How do I determine the divergence/convergence of $sum_n frac{1}{log(log(n))}$?












0












$begingroup$


I am working through some problems in Durrett's probability book, and one of them involves a variant of the law of iterated logarithm.



I've managed to reduce the result to showing that



$$sum_n frac{1}{log log (n)}exp(-log log(n)) < infty$$



Using upperbounds for tail probabilities of standard normal. But, as I'm really not good with this stuff, I'm unsure how I am supposed to show this (if it's indeed true?).



Clearly without the exponential, the series would diverge since $log (n) leq n$. But, with the exponential it seems as though this could converge.



Could someone advise me how to complete this last step?






convergence







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    I am working through some problems in Durrett's probability book, and one of them involves a variant of the law of iterated logarithm.



    I've managed to reduce the result to showing that



    $$sum_n frac{1}{log log (n)}exp(-log log(n)) < infty$$



    Using upperbounds for tail probabilities of standard normal. But, as I'm really not good with this stuff, I'm unsure how I am supposed to show this (if it's indeed true?).



    Clearly without the exponential, the series would diverge since $log (n) leq n$. But, with the exponential it seems as though this could converge.



    Could someone advise me how to complete this last step?






    convergence







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I am working through some problems in Durrett's probability book, and one of them involves a variant of the law of iterated logarithm.



      I've managed to reduce the result to showing that



      $$sum_n frac{1}{log log (n)}exp(-log log(n)) < infty$$



      Using upperbounds for tail probabilities of standard normal. But, as I'm really not good with this stuff, I'm unsure how I am supposed to show this (if it's indeed true?).



      Clearly without the exponential, the series would diverge since $log (n) leq n$. But, with the exponential it seems as though this could converge.



      Could someone advise me how to complete this last step?






      convergence







      share|cite|improve this question











      $endgroup$




      I am working through some problems in Durrett's probability book, and one of them involves a variant of the law of iterated logarithm.



      I've managed to reduce the result to showing that



      $$sum_n frac{1}{log log (n)}exp(-log log(n)) < infty$$



      Using upperbounds for tail probabilities of standard normal. But, as I'm really not good with this stuff, I'm unsure how I am supposed to show this (if it's indeed true?).



      Clearly without the exponential, the series would diverge since $log (n) leq n$. But, with the exponential it seems as though this could converge.



      Could someone advise me how to complete this last step?






      convergence




      convergence






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 1 '18 at 6:15









      Chinnapparaj R

      5,4872928




      5,4872928










      asked Dec 1 '18 at 6:12









      XiaomiXiaomi

      1,057115




      1,057115






















          5 Answers
          5






          active

          oldest

          votes


















          4












          $begingroup$

          We can simplify
          $$exp(- log log n) = frac{1}{e^{log log n}} = frac{1}{log n}$$
          Since $(log log n) log n leq n$, this diverges.






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            $sum_n frac{1}{n}$ is divergent. So is $sum_n frac{1}{log(log(n))}$



            But I feel $sum_n frac{exp(-log(log(n))}{log(log(n))}$ might still be convergent, you just need to find another way to prove it.






            share|cite|improve this answer









            $endgroup$





















              1












              $begingroup$

              $$
              frac 1{(log log n )exp(log log n)} = frac 1 {log log n cdot log n} geqslant frac 1{log ^2 n} geqslant frac 1{n^{1/2 times 2}} = frac 1n,
              $$
              so it still diverges.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Oh, ok. Thanks a lot. Do you have any advice on how to show $sum_n P(Z>sqrt{2log log n}(1+epsilon)) < infty$ for $Z sim N(0,1)$?
                $endgroup$
                – Xiaomi
                Dec 1 '18 at 6:21












              • $begingroup$
                @Xiaomi Sorry, I suck at probability theory.
                $endgroup$
                – xbh
                Dec 1 '18 at 7:15



















              1












              $begingroup$

              Cauchy condensation shows
              $$sum_n frac{1}{log(log(n))} sim sum_n frac{2^n}{log(log(2^n))} = sum_n frac{2^n}{log(nlog(2))} = sum_n frac{2^n}{log(n) + log(log(2))} $$
              So, it is divergent.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Thanks a lot. Do you have any advice on how to show $sum_n P(Z>sqrt{2log log n}(1+epsilon)) < infty$ for $Z sim N(0,1)$? The common probability inequalities do not seem tight enough to show this..
                $endgroup$
                – Xiaomi
                Dec 1 '18 at 6:24










              • $begingroup$
                You may use the integral for $N(0,1)$ for each member of the series and estimate it: $$P(Z>sqrt{2log log n}(1+epsilon)) sim int_{sqrt{2log log n}(1+epsilon)}^{infty} e^{frac{-x^2}{2}}dx$$
                $endgroup$
                – trancelocation
                Dec 1 '18 at 6:48





















              1












              $begingroup$

              Credits to user MoonKnight.



              $log n <n ; $



              And once more:



              $log (log n) lt log n lt n.$



              $dfrac {1}{n} lt dfrac{1}{log n} lt dfrac{1}{log (log n)}.$



              Comparison test.






              share|cite|improve this answer









              $endgroup$













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                5 Answers
                5






                active

                oldest

                votes








                5 Answers
                5






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

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                4












                $begingroup$

                We can simplify
                $$exp(- log log n) = frac{1}{e^{log log n}} = frac{1}{log n}$$
                Since $(log log n) log n leq n$, this diverges.






                share|cite|improve this answer









                $endgroup$


















                  4












                  $begingroup$

                  We can simplify
                  $$exp(- log log n) = frac{1}{e^{log log n}} = frac{1}{log n}$$
                  Since $(log log n) log n leq n$, this diverges.






                  share|cite|improve this answer









                  $endgroup$
















                    4












                    4








                    4





                    $begingroup$

                    We can simplify
                    $$exp(- log log n) = frac{1}{e^{log log n}} = frac{1}{log n}$$
                    Since $(log log n) log n leq n$, this diverges.






                    share|cite|improve this answer









                    $endgroup$



                    We can simplify
                    $$exp(- log log n) = frac{1}{e^{log log n}} = frac{1}{log n}$$
                    Since $(log log n) log n leq n$, this diverges.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 1 '18 at 6:17









                    plattyplatty

                    3,370320




                    3,370320























                        2












                        $begingroup$

                        $sum_n frac{1}{n}$ is divergent. So is $sum_n frac{1}{log(log(n))}$



                        But I feel $sum_n frac{exp(-log(log(n))}{log(log(n))}$ might still be convergent, you just need to find another way to prove it.






                        share|cite|improve this answer









                        $endgroup$


















                          2












                          $begingroup$

                          $sum_n frac{1}{n}$ is divergent. So is $sum_n frac{1}{log(log(n))}$



                          But I feel $sum_n frac{exp(-log(log(n))}{log(log(n))}$ might still be convergent, you just need to find another way to prove it.






                          share|cite|improve this answer









                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            $sum_n frac{1}{n}$ is divergent. So is $sum_n frac{1}{log(log(n))}$



                            But I feel $sum_n frac{exp(-log(log(n))}{log(log(n))}$ might still be convergent, you just need to find another way to prove it.






                            share|cite|improve this answer









                            $endgroup$



                            $sum_n frac{1}{n}$ is divergent. So is $sum_n frac{1}{log(log(n))}$



                            But I feel $sum_n frac{exp(-log(log(n))}{log(log(n))}$ might still be convergent, you just need to find another way to prove it.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 1 '18 at 6:19









                            MoonKnightMoonKnight

                            1,369611




                            1,369611























                                1












                                $begingroup$

                                $$
                                frac 1{(log log n )exp(log log n)} = frac 1 {log log n cdot log n} geqslant frac 1{log ^2 n} geqslant frac 1{n^{1/2 times 2}} = frac 1n,
                                $$
                                so it still diverges.






                                share|cite|improve this answer









                                $endgroup$













                                • $begingroup$
                                  Oh, ok. Thanks a lot. Do you have any advice on how to show $sum_n P(Z>sqrt{2log log n}(1+epsilon)) < infty$ for $Z sim N(0,1)$?
                                  $endgroup$
                                  – Xiaomi
                                  Dec 1 '18 at 6:21












                                • $begingroup$
                                  @Xiaomi Sorry, I suck at probability theory.
                                  $endgroup$
                                  – xbh
                                  Dec 1 '18 at 7:15
















                                1












                                $begingroup$

                                $$
                                frac 1{(log log n )exp(log log n)} = frac 1 {log log n cdot log n} geqslant frac 1{log ^2 n} geqslant frac 1{n^{1/2 times 2}} = frac 1n,
                                $$
                                so it still diverges.






                                share|cite|improve this answer









                                $endgroup$













                                • $begingroup$
                                  Oh, ok. Thanks a lot. Do you have any advice on how to show $sum_n P(Z>sqrt{2log log n}(1+epsilon)) < infty$ for $Z sim N(0,1)$?
                                  $endgroup$
                                  – Xiaomi
                                  Dec 1 '18 at 6:21












                                • $begingroup$
                                  @Xiaomi Sorry, I suck at probability theory.
                                  $endgroup$
                                  – xbh
                                  Dec 1 '18 at 7:15














                                1












                                1








                                1





                                $begingroup$

                                $$
                                frac 1{(log log n )exp(log log n)} = frac 1 {log log n cdot log n} geqslant frac 1{log ^2 n} geqslant frac 1{n^{1/2 times 2}} = frac 1n,
                                $$
                                so it still diverges.






                                share|cite|improve this answer









                                $endgroup$



                                $$
                                frac 1{(log log n )exp(log log n)} = frac 1 {log log n cdot log n} geqslant frac 1{log ^2 n} geqslant frac 1{n^{1/2 times 2}} = frac 1n,
                                $$
                                so it still diverges.







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered Dec 1 '18 at 6:18









                                xbhxbh

                                6,1351522




                                6,1351522












                                • $begingroup$
                                  Oh, ok. Thanks a lot. Do you have any advice on how to show $sum_n P(Z>sqrt{2log log n}(1+epsilon)) < infty$ for $Z sim N(0,1)$?
                                  $endgroup$
                                  – Xiaomi
                                  Dec 1 '18 at 6:21












                                • $begingroup$
                                  @Xiaomi Sorry, I suck at probability theory.
                                  $endgroup$
                                  – xbh
                                  Dec 1 '18 at 7:15


















                                • $begingroup$
                                  Oh, ok. Thanks a lot. Do you have any advice on how to show $sum_n P(Z>sqrt{2log log n}(1+epsilon)) < infty$ for $Z sim N(0,1)$?
                                  $endgroup$
                                  – Xiaomi
                                  Dec 1 '18 at 6:21












                                • $begingroup$
                                  @Xiaomi Sorry, I suck at probability theory.
                                  $endgroup$
                                  – xbh
                                  Dec 1 '18 at 7:15
















                                $begingroup$
                                Oh, ok. Thanks a lot. Do you have any advice on how to show $sum_n P(Z>sqrt{2log log n}(1+epsilon)) < infty$ for $Z sim N(0,1)$?
                                $endgroup$
                                – Xiaomi
                                Dec 1 '18 at 6:21






                                $begingroup$
                                Oh, ok. Thanks a lot. Do you have any advice on how to show $sum_n P(Z>sqrt{2log log n}(1+epsilon)) < infty$ for $Z sim N(0,1)$?
                                $endgroup$
                                – Xiaomi
                                Dec 1 '18 at 6:21














                                $begingroup$
                                @Xiaomi Sorry, I suck at probability theory.
                                $endgroup$
                                – xbh
                                Dec 1 '18 at 7:15




                                $begingroup$
                                @Xiaomi Sorry, I suck at probability theory.
                                $endgroup$
                                – xbh
                                Dec 1 '18 at 7:15











                                1












                                $begingroup$

                                Cauchy condensation shows
                                $$sum_n frac{1}{log(log(n))} sim sum_n frac{2^n}{log(log(2^n))} = sum_n frac{2^n}{log(nlog(2))} = sum_n frac{2^n}{log(n) + log(log(2))} $$
                                So, it is divergent.






                                share|cite|improve this answer









                                $endgroup$













                                • $begingroup$
                                  Thanks a lot. Do you have any advice on how to show $sum_n P(Z>sqrt{2log log n}(1+epsilon)) < infty$ for $Z sim N(0,1)$? The common probability inequalities do not seem tight enough to show this..
                                  $endgroup$
                                  – Xiaomi
                                  Dec 1 '18 at 6:24










                                • $begingroup$
                                  You may use the integral for $N(0,1)$ for each member of the series and estimate it: $$P(Z>sqrt{2log log n}(1+epsilon)) sim int_{sqrt{2log log n}(1+epsilon)}^{infty} e^{frac{-x^2}{2}}dx$$
                                  $endgroup$
                                  – trancelocation
                                  Dec 1 '18 at 6:48


















                                1












                                $begingroup$

                                Cauchy condensation shows
                                $$sum_n frac{1}{log(log(n))} sim sum_n frac{2^n}{log(log(2^n))} = sum_n frac{2^n}{log(nlog(2))} = sum_n frac{2^n}{log(n) + log(log(2))} $$
                                So, it is divergent.






                                share|cite|improve this answer









                                $endgroup$













                                • $begingroup$
                                  Thanks a lot. Do you have any advice on how to show $sum_n P(Z>sqrt{2log log n}(1+epsilon)) < infty$ for $Z sim N(0,1)$? The common probability inequalities do not seem tight enough to show this..
                                  $endgroup$
                                  – Xiaomi
                                  Dec 1 '18 at 6:24










                                • $begingroup$
                                  You may use the integral for $N(0,1)$ for each member of the series and estimate it: $$P(Z>sqrt{2log log n}(1+epsilon)) sim int_{sqrt{2log log n}(1+epsilon)}^{infty} e^{frac{-x^2}{2}}dx$$
                                  $endgroup$
                                  – trancelocation
                                  Dec 1 '18 at 6:48
















                                1












                                1








                                1





                                $begingroup$

                                Cauchy condensation shows
                                $$sum_n frac{1}{log(log(n))} sim sum_n frac{2^n}{log(log(2^n))} = sum_n frac{2^n}{log(nlog(2))} = sum_n frac{2^n}{log(n) + log(log(2))} $$
                                So, it is divergent.






                                share|cite|improve this answer









                                $endgroup$



                                Cauchy condensation shows
                                $$sum_n frac{1}{log(log(n))} sim sum_n frac{2^n}{log(log(2^n))} = sum_n frac{2^n}{log(nlog(2))} = sum_n frac{2^n}{log(n) + log(log(2))} $$
                                So, it is divergent.







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered Dec 1 '18 at 6:20









                                trancelocationtrancelocation

                                11.3k1724




                                11.3k1724












                                • $begingroup$
                                  Thanks a lot. Do you have any advice on how to show $sum_n P(Z>sqrt{2log log n}(1+epsilon)) < infty$ for $Z sim N(0,1)$? The common probability inequalities do not seem tight enough to show this..
                                  $endgroup$
                                  – Xiaomi
                                  Dec 1 '18 at 6:24










                                • $begingroup$
                                  You may use the integral for $N(0,1)$ for each member of the series and estimate it: $$P(Z>sqrt{2log log n}(1+epsilon)) sim int_{sqrt{2log log n}(1+epsilon)}^{infty} e^{frac{-x^2}{2}}dx$$
                                  $endgroup$
                                  – trancelocation
                                  Dec 1 '18 at 6:48




















                                • $begingroup$
                                  Thanks a lot. Do you have any advice on how to show $sum_n P(Z>sqrt{2log log n}(1+epsilon)) < infty$ for $Z sim N(0,1)$? The common probability inequalities do not seem tight enough to show this..
                                  $endgroup$
                                  – Xiaomi
                                  Dec 1 '18 at 6:24










                                • $begingroup$
                                  You may use the integral for $N(0,1)$ for each member of the series and estimate it: $$P(Z>sqrt{2log log n}(1+epsilon)) sim int_{sqrt{2log log n}(1+epsilon)}^{infty} e^{frac{-x^2}{2}}dx$$
                                  $endgroup$
                                  – trancelocation
                                  Dec 1 '18 at 6:48


















                                $begingroup$
                                Thanks a lot. Do you have any advice on how to show $sum_n P(Z>sqrt{2log log n}(1+epsilon)) < infty$ for $Z sim N(0,1)$? The common probability inequalities do not seem tight enough to show this..
                                $endgroup$
                                – Xiaomi
                                Dec 1 '18 at 6:24




                                $begingroup$
                                Thanks a lot. Do you have any advice on how to show $sum_n P(Z>sqrt{2log log n}(1+epsilon)) < infty$ for $Z sim N(0,1)$? The common probability inequalities do not seem tight enough to show this..
                                $endgroup$
                                – Xiaomi
                                Dec 1 '18 at 6:24












                                $begingroup$
                                You may use the integral for $N(0,1)$ for each member of the series and estimate it: $$P(Z>sqrt{2log log n}(1+epsilon)) sim int_{sqrt{2log log n}(1+epsilon)}^{infty} e^{frac{-x^2}{2}}dx$$
                                $endgroup$
                                – trancelocation
                                Dec 1 '18 at 6:48






                                $begingroup$
                                You may use the integral for $N(0,1)$ for each member of the series and estimate it: $$P(Z>sqrt{2log log n}(1+epsilon)) sim int_{sqrt{2log log n}(1+epsilon)}^{infty} e^{frac{-x^2}{2}}dx$$
                                $endgroup$
                                – trancelocation
                                Dec 1 '18 at 6:48













                                1












                                $begingroup$

                                Credits to user MoonKnight.



                                $log n <n ; $



                                And once more:



                                $log (log n) lt log n lt n.$



                                $dfrac {1}{n} lt dfrac{1}{log n} lt dfrac{1}{log (log n)}.$



                                Comparison test.






                                share|cite|improve this answer









                                $endgroup$


















                                  1












                                  $begingroup$

                                  Credits to user MoonKnight.



                                  $log n <n ; $



                                  And once more:



                                  $log (log n) lt log n lt n.$



                                  $dfrac {1}{n} lt dfrac{1}{log n} lt dfrac{1}{log (log n)}.$



                                  Comparison test.






                                  share|cite|improve this answer









                                  $endgroup$
















                                    1












                                    1








                                    1





                                    $begingroup$

                                    Credits to user MoonKnight.



                                    $log n <n ; $



                                    And once more:



                                    $log (log n) lt log n lt n.$



                                    $dfrac {1}{n} lt dfrac{1}{log n} lt dfrac{1}{log (log n)}.$



                                    Comparison test.






                                    share|cite|improve this answer









                                    $endgroup$



                                    Credits to user MoonKnight.



                                    $log n <n ; $



                                    And once more:



                                    $log (log n) lt log n lt n.$



                                    $dfrac {1}{n} lt dfrac{1}{log n} lt dfrac{1}{log (log n)}.$



                                    Comparison test.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Dec 1 '18 at 7:15









                                    Peter SzilasPeter Szilas

                                    11.3k2822




                                    11.3k2822






























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                                        Can I use Tabulator js library in my java Spring + Thymeleaf project?

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                                        0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
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                                        Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

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                                        up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx...
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