}=\infty & \sum_{p\text{ prime}}\frac{1}{p}=\infty\\ \hline \sum_{n=1}^{\infty}\frac{1}{n}=\ln(\infty)+\gamma & \sum_{p\text{ prime}}\frac{1}{p}=\ln(\ln(\infty))+M\\ \hline \sum_{n=1}^{\infty}\frac{x^n}{n}=\ln(\frac{1}{1-x}) & ¿\sum_{p\text{ prime}}\frac{x^p}{p}=\ln(\ln(\frac{1}{1-x}))?\\ \hline \end{array} From what I understand $\sum_{n=1}^{\infty}\frac{1}{n}=\infty$ from the fact that $\lim_{N\to\infty}\sum_{n=1}^{N}\frac{1}{n}=\lim_{N\to\infty}\ln(N)+\gamma$ but also from evaluating at x=1 the polynomial series $\sum_{n=1}^{\infty} \frac{x^n}{n}=\ln(\frac{1}{1-x})$.

Similarly $\sum_{p\text{ prime}}\frac{1}{p}=\infty$ from the fact that $\lim_{N\to\infty}\sum_{p \text{ prime}}^{N}\frac{1}{p}=\lim_{N\to\infty}\ln(\ln(N))+M$, but also (if it were true) from evaluating at x=1 the polynomial series $\sum_{p\text{ prime}}\frac{x^p}{p}=\ln(\ln(\frac{1}{1-x}))$.

I don't know if the comparison is too naive or if it can be justified, looked it up but couldn't find if it was true or not, the only idea I have it is to reexpress the series so that $\ln(\ln(\frac{1}{1-x}))=-\ln(\frac{1}{1-(1-\ln(\frac{1}{1-x}))})=\sum_{n=1}^{\infty}-\frac{(1-\ln(\frac{1}{1-x}))^n}{n}=\sum_{n=1}^{\infty}-\frac{(1-\sum_{n=1}^{\infty} \frac{x^n}{n})^n}{n}$ but doesn't seems to work.

I would appreciate any help, thanks

$\endgroup$
  • 4
    $\begingroup$ $$\ln\ln\frac1{1-x}\to-\infty$$ as $x\to0^+$. $\endgroup$ – Angina Seng Jun 19 '19 at 6:27
  • 1
    $\begingroup$ Your notation is confusing. What is p? Is it meant to be a prime number? The summations on the right-hand side do not make any sense at the moment. $\endgroup$ – MachineLearner Jun 19 '19 at 7:09
  • $\begingroup$ @MachineLearner, I changed the notation hopefully is better now, thanks $\endgroup$ – Daniel D. Jun 19 '19 at 7:34
  • $\begingroup$ @Lord Shark the Unknown, I should put more attention to your comment, very simple observation and in fact proves the comparison is wrong, so simple I'm unsure this post is worth keeping but metamorphy has answered so I have accepted his answer and will leave it like that, thanks $\endgroup$ – Daniel D. Jun 19 '19 at 13:59
2
$\begingroup$

Let $\mathcal{P}$ denote the set of all primes, and $f(x)=\sum_{p\in\mathcal{P}}x^p/p$ for $|x|<1$. As noted in the comments, $f(x)$ cannot be equal to $\ln\ln\big(1/(1-x)\big)$ simply because the latter tends to $-\infty$ with $x\downarrow 0$.

But the asymptotics of $f(x)$ with $x\uparrow 1$ is worth looking at. As you've already noted, $$f_n:=\sum_{p\in\mathcal{P},\ p\leqslant n}\frac{1}{p}=\ln\ln n+O(1),\qquad n\to\infty$$ so we must have $$f(x)=(1-x)\sum_{n=2}^{\infty}f_n x^n=(1-x)\int_{u_0}^{\infty}x^u\ln\ln u\,du+O(1),\qquad x\uparrow 1$$ where $u_0>1$ is (any fixed) constant. Now let $x=e^{-t}$ and note that $$t\int_{u_0}^{\infty}e^{-tu}\ln\ln u\,du=\int_{tu_0}^{\infty}e^{-v}\ln\ln(v/t)\,dv\\=\ln(-\ln t)\int_{tu_0}^{\infty}e^{-v}\,dv+\int_{tu_0}^{\infty}e^{-v}\ln\left(1+\frac{\ln v}{-\ln t}\right)\,dv$$ behaves like $\ln(-\ln t)$ when $t\downarrow 0$. This allows to conclude that $$f(x)=\ln\big(-\ln(-\ln x)\big)+O(1)=\ln\ln\frac{1}{1-x}+O(1),\qquad x\uparrow 1$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Hi, you're correct so I have accepted your answer. I'm thankful also for the analysis when $x\to1$ very clever, a little advanced to me in fact because I don't get it all, from $(1-x)\int_{u_0}^{\infty}x^u\ln\ln u\,du$ to down I think I get it and it seems correct, what I don't understand is that you define $f(x)=\sum_{p\in\mathcal{P}}x^p/p$ but next you write $f(x)=(1-x)\sum_{n=2}^{\infty}f_n x^n$. [Also $\sum\frac{1}{n\ln n}$ behaves like $\sum\frac{1}{p}$ but $\sum\frac{1}{p\ln p }$ doesn't seems to behave like anything so I don't see why to apply the analogy here], thanks $\endgroup$ – Daniel D. Jun 19 '19 at 13:53
  • $\begingroup$ We have $(1-x)\sum_{n=2}^{\infty}f_n x^n=\sum_{n=2}^{\infty}f_n x^n-\sum_{n=2}^{\infty}f_{n}x^{n+1}$. Introducing $f_1:=0$ and replacing $n$ by $n-1$ in the second sum, we get $$(1-x)\sum_{n=2}^{\infty}f_n x^n=\sum_{n=2}^{\infty}(f_n-f_{n-1})x^n.$$ Now recall what $f_n-f_{n-1}$ is. (Didn't want to be that wordy.) As for the last note - ok, removed ;) $\endgroup$ – metamorphy Jun 19 '19 at 14:02
  • $\begingroup$ Wow so fast, I get it now what you meant, I should have looked at it more carefully, very nice indeed no that I catch on, I will keep this way to express the series in my mind, thank you $\endgroup$ – Daniel D. Jun 19 '19 at 14:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.