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!
sequences-and-series convergence
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add a comment |
$begingroup$
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!
sequences-and-series convergence
$endgroup$
add a comment |
$begingroup$
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!
sequences-and-series convergence
$endgroup$
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
sequences-and-series convergence
asked Jan 21 at 23:16
B.SwanB.Swan
1,0541720
1,0541720
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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
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1 Answer
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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 $sum frac 1 {n (ln, n)}=infty$].
$endgroup$
add a comment |
$begingroup$
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 $sum frac 1 {n (ln, n)}=infty$].
$endgroup$
add a comment |
$begingroup$
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 $sum frac 1 {n (ln, n)}=infty$].
$endgroup$
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 $sum frac 1 {n (ln, n)}=infty$].
answered Jan 21 at 23:24
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
62.7k42262
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