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












0














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







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    0














    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



























      0












      0








      0







      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















      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

      5,2481828




      5,2481828










      asked Dec 1 '18 at 6:12









      XiaomiXiaomi

      1,031115




      1,031115






















          5 Answers
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          4














          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





























            2














            $sum_n frac{1}{n}$ 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





























              1














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






              share|cite|improve this answer





















              • 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 $Z sim N(0,1)$?
                – Xiaomi
                Dec 1 '18 at 6:21












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



















              1














              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





















              • 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..
                – Xiaomi
                Dec 1 '18 at 6:24










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





















              1














              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





















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                5 Answers
                5






                active

                oldest

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                5 Answers
                5






                active

                oldest

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                active

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                4














                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


























                  4














                  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
























                    4












                    4








                    4






                    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












                    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














                        $sum_n frac{1}{n}$ 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


























                          2














                          $sum_n frac{1}{n}$ 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
























                            2












                            2








                            2






                            $sum_n frac{1}{n}$ 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












                            $sum_n frac{1}{n}$ 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

                            1,324611




                            1,324611























                                1














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






                                share|cite|improve this answer





















                                • 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 $Z sim N(0,1)$?
                                  – Xiaomi
                                  Dec 1 '18 at 6:21












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
















                                1














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






                                share|cite|improve this answer





















                                • 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 $Z sim N(0,1)$?
                                  – Xiaomi
                                  Dec 1 '18 at 6:21












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














                                1












                                1








                                1






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






                                share|cite|improve this answer












                                $$
                                frac 1{(log log n )exp(log log n)} = frac 1 {log log n cdot log n} geqslant frac 1{log ^2 n} geqslant frac 1{n^{1/2 times 2}} = frac 1n,
                                $$
                                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

                                5,8281522




                                5,8281522












                                • 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 $Z sim N(0,1)$?
                                  – Xiaomi
                                  Dec 1 '18 at 6:21












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


















                                • 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 $Z sim N(0,1)$?
                                  – Xiaomi
                                  Dec 1 '18 at 6:21












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
















                                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 $Z sim N(0,1)$?
                                – Xiaomi
                                Dec 1 '18 at 6:21






                                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 $Z sim N(0,1)$?
                                – Xiaomi
                                Dec 1 '18 at 6:21














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




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











                                1














                                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





















                                • 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..
                                  – Xiaomi
                                  Dec 1 '18 at 6:24










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


















                                1














                                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





















                                • 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..
                                  – Xiaomi
                                  Dec 1 '18 at 6:24










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
















                                1












                                1








                                1






                                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












                                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

                                9,5601722




                                9,5601722












                                • 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..
                                  – Xiaomi
                                  Dec 1 '18 at 6:24










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




















                                • 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..
                                  – Xiaomi
                                  Dec 1 '18 at 6:24










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


















                                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..
                                – Xiaomi
                                Dec 1 '18 at 6:24




                                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..
                                – Xiaomi
                                Dec 1 '18 at 6:24












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






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













                                1














                                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


























                                  1














                                  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
























                                    1












                                    1








                                    1






                                    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












                                    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

                                    10.9k2720




                                    10.9k2720






























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