Show that if $sum_{k=1}^{infty}(u_k)^2 lt 1$ then $sum_{k=1}^{infty}frac{2^{-k}}{1+3^{-k}-u_k} lt infty$












3















I need to show that if $sum_{k=1}^{infty}(u_k)^2 lt 1$ then $sum_{k=1}^{infty}frac{2^{-k}}{1+3^{-k}-u_k} lt infty$.




I tried using all convergence tests, but to no avail. There's probably some way to decompose the fraction so that you can directly use the convergence of the $u_i$ series, but I can't find that.



Can someone maybe help me please?






calculus real-analysis sequences-and-series







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  • The limit comparison test works: Compare with $2^{-k}$ and use $u_kto 0.$
    – zhw.
    Nov 20 '18 at 17:31
















3















I need to show that if $sum_{k=1}^{infty}(u_k)^2 lt 1$ then $sum_{k=1}^{infty}frac{2^{-k}}{1+3^{-k}-u_k} lt infty$.




I tried using all convergence tests, but to no avail. There's probably some way to decompose the fraction so that you can directly use the convergence of the $u_i$ series, but I can't find that.



Can someone maybe help me please?






calculus real-analysis sequences-and-series







share|cite|improve this question
























  • The limit comparison test works: Compare with $2^{-k}$ and use $u_kto 0.$
    – zhw.
    Nov 20 '18 at 17:31














3












3








3


1






I need to show that if $sum_{k=1}^{infty}(u_k)^2 lt 1$ then $sum_{k=1}^{infty}frac{2^{-k}}{1+3^{-k}-u_k} lt infty$.




I tried using all convergence tests, but to no avail. There's probably some way to decompose the fraction so that you can directly use the convergence of the $u_i$ series, but I can't find that.



Can someone maybe help me please?






calculus real-analysis sequences-and-series







share|cite|improve this question
















I need to show that if $sum_{k=1}^{infty}(u_k)^2 lt 1$ then $sum_{k=1}^{infty}frac{2^{-k}}{1+3^{-k}-u_k} lt infty$.




I tried using all convergence tests, but to no avail. There's probably some way to decompose the fraction so that you can directly use the convergence of the $u_i$ series, but I can't find that.



Can someone maybe help me please?






calculus real-analysis sequences-and-series




calculus real-analysis sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Nov 20 '18 at 18:39









Robert Z

93.3k1061132




93.3k1061132










asked Nov 20 '18 at 16:59









eager2learn

1,23211430




1,23211430












  • The limit comparison test works: Compare with $2^{-k}$ and use $u_kto 0.$
    – zhw.
    Nov 20 '18 at 17:31


















  • The limit comparison test works: Compare with $2^{-k}$ and use $u_kto 0.$
    – zhw.
    Nov 20 '18 at 17:31
















The limit comparison test works: Compare with $2^{-k}$ and use $u_kto 0.$
– zhw.
Nov 20 '18 at 17:31




The limit comparison test works: Compare with $2^{-k}$ and use $u_kto 0.$
– zhw.
Nov 20 '18 at 17:31










1 Answer
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4














Note that the convergence of $sum_{k=1}^{infty}(u_k)^2$ implies that $u_kto 0$. Hence $3^{-k}-u_kto 0$ and there is $N>0$ such that for all $kgeq N$, $|3^{-k}-u_k|<1/2$. Then
$$frac{2^{-k}}{1+1/2}leq frac{2^{-k}}{1+3^{-k}-u_k}leq frac{2^{-k}}{1-1/2}.$$
Can you take it from here?



P.S. Since $sum_{k=1}^{infty}(u_k)^2<1$ then $|u_k|<1$ for each $kgeq1$, and it follows that the denominator $1+3^{-k}-u_k$ is always positive.






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    Note that the convergence of $sum_{k=1}^{infty}(u_k)^2$ implies that $u_kto 0$. Hence $3^{-k}-u_kto 0$ and there is $N>0$ such that for all $kgeq N$, $|3^{-k}-u_k|<1/2$. Then
    $$frac{2^{-k}}{1+1/2}leq frac{2^{-k}}{1+3^{-k}-u_k}leq frac{2^{-k}}{1-1/2}.$$
    Can you take it from here?



    P.S. Since $sum_{k=1}^{infty}(u_k)^2<1$ then $|u_k|<1$ for each $kgeq1$, and it follows that the denominator $1+3^{-k}-u_k$ is always positive.






    share|cite|improve this answer




























      4














      Note that the convergence of $sum_{k=1}^{infty}(u_k)^2$ implies that $u_kto 0$. Hence $3^{-k}-u_kto 0$ and there is $N>0$ such that for all $kgeq N$, $|3^{-k}-u_k|<1/2$. Then
      $$frac{2^{-k}}{1+1/2}leq frac{2^{-k}}{1+3^{-k}-u_k}leq frac{2^{-k}}{1-1/2}.$$
      Can you take it from here?



      P.S. Since $sum_{k=1}^{infty}(u_k)^2<1$ then $|u_k|<1$ for each $kgeq1$, and it follows that the denominator $1+3^{-k}-u_k$ is always positive.






      share|cite|improve this answer


























        4












        4








        4






        Note that the convergence of $sum_{k=1}^{infty}(u_k)^2$ implies that $u_kto 0$. Hence $3^{-k}-u_kto 0$ and there is $N>0$ such that for all $kgeq N$, $|3^{-k}-u_k|<1/2$. Then
        $$frac{2^{-k}}{1+1/2}leq frac{2^{-k}}{1+3^{-k}-u_k}leq frac{2^{-k}}{1-1/2}.$$
        Can you take it from here?



        P.S. Since $sum_{k=1}^{infty}(u_k)^2<1$ then $|u_k|<1$ for each $kgeq1$, and it follows that the denominator $1+3^{-k}-u_k$ is always positive.






        share|cite|improve this answer














        Note that the convergence of $sum_{k=1}^{infty}(u_k)^2$ implies that $u_kto 0$. Hence $3^{-k}-u_kto 0$ and there is $N>0$ such that for all $kgeq N$, $|3^{-k}-u_k|<1/2$. Then
        $$frac{2^{-k}}{1+1/2}leq frac{2^{-k}}{1+3^{-k}-u_k}leq frac{2^{-k}}{1-1/2}.$$
        Can you take it from here?



        P.S. Since $sum_{k=1}^{infty}(u_k)^2<1$ then $|u_k|<1$ for each $kgeq1$, and it follows that the denominator $1+3^{-k}-u_k$ is always positive.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 20 '18 at 17:11

























        answered Nov 20 '18 at 17:05









        Robert Z

        93.3k1061132




        93.3k1061132






























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