How I can to prove the succession $x_n$ converge to 1?
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Let $0<x_0<1$ if $x_{n+1} = sin(x_n)$ show that $lim_{ntoinfty} frac{x_n}{sqrt{3}/n} = 1$
sequences-and-series numerical-calculus
asked Nov 18 at 4:26
Sergio MNZ
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up vote
-1
down vote
favorite
Let $0<x_0<1$ if $x_{n+1} = sin(x_n)$ show that $lim_{ntoinfty} frac{x_n}{sqrt{3}/n} = 1$
sequences-and-series numerical-calculus
asked Nov 18 at 4:26
Sergio MNZ
91
thanks, I already solved the problem with Stolz–Cesàro theorem. .
– Sergio MNZ
Nov 18 at 4:39
The first appearance of this question seems to be MSE question "Convergence of $sqrt{n}x_{n}$ where $x_{n+1} = sin(x_{n})$" from 2010.
– Somos
Nov 18 at 5:29
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Let $0<x_0<1$ if $x_{n+1} = sin(x_n)$ show that $lim_{ntoinfty} frac{x_n}{sqrt{3}/n} = 1$
sequences-and-series numerical-calculus
asked Nov 18 at 4:26
Sergio MNZ
91
Let $0<x_0<1$ if $x_{n+1} = sin(x_n)$ show that $lim_{ntoinfty} frac{x_n}{sqrt{3}/n} = 1$
sequences-and-series numerical-calculus
sequences-and-series numerical-calculus
asked Nov 18 at 4:26
Sergio MNZ
91
asked Nov 18 at 4:26
Sergio MNZ
91
edited Nov 18 at 4:32
asked Nov 18 at 4:26
Sergio MNZ
91
asked Nov 18 at 4:26
Sergio MNZ
91
asked Nov 18 at 4:26
Sergio MNZ
91
91
thanks, I already solved the problem with Stolz–Cesàro theorem. .
– Sergio MNZ
Nov 18 at 4:39
The first appearance of this question seems to be MSE question "Convergence of $sqrt{n}x_{n}$ where $x_{n+1} = sin(x_{n})$" from 2010.
– Somos
Nov 18 at 5:29
add a comment |
thanks, I already solved the problem with Stolz–Cesàro theorem. .
– Sergio MNZ
Nov 18 at 4:39
The first appearance of this question seems to be MSE question "Convergence of $sqrt{n}x_{n}$ where $x_{n+1} = sin(x_{n})$" from 2010.
– Somos
Nov 18 at 5:29
thanks, I already solved the problem with Stolz–Cesàro theorem. .
– Sergio MNZ
Nov 18 at 4:39
thanks, I already solved the problem with Stolz–Cesàro theorem. .
– Sergio MNZ
Nov 18 at 4:39
The first appearance of this question seems to be MSE question "Convergence of $sqrt{n}x_{n}$ where $x_{n+1} = sin(x_{n})$" from 2010.
– Somos
Nov 18 at 5:29
The first appearance of this question seems to be MSE question "Convergence of $sqrt{n}x_{n}$ where $x_{n+1} = sin(x_{n})$" from 2010.
– Somos
Nov 18 at 5:29
add a comment |
1 Answer
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We have that:
$$frac{x_n}{(frac {sqrt3}{n})}=frac{nx_n}{sqrt3}$$
Also, since for $0 <x <1$ we have that $x >sin x$, $$lim_{nto infty}{x_n}=0$$
You can do the rest.
answered Nov 18 at 4:43
Rhys Hughes
4,5851327
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
We have that:
$$frac{x_n}{(frac {sqrt3}{n})}=frac{nx_n}{sqrt3}$$
Also, since for $0 <x <1$ we have that $x >sin x$, $$lim_{nto infty}{x_n}=0$$
You can do the rest.
answered Nov 18 at 4:43
Rhys Hughes
4,5851327
add a comment |
up vote
0
down vote
We have that:
$$frac{x_n}{(frac {sqrt3}{n})}=frac{nx_n}{sqrt3}$$
Also, since for $0 <x <1$ we have that $x >sin x$, $$lim_{nto infty}{x_n}=0$$
You can do the rest.
answered Nov 18 at 4:43
Rhys Hughes
4,5851327
add a comment |
up vote
0
down vote
up vote
0
down vote
We have that:
$$frac{x_n}{(frac {sqrt3}{n})}=frac{nx_n}{sqrt3}$$
Also, since for $0 <x <1$ we have that $x >sin x$, $$lim_{nto infty}{x_n}=0$$
You can do the rest.
answered Nov 18 at 4:43
Rhys Hughes
4,5851327
We have that:
$$frac{x_n}{(frac {sqrt3}{n})}=frac{nx_n}{sqrt3}$$
Also, since for $0 <x <1$ we have that $x >sin x$, $$lim_{nto infty}{x_n}=0$$
You can do the rest.
answered Nov 18 at 4:43
Rhys Hughes
4,5851327
answered Nov 18 at 4:43
Rhys Hughes
4,5851327
answered Nov 18 at 4:43
Rhys Hughes
4,5851327
answered Nov 18 at 4:43
Rhys Hughes
4,5851327
4,5851327
add a comment |
add a comment |
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thanks, I already solved the problem with Stolz–Cesàro theorem. .
– Sergio MNZ
Nov 18 at 4:39
The first appearance of this question seems to be MSE question "Convergence of $sqrt{n}x_{n}$ where $x_{n+1} = sin(x_{n})$" from 2010.
– Somos
Nov 18 at 5:29