If $sum_{n=0}^infty|x_n|^2 $ converges then $sum_{n=0}^infty frac{x_n}{sqrt{n+1}}$ converges?












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Does $(x_n) in ell^2$ imply that $sum_{n=0}^infty frac{x_n}{sqrt{n+1}}$ converges?



I have tried finding an upper bound to the series but have failed so far. I have tried Cauchy-Schwarz, some other futile trials and I tried to find counter examples by hand, but I cannot prove it. I have a strong feeling it is true because it even holds for $x_n=1/n$.



Any hints welcome!










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    0












    $begingroup$


    Does $(x_n) in ell^2$ imply that $sum_{n=0}^infty frac{x_n}{sqrt{n+1}}$ converges?



    I have tried finding an upper bound to the series but have failed so far. I have tried Cauchy-Schwarz, some other futile trials and I tried to find counter examples by hand, but I cannot prove it. I have a strong feeling it is true because it even holds for $x_n=1/n$.



    Any hints welcome!










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Does $(x_n) in ell^2$ imply that $sum_{n=0}^infty frac{x_n}{sqrt{n+1}}$ converges?



      I have tried finding an upper bound to the series but have failed so far. I have tried Cauchy-Schwarz, some other futile trials and I tried to find counter examples by hand, but I cannot prove it. I have a strong feeling it is true because it even holds for $x_n=1/n$.



      Any hints welcome!










      share|cite|improve this question









      $endgroup$




      Does $(x_n) in ell^2$ imply that $sum_{n=0}^infty frac{x_n}{sqrt{n+1}}$ converges?



      I have tried finding an upper bound to the series but have failed so far. I have tried Cauchy-Schwarz, some other futile trials and I tried to find counter examples by hand, but I cannot prove it. I have a strong feeling it is true because it even holds for $x_n=1/n$.



      Any hints welcome!







      sequences-and-series convergence






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      asked Jan 21 at 23:16









      B.SwanB.Swan

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          This is not true. Let $x_n=frac 1 {sqrt n ln , n}$ for $n >1$. Then ${x_n} in ell^{2}$ but $sum frac {x_n} {sqrt {n+1} }=infty$. [Use the fact that $sum frac 1 {n (ln, n)^{2}} <infty$ but $sum frac 1 {n (ln, n)}=infty$].






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            1 Answer
            1






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            $begingroup$

            This is not true. Let $x_n=frac 1 {sqrt n ln , n}$ for $n >1$. Then ${x_n} in ell^{2}$ but $sum frac {x_n} {sqrt {n+1} }=infty$. [Use the fact that $sum frac 1 {n (ln, n)^{2}} <infty$ but $sum frac 1 {n (ln, n)}=infty$].






            share|cite|improve this answer









            $endgroup$


















              5












              $begingroup$

              This is not true. Let $x_n=frac 1 {sqrt n ln , n}$ for $n >1$. Then ${x_n} in ell^{2}$ but $sum frac {x_n} {sqrt {n+1} }=infty$. [Use the fact that $sum frac 1 {n (ln, n)^{2}} <infty$ but $sum frac 1 {n (ln, n)}=infty$].






              share|cite|improve this answer









              $endgroup$
















                5












                5








                5





                $begingroup$

                This is not true. Let $x_n=frac 1 {sqrt n ln , n}$ for $n >1$. Then ${x_n} in ell^{2}$ but $sum frac {x_n} {sqrt {n+1} }=infty$. [Use the fact that $sum frac 1 {n (ln, n)^{2}} <infty$ but $sum frac 1 {n (ln, n)}=infty$].






                share|cite|improve this answer









                $endgroup$



                This is not true. Let $x_n=frac 1 {sqrt n ln , n}$ for $n >1$. Then ${x_n} in ell^{2}$ but $sum frac {x_n} {sqrt {n+1} }=infty$. [Use the fact that $sum frac 1 {n (ln, n)^{2}} <infty$ but $sum frac 1 {n (ln, n)}=infty$].







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 21 at 23:24









                Kavi Rama MurthyKavi Rama Murthy

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                62.7k42262






























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