limsup of a series












0












$begingroup$


The series is from Rudin's *Principles of Mathematical Analysis$ ("Baby Rudin"), p.67.



$$frac{1}{2}+frac{1}{3}+frac{1}{2^2}+frac{1}{3^2}+... $$



and Rudin claims that
$$limsup_{nto infty} (a_n)^frac{1}{n} = frac{1}{sqrt{2}}.$$



I do not get why this is the case. Can't we pick
$$frac{1}{2^3}, frac{1}{2^6},frac{1}{2^9}ldots $$
or higher powers of $frac{1}{2}$ in which case we get $$limsup (a_{n_k})^{1/n} = frac{1}{3^frac{1}{3}} > frac{1}{sqrt2}.$$






real-analysis sequences-and-series real-numbers







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Because each $3^{-3n}$ is $a_{6n}$ but you apply the root test as if $3^{-3n}=a_{9n}$.
    $endgroup$
    – Did
    Dec 31 '18 at 21:40












  • $begingroup$
    @Did makes a lot of sense. thanks!
    $endgroup$
    – Kaan Yolsever
    Dec 31 '18 at 21:41






  • 3




    $begingroup$
    Reatedly, note that, in your post, $$limsup (a_{n_k})^{1/n}$$ is intrinsically faulty, one should consider $$limsup (a_{n_k})^{1/n_k}$$
    $endgroup$
    – Did
    Dec 31 '18 at 21:42








  • 1




    $begingroup$
    Please do not change thoroughly your post after you received comments (especially when, as here, basically the same idea applies).
    $endgroup$
    – Did
    Dec 31 '18 at 21:47






  • 1




    $begingroup$
    Yu could first and foremost "make sure" to cancel the unfortunate edit on this very page.
    $endgroup$
    – Did
    Dec 31 '18 at 21:48
















0












$begingroup$


The series is from Rudin's *Principles of Mathematical Analysis$ ("Baby Rudin"), p.67.



$$frac{1}{2}+frac{1}{3}+frac{1}{2^2}+frac{1}{3^2}+... $$



and Rudin claims that
$$limsup_{nto infty} (a_n)^frac{1}{n} = frac{1}{sqrt{2}}.$$



I do not get why this is the case. Can't we pick
$$frac{1}{2^3}, frac{1}{2^6},frac{1}{2^9}ldots $$
or higher powers of $frac{1}{2}$ in which case we get $$limsup (a_{n_k})^{1/n} = frac{1}{3^frac{1}{3}} > frac{1}{sqrt2}.$$






real-analysis sequences-and-series real-numbers







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Because each $3^{-3n}$ is $a_{6n}$ but you apply the root test as if $3^{-3n}=a_{9n}$.
    $endgroup$
    – Did
    Dec 31 '18 at 21:40












  • $begingroup$
    @Did makes a lot of sense. thanks!
    $endgroup$
    – Kaan Yolsever
    Dec 31 '18 at 21:41






  • 3




    $begingroup$
    Reatedly, note that, in your post, $$limsup (a_{n_k})^{1/n}$$ is intrinsically faulty, one should consider $$limsup (a_{n_k})^{1/n_k}$$
    $endgroup$
    – Did
    Dec 31 '18 at 21:42








  • 1




    $begingroup$
    Please do not change thoroughly your post after you received comments (especially when, as here, basically the same idea applies).
    $endgroup$
    – Did
    Dec 31 '18 at 21:47






  • 1




    $begingroup$
    Yu could first and foremost "make sure" to cancel the unfortunate edit on this very page.
    $endgroup$
    – Did
    Dec 31 '18 at 21:48














0












0








0





$begingroup$


The series is from Rudin's *Principles of Mathematical Analysis$ ("Baby Rudin"), p.67.



$$frac{1}{2}+frac{1}{3}+frac{1}{2^2}+frac{1}{3^2}+... $$



and Rudin claims that
$$limsup_{nto infty} (a_n)^frac{1}{n} = frac{1}{sqrt{2}}.$$



I do not get why this is the case. Can't we pick
$$frac{1}{2^3}, frac{1}{2^6},frac{1}{2^9}ldots $$
or higher powers of $frac{1}{2}$ in which case we get $$limsup (a_{n_k})^{1/n} = frac{1}{3^frac{1}{3}} > frac{1}{sqrt2}.$$






real-analysis sequences-and-series real-numbers







share|cite|improve this question











$endgroup$




The series is from Rudin's *Principles of Mathematical Analysis$ ("Baby Rudin"), p.67.



$$frac{1}{2}+frac{1}{3}+frac{1}{2^2}+frac{1}{3^2}+... $$



and Rudin claims that
$$limsup_{nto infty} (a_n)^frac{1}{n} = frac{1}{sqrt{2}}.$$



I do not get why this is the case. Can't we pick
$$frac{1}{2^3}, frac{1}{2^6},frac{1}{2^9}ldots $$
or higher powers of $frac{1}{2}$ in which case we get $$limsup (a_{n_k})^{1/n} = frac{1}{3^frac{1}{3}} > frac{1}{sqrt2}.$$






real-analysis sequences-and-series real-numbers




real-analysis sequences-and-series real-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 21:49







Kaan Yolsever

















asked Dec 31 '18 at 21:26









Kaan YolseverKaan Yolsever

1309




1309








  • 2




    $begingroup$
    Because each $3^{-3n}$ is $a_{6n}$ but you apply the root test as if $3^{-3n}=a_{9n}$.
    $endgroup$
    – Did
    Dec 31 '18 at 21:40












  • $begingroup$
    @Did makes a lot of sense. thanks!
    $endgroup$
    – Kaan Yolsever
    Dec 31 '18 at 21:41






  • 3




    $begingroup$
    Reatedly, note that, in your post, $$limsup (a_{n_k})^{1/n}$$ is intrinsically faulty, one should consider $$limsup (a_{n_k})^{1/n_k}$$
    $endgroup$
    – Did
    Dec 31 '18 at 21:42








  • 1




    $begingroup$
    Please do not change thoroughly your post after you received comments (especially when, as here, basically the same idea applies).
    $endgroup$
    – Did
    Dec 31 '18 at 21:47






  • 1




    $begingroup$
    Yu could first and foremost "make sure" to cancel the unfortunate edit on this very page.
    $endgroup$
    – Did
    Dec 31 '18 at 21:48














  • 2




    $begingroup$
    Because each $3^{-3n}$ is $a_{6n}$ but you apply the root test as if $3^{-3n}=a_{9n}$.
    $endgroup$
    – Did
    Dec 31 '18 at 21:40












  • $begingroup$
    @Did makes a lot of sense. thanks!
    $endgroup$
    – Kaan Yolsever
    Dec 31 '18 at 21:41






  • 3




    $begingroup$
    Reatedly, note that, in your post, $$limsup (a_{n_k})^{1/n}$$ is intrinsically faulty, one should consider $$limsup (a_{n_k})^{1/n_k}$$
    $endgroup$
    – Did
    Dec 31 '18 at 21:42








  • 1




    $begingroup$
    Please do not change thoroughly your post after you received comments (especially when, as here, basically the same idea applies).
    $endgroup$
    – Did
    Dec 31 '18 at 21:47






  • 1




    $begingroup$
    Yu could first and foremost "make sure" to cancel the unfortunate edit on this very page.
    $endgroup$
    – Did
    Dec 31 '18 at 21:48








2




2




$begingroup$
Because each $3^{-3n}$ is $a_{6n}$ but you apply the root test as if $3^{-3n}=a_{9n}$.
$endgroup$
– Did
Dec 31 '18 at 21:40






$begingroup$
Because each $3^{-3n}$ is $a_{6n}$ but you apply the root test as if $3^{-3n}=a_{9n}$.
$endgroup$
– Did
Dec 31 '18 at 21:40














$begingroup$
@Did makes a lot of sense. thanks!
$endgroup$
– Kaan Yolsever
Dec 31 '18 at 21:41




$begingroup$
@Did makes a lot of sense. thanks!
$endgroup$
– Kaan Yolsever
Dec 31 '18 at 21:41




3




3




$begingroup$
Reatedly, note that, in your post, $$limsup (a_{n_k})^{1/n}$$ is intrinsically faulty, one should consider $$limsup (a_{n_k})^{1/n_k}$$
$endgroup$
– Did
Dec 31 '18 at 21:42






$begingroup$
Reatedly, note that, in your post, $$limsup (a_{n_k})^{1/n}$$ is intrinsically faulty, one should consider $$limsup (a_{n_k})^{1/n_k}$$
$endgroup$
– Did
Dec 31 '18 at 21:42






1




1




$begingroup$
Please do not change thoroughly your post after you received comments (especially when, as here, basically the same idea applies).
$endgroup$
– Did
Dec 31 '18 at 21:47




$begingroup$
Please do not change thoroughly your post after you received comments (especially when, as here, basically the same idea applies).
$endgroup$
– Did
Dec 31 '18 at 21:47




1




1




$begingroup$
Yu could first and foremost "make sure" to cancel the unfortunate edit on this very page.
$endgroup$
– Did
Dec 31 '18 at 21:48




$begingroup$
Yu could first and foremost "make sure" to cancel the unfortunate edit on this very page.
$endgroup$
– Did
Dec 31 '18 at 21:48










1 Answer
1






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oldest

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1












$begingroup$

One way, perhaps easier, to see what's going on, is to observe that:0



$$a_1=frac12;,;a_2=frac13;,;a_3=frac1{2^2};,;a_4=frac1{3^2};...$$



Clearly, to get the supremum of the above we have to take the odd numbered elements:



$$a_1,,a_3,,a_5,,....implies limsup_{ntoinfty}sqrt[n]{a_n}=limsup_{ntoinfty}sqrt[2n-1]{frac1{2^n}}=frac1{sqrt2}$$



Because at index $;1=2cdot1-1;$, we have $;frac12;$ , at index $;2cdot2-1=3;$, we have $;frac1{2^2};$, ...at index $;2n-1;$ , we have $;frac1{2^n};$ ...



or taking $;k;$ to be a general odd index, the corresponding element here is $;cfrac1{2^{frac{k+1}2}};$ , so again



$$limsup_{ktoinfty}sqrt[k]{frac1{2^{frac k2+frac12}}}=limsup_{ktoinfty}frac1{2^{1/2}sqrt[k]{sqrt2}}=frac1{sqrt2}$$






share|cite|improve this answer









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    1












    $begingroup$

    One way, perhaps easier, to see what's going on, is to observe that:0



    $$a_1=frac12;,;a_2=frac13;,;a_3=frac1{2^2};,;a_4=frac1{3^2};...$$



    Clearly, to get the supremum of the above we have to take the odd numbered elements:



    $$a_1,,a_3,,a_5,,....implies limsup_{ntoinfty}sqrt[n]{a_n}=limsup_{ntoinfty}sqrt[2n-1]{frac1{2^n}}=frac1{sqrt2}$$



    Because at index $;1=2cdot1-1;$, we have $;frac12;$ , at index $;2cdot2-1=3;$, we have $;frac1{2^2};$, ...at index $;2n-1;$ , we have $;frac1{2^n};$ ...



    or taking $;k;$ to be a general odd index, the corresponding element here is $;cfrac1{2^{frac{k+1}2}};$ , so again



    $$limsup_{ktoinfty}sqrt[k]{frac1{2^{frac k2+frac12}}}=limsup_{ktoinfty}frac1{2^{1/2}sqrt[k]{sqrt2}}=frac1{sqrt2}$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      One way, perhaps easier, to see what's going on, is to observe that:0



      $$a_1=frac12;,;a_2=frac13;,;a_3=frac1{2^2};,;a_4=frac1{3^2};...$$



      Clearly, to get the supremum of the above we have to take the odd numbered elements:



      $$a_1,,a_3,,a_5,,....implies limsup_{ntoinfty}sqrt[n]{a_n}=limsup_{ntoinfty}sqrt[2n-1]{frac1{2^n}}=frac1{sqrt2}$$



      Because at index $;1=2cdot1-1;$, we have $;frac12;$ , at index $;2cdot2-1=3;$, we have $;frac1{2^2};$, ...at index $;2n-1;$ , we have $;frac1{2^n};$ ...



      or taking $;k;$ to be a general odd index, the corresponding element here is $;cfrac1{2^{frac{k+1}2}};$ , so again



      $$limsup_{ktoinfty}sqrt[k]{frac1{2^{frac k2+frac12}}}=limsup_{ktoinfty}frac1{2^{1/2}sqrt[k]{sqrt2}}=frac1{sqrt2}$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        One way, perhaps easier, to see what's going on, is to observe that:0



        $$a_1=frac12;,;a_2=frac13;,;a_3=frac1{2^2};,;a_4=frac1{3^2};...$$



        Clearly, to get the supremum of the above we have to take the odd numbered elements:



        $$a_1,,a_3,,a_5,,....implies limsup_{ntoinfty}sqrt[n]{a_n}=limsup_{ntoinfty}sqrt[2n-1]{frac1{2^n}}=frac1{sqrt2}$$



        Because at index $;1=2cdot1-1;$, we have $;frac12;$ , at index $;2cdot2-1=3;$, we have $;frac1{2^2};$, ...at index $;2n-1;$ , we have $;frac1{2^n};$ ...



        or taking $;k;$ to be a general odd index, the corresponding element here is $;cfrac1{2^{frac{k+1}2}};$ , so again



        $$limsup_{ktoinfty}sqrt[k]{frac1{2^{frac k2+frac12}}}=limsup_{ktoinfty}frac1{2^{1/2}sqrt[k]{sqrt2}}=frac1{sqrt2}$$






        share|cite|improve this answer









        $endgroup$



        One way, perhaps easier, to see what's going on, is to observe that:0



        $$a_1=frac12;,;a_2=frac13;,;a_3=frac1{2^2};,;a_4=frac1{3^2};...$$



        Clearly, to get the supremum of the above we have to take the odd numbered elements:



        $$a_1,,a_3,,a_5,,....implies limsup_{ntoinfty}sqrt[n]{a_n}=limsup_{ntoinfty}sqrt[2n-1]{frac1{2^n}}=frac1{sqrt2}$$



        Because at index $;1=2cdot1-1;$, we have $;frac12;$ , at index $;2cdot2-1=3;$, we have $;frac1{2^2};$, ...at index $;2n-1;$ , we have $;frac1{2^n};$ ...



        or taking $;k;$ to be a general odd index, the corresponding element here is $;cfrac1{2^{frac{k+1}2}};$ , so again



        $$limsup_{ktoinfty}sqrt[k]{frac1{2^{frac k2+frac12}}}=limsup_{ktoinfty}frac1{2^{1/2}sqrt[k]{sqrt2}}=frac1{sqrt2}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 31 '18 at 22:38









        DonAntonioDonAntonio

        180k1494233




        180k1494233






























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