For which positive real numbers $a$ does $sum_{n=1}^infty a^{log n}$ converge?












-1












$begingroup$


I know the answer is $a<1$. I tried the ratio test which was inconclusive, and also the root test which did not work. Any tips on how to proceed?



Any help would be appreciated, thank you






sequences-and-series convergence







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$endgroup$












  • $begingroup$
    Hint: $a^{log n}=e^{log acdotlog n}=dots$
    $endgroup$
    – Wojowu
    Jan 14 at 23:03










  • $begingroup$
    The answer $a<1$ is actually wrong. The correct one is $a<1/e$.
    $endgroup$
    – Wojowu
    Jan 14 at 23:08
















-1












$begingroup$


I know the answer is $a<1$. I tried the ratio test which was inconclusive, and also the root test which did not work. Any tips on how to proceed?



Any help would be appreciated, thank you






sequences-and-series convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hint: $a^{log n}=e^{log acdotlog n}=dots$
    $endgroup$
    – Wojowu
    Jan 14 at 23:03










  • $begingroup$
    The answer $a<1$ is actually wrong. The correct one is $a<1/e$.
    $endgroup$
    – Wojowu
    Jan 14 at 23:08














-1












-1








-1





$begingroup$


I know the answer is $a<1$. I tried the ratio test which was inconclusive, and also the root test which did not work. Any tips on how to proceed?



Any help would be appreciated, thank you






sequences-and-series convergence







share|cite|improve this question











$endgroup$




I know the answer is $a<1$. I tried the ratio test which was inconclusive, and also the root test which did not work. Any tips on how to proceed?



Any help would be appreciated, thank you






sequences-and-series convergence




sequences-and-series convergence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 14 at 23:00









Math1000

19.4k31746




19.4k31746










asked Jan 14 at 22:52









user101user101

304




304












  • $begingroup$
    Hint: $a^{log n}=e^{log acdotlog n}=dots$
    $endgroup$
    – Wojowu
    Jan 14 at 23:03










  • $begingroup$
    The answer $a<1$ is actually wrong. The correct one is $a<1/e$.
    $endgroup$
    – Wojowu
    Jan 14 at 23:08


















  • $begingroup$
    Hint: $a^{log n}=e^{log acdotlog n}=dots$
    $endgroup$
    – Wojowu
    Jan 14 at 23:03










  • $begingroup$
    The answer $a<1$ is actually wrong. The correct one is $a<1/e$.
    $endgroup$
    – Wojowu
    Jan 14 at 23:08
















$begingroup$
Hint: $a^{log n}=e^{log acdotlog n}=dots$
$endgroup$
– Wojowu
Jan 14 at 23:03




$begingroup$
Hint: $a^{log n}=e^{log acdotlog n}=dots$
$endgroup$
– Wojowu
Jan 14 at 23:03












$begingroup$
The answer $a<1$ is actually wrong. The correct one is $a<1/e$.
$endgroup$
– Wojowu
Jan 14 at 23:08




$begingroup$
The answer $a<1$ is actually wrong. The correct one is $a<1/e$.
$endgroup$
– Wojowu
Jan 14 at 23:08










2 Answers
2






active

oldest

votes


















2












$begingroup$

Notice that



$$ a^{log n} = exp( log a log n) = n^{log a } $$



Notice that $log a > 0$ iff $a > 1 $. Thus $sum a^{log n }$ diverges if $a geq 1$ and converges when $ log a < -1 $ or when $ a < 1/e $






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Series $n^c$ converges for $c<-1$, not $c<0$.
    $endgroup$
    – Wojowu
    Jan 14 at 23:09










  • $begingroup$
    fixed it now...
    $endgroup$
    – James
    Jan 14 at 23:14



















0












$begingroup$

Note that
$$begin{eqnarray}
sum_{n=2^k}^{2^{k+1}-1} a^{log n}&ge&sum_{n=2^k}^{2^{k+1}-1}(a^{log 2})^{k}\&ge&2^kcdot (a^{log 2})^k\&=& (2cdot a^{log 2})^k,
end{eqnarray}$$and
$$begin{eqnarray}
sum_{n=2^k}^{2^{k+1}-1} a^{log n}&le&sum_{n=2^k}^{2^{k+1}-1}(a^{log 2})^{k+1}\&=&2^kcdot( a^{log2})^{k+1}\&le& (2cdot a^{log2})^{k+1}
end{eqnarray}$$ for all $kge 0$. This shows
$$
lim_{Ntoinfty}sum_{n=1}^N a^{log n}=lim_{Ktoinfty}sum_{k=0}^K sum_{n=2^k}^{2^{k+1}-1} a^{log n}
$$ converges if and only if $sum_{k=0}^infty (2cdot a^{log 2})^k$ converges (by comparison test). This gives the condition
$
2cdot a^{log 2}<1,
$ or equivalently
$
2^{1+log a}<2^0=1.
$ Hence it is sufficient and necessary that $$log a<-1Leftrightarrow 0< a<frac{1}{e}.$$






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Notice that



    $$ a^{log n} = exp( log a log n) = n^{log a } $$



    Notice that $log a > 0$ iff $a > 1 $. Thus $sum a^{log n }$ diverges if $a geq 1$ and converges when $ log a < -1 $ or when $ a < 1/e $






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Series $n^c$ converges for $c<-1$, not $c<0$.
      $endgroup$
      – Wojowu
      Jan 14 at 23:09










    • $begingroup$
      fixed it now...
      $endgroup$
      – James
      Jan 14 at 23:14
















    2












    $begingroup$

    Notice that



    $$ a^{log n} = exp( log a log n) = n^{log a } $$



    Notice that $log a > 0$ iff $a > 1 $. Thus $sum a^{log n }$ diverges if $a geq 1$ and converges when $ log a < -1 $ or when $ a < 1/e $






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Series $n^c$ converges for $c<-1$, not $c<0$.
      $endgroup$
      – Wojowu
      Jan 14 at 23:09










    • $begingroup$
      fixed it now...
      $endgroup$
      – James
      Jan 14 at 23:14














    2












    2








    2





    $begingroup$

    Notice that



    $$ a^{log n} = exp( log a log n) = n^{log a } $$



    Notice that $log a > 0$ iff $a > 1 $. Thus $sum a^{log n }$ diverges if $a geq 1$ and converges when $ log a < -1 $ or when $ a < 1/e $






    share|cite|improve this answer











    $endgroup$



    Notice that



    $$ a^{log n} = exp( log a log n) = n^{log a } $$



    Notice that $log a > 0$ iff $a > 1 $. Thus $sum a^{log n }$ diverges if $a geq 1$ and converges when $ log a < -1 $ or when $ a < 1/e $







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 14 at 23:14

























    answered Jan 14 at 23:05









    JamesJames

    2,636425




    2,636425












    • $begingroup$
      Series $n^c$ converges for $c<-1$, not $c<0$.
      $endgroup$
      – Wojowu
      Jan 14 at 23:09










    • $begingroup$
      fixed it now...
      $endgroup$
      – James
      Jan 14 at 23:14


















    • $begingroup$
      Series $n^c$ converges for $c<-1$, not $c<0$.
      $endgroup$
      – Wojowu
      Jan 14 at 23:09










    • $begingroup$
      fixed it now...
      $endgroup$
      – James
      Jan 14 at 23:14
















    $begingroup$
    Series $n^c$ converges for $c<-1$, not $c<0$.
    $endgroup$
    – Wojowu
    Jan 14 at 23:09




    $begingroup$
    Series $n^c$ converges for $c<-1$, not $c<0$.
    $endgroup$
    – Wojowu
    Jan 14 at 23:09












    $begingroup$
    fixed it now...
    $endgroup$
    – James
    Jan 14 at 23:14




    $begingroup$
    fixed it now...
    $endgroup$
    – James
    Jan 14 at 23:14











    0












    $begingroup$

    Note that
    $$begin{eqnarray}
    sum_{n=2^k}^{2^{k+1}-1} a^{log n}&ge&sum_{n=2^k}^{2^{k+1}-1}(a^{log 2})^{k}\&ge&2^kcdot (a^{log 2})^k\&=& (2cdot a^{log 2})^k,
    end{eqnarray}$$    and
    $$begin{eqnarray}
    sum_{n=2^k}^{2^{k+1}-1} a^{log n}&le&sum_{n=2^k}^{2^{k+1}-1}(a^{log 2})^{k+1}\&=&2^kcdot( a^{log2})^{k+1}\&le& (2cdot a^{log2})^{k+1}
    end{eqnarray}$$     for all $kge 0$. This shows
    $$
    lim_{Ntoinfty}sum_{n=1}^N a^{log n}=lim_{Ktoinfty}sum_{k=0}^K sum_{n=2^k}^{2^{k+1}-1} a^{log n}
    $$     converges if and only if $sum_{k=0}^infty (2cdot a^{log 2})^k$ converges (by comparison test). This gives the condition
    $
    2cdot a^{log 2}<1,
    $     or equivalently
    $
    2^{1+log a}<2^0=1.
    $     Hence it is sufficient and necessary that $$log a<-1Leftrightarrow 0< a<frac{1}{e}.$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Note that
      $$begin{eqnarray}
      sum_{n=2^k}^{2^{k+1}-1} a^{log n}&ge&sum_{n=2^k}^{2^{k+1}-1}(a^{log 2})^{k}\&ge&2^kcdot (a^{log 2})^k\&=& (2cdot a^{log 2})^k,
      end{eqnarray}$$      and
      $$begin{eqnarray}
      sum_{n=2^k}^{2^{k+1}-1} a^{log n}&le&sum_{n=2^k}^{2^{k+1}-1}(a^{log 2})^{k+1}\&=&2^kcdot( a^{log2})^{k+1}\&le& (2cdot a^{log2})^{k+1}
      end{eqnarray}$$       for all $kge 0$. This shows
      $$
      lim_{Ntoinfty}sum_{n=1}^N a^{log n}=lim_{Ktoinfty}sum_{k=0}^K sum_{n=2^k}^{2^{k+1}-1} a^{log n}
      $$       converges if and only if $sum_{k=0}^infty (2cdot a^{log 2})^k$ converges (by comparison test). This gives the condition
      $
      2cdot a^{log 2}<1,
      $       or equivalently
      $
      2^{1+log a}<2^0=1.
      $       Hence it is sufficient and necessary that $$log a<-1Leftrightarrow 0< a<frac{1}{e}.$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Note that
        $$begin{eqnarray}
        sum_{n=2^k}^{2^{k+1}-1} a^{log n}&ge&sum_{n=2^k}^{2^{k+1}-1}(a^{log 2})^{k}\&ge&2^kcdot (a^{log 2})^k\&=& (2cdot a^{log 2})^k,
        end{eqnarray}$$        and
        $$begin{eqnarray}
        sum_{n=2^k}^{2^{k+1}-1} a^{log n}&le&sum_{n=2^k}^{2^{k+1}-1}(a^{log 2})^{k+1}\&=&2^kcdot( a^{log2})^{k+1}\&le& (2cdot a^{log2})^{k+1}
        end{eqnarray}$$         for all $kge 0$. This shows
        $$
        lim_{Ntoinfty}sum_{n=1}^N a^{log n}=lim_{Ktoinfty}sum_{k=0}^K sum_{n=2^k}^{2^{k+1}-1} a^{log n}
        $$         converges if and only if $sum_{k=0}^infty (2cdot a^{log 2})^k$ converges (by comparison test). This gives the condition
        $
        2cdot a^{log 2}<1,
        $         or equivalently
        $
        2^{1+log a}<2^0=1.
        $         Hence it is sufficient and necessary that $$log a<-1Leftrightarrow 0< a<frac{1}{e}.$$






        share|cite|improve this answer









        $endgroup$



        Note that
        $$begin{eqnarray}
        sum_{n=2^k}^{2^{k+1}-1} a^{log n}&ge&sum_{n=2^k}^{2^{k+1}-1}(a^{log 2})^{k}\&ge&2^kcdot (a^{log 2})^k\&=& (2cdot a^{log 2})^k,
        end{eqnarray}$$        and
        $$begin{eqnarray}
        sum_{n=2^k}^{2^{k+1}-1} a^{log n}&le&sum_{n=2^k}^{2^{k+1}-1}(a^{log 2})^{k+1}\&=&2^kcdot( a^{log2})^{k+1}\&le& (2cdot a^{log2})^{k+1}
        end{eqnarray}$$         for all $kge 0$. This shows
        $$
        lim_{Ntoinfty}sum_{n=1}^N a^{log n}=lim_{Ktoinfty}sum_{k=0}^K sum_{n=2^k}^{2^{k+1}-1} a^{log n}
        $$         converges if and only if $sum_{k=0}^infty (2cdot a^{log 2})^k$ converges (by comparison test). This gives the condition
        $
        2cdot a^{log 2}<1,
        $         or equivalently
        $
        2^{1+log a}<2^0=1.
        $         Hence it is sufficient and necessary that $$log a<-1Leftrightarrow 0< a<frac{1}{e}.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 15 at 15:07









        SongSong

        18.6k21651




        18.6k21651






























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