Mixing
Limit of sequence $x_n = x_{n-1} cdot(2-frac{tan^{-1}(n) cdot n^2 - 3}{3 + n^2})$
$begingroup$
Find limit of a sequence:
$$x_n = left(2-frac{tan^{-1}(1) cdot 1^2 - 3}{3 + 1^2}right)left(2-frac{tan^{-1}(2) cdot 2^2 - 3}{3 + 2^2}right) cdot ldots cdot left(2-frac{tan^{-1}(n) cdot n^2 - 3}{3 + n^2}right)$$
Recursively:
$x_n = x_{n-1} cdot left(2-frac{tan^{-1}(n) cdot n^2 - 3}{3 + n^2}right)$.
I think that for big enough $n$, expression in the parentheses becomes smaller than $1$, so $(x_n)$ should be decreasing.
And after I move on to limit in recurrence formula, I get that limit is $0$. Is this right ?
real-analysis sequences-and-series limits
edited Jan 29 at 14:36
Viktor Glombik
1,3062628
$endgroup$
add a comment |
$begingroup$
Find limit of a sequence:
$$x_n = left(2-frac{tan^{-1}(1) cdot 1^2 - 3}{3 + 1^2}right)left(2-frac{tan^{-1}(2) cdot 2^2 - 3}{3 + 2^2}right) cdot ldots cdot left(2-frac{tan^{-1}(n) cdot n^2 - 3}{3 + n^2}right)$$
Recursively:
$x_n = x_{n-1} cdot left(2-frac{tan^{-1}(n) cdot n^2 - 3}{3 + n^2}right)$.
I think that for big enough $n$, expression in the parentheses becomes smaller than $1$, so $(x_n)$ should be decreasing.
And after I move on to limit in recurrence formula, I get that limit is $0$. Is this right ?
real-analysis sequences-and-series limits
edited Jan 29 at 14:36
Viktor Glombik
1,3062628
$endgroup$
1
$begingroup$
That the factors are eventually smaller than $1$ is not enough and only shows that $x_n$ is decreasing. However, the limit of the factors is $2-frac pi2$, so $<1$, and that implies $x_nto 0$.
$endgroup$
– Hagen von Eitzen
Jan 29 at 14:33
$begingroup$
Yes, but from the fact that $x_n$ is decreasing and bounded below, by zero, it follows that it converges ? And if limit of $x_n$ is $a$, from $a = a timesleft(2-frac{arctg(n) times n^2 - 3}{3 + n^2}right)$, it follows that $a$ is zero. Is this reasoning ok ?
$endgroup$
– user639107
Jan 29 at 14:38
add a comment |
$begingroup$
Find limit of a sequence:
$$x_n = left(2-frac{tan^{-1}(1) cdot 1^2 - 3}{3 + 1^2}right)left(2-frac{tan^{-1}(2) cdot 2^2 - 3}{3 + 2^2}right) cdot ldots cdot left(2-frac{tan^{-1}(n) cdot n^2 - 3}{3 + n^2}right)$$
Recursively:
$x_n = x_{n-1} cdot left(2-frac{tan^{-1}(n) cdot n^2 - 3}{3 + n^2}right)$.
I think that for big enough $n$, expression in the parentheses becomes smaller than $1$, so $(x_n)$ should be decreasing.
And after I move on to limit in recurrence formula, I get that limit is $0$. Is this right ?
real-analysis sequences-and-series limits
edited Jan 29 at 14:36
Viktor Glombik
1,3062628
$endgroup$
Find limit of a sequence:
$$x_n = left(2-frac{tan^{-1}(1) cdot 1^2 - 3}{3 + 1^2}right)left(2-frac{tan^{-1}(2) cdot 2^2 - 3}{3 + 2^2}right) cdot ldots cdot left(2-frac{tan^{-1}(n) cdot n^2 - 3}{3 + n^2}right)$$
Recursively:
$x_n = x_{n-1} cdot left(2-frac{tan^{-1}(n) cdot n^2 - 3}{3 + n^2}right)$.
I think that for big enough $n$, expression in the parentheses becomes smaller than $1$, so $(x_n)$ should be decreasing.
And after I move on to limit in recurrence formula, I get that limit is $0$. Is this right ?
real-analysis sequences-and-series limits
real-analysis sequences-and-series limits
edited Jan 29 at 14:36
Viktor Glombik
1,3062628
edited Jan 29 at 14:36
Viktor Glombik
1,3062628
edited Jan 29 at 14:36
Viktor Glombik
1,3062628
edited Jan 29 at 14:36
Viktor Glombik
1,3062628
edited Jan 29 at 14:36
Viktor Glombik
1,3062628
1,3062628
asked Jan 29 at 14:24
user639107
1
$begingroup$
That the factors are eventually smaller than $1$ is not enough and only shows that $x_n$ is decreasing. However, the limit of the factors is $2-frac pi2$, so $<1$, and that implies $x_nto 0$.
$endgroup$
– Hagen von Eitzen
Jan 29 at 14:33
$begingroup$
Yes, but from the fact that $x_n$ is decreasing and bounded below, by zero, it follows that it converges ? And if limit of $x_n$ is $a$, from $a = a timesleft(2-frac{arctg(n) times n^2 - 3}{3 + n^2}right)$, it follows that $a$ is zero. Is this reasoning ok ?
$endgroup$
– user639107
Jan 29 at 14:38
add a comment |
1
$begingroup$
That the factors are eventually smaller than $1$ is not enough and only shows that $x_n$ is decreasing. However, the limit of the factors is $2-frac pi2$, so $<1$, and that implies $x_nto 0$.
$endgroup$
– Hagen von Eitzen
Jan 29 at 14:33
$begingroup$
Yes, but from the fact that $x_n$ is decreasing and bounded below, by zero, it follows that it converges ? And if limit of $x_n$ is $a$, from $a = a timesleft(2-frac{arctg(n) times n^2 - 3}{3 + n^2}right)$, it follows that $a$ is zero. Is this reasoning ok ?
$endgroup$
– user639107
Jan 29 at 14:38
1
1
$begingroup$
That the factors are eventually smaller than $1$ is not enough and only shows that $x_n$ is decreasing. However, the limit of the factors is $2-frac pi2$, so $<1$, and that implies $x_nto 0$.
$endgroup$
– Hagen von Eitzen
Jan 29 at 14:33
$begingroup$
That the factors are eventually smaller than $1$ is not enough and only shows that $x_n$ is decreasing. However, the limit of the factors is $2-frac pi2$, so $<1$, and that implies $x_nto 0$.
$endgroup$
– Hagen von Eitzen
Jan 29 at 14:33
$begingroup$
Yes, but from the fact that $x_n$ is decreasing and bounded below, by zero, it follows that it converges ? And if limit of $x_n$ is $a$, from $a = a timesleft(2-frac{arctg(n) times n^2 - 3}{3 + n^2}right)$, it follows that $a$ is zero. Is this reasoning ok ?
$endgroup$
– user639107
Jan 29 at 14:38
$begingroup$
Yes, but from the fact that $x_n$ is decreasing and bounded below, by zero, it follows that it converges ? And if limit of $x_n$ is $a$, from $a = a timesleft(2-frac{arctg(n) times n^2 - 3}{3 + n^2}right)$, it follows that $a$ is zero. Is this reasoning ok ?
$endgroup$
– user639107
Jan 29 at 14:38
add a comment |
1 Answer
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$begingroup$
For $n$ large enough $arctan(n)in (pi/2-0.1,pi/2)$, let $d=pi/2-0.1$, then $arctan(n)>d$ for $n$ large enough. Some algebra gives
begin{equation}begin{aligned}
2-frac{arctan(n) n^2-3}{3+n^2}&leqslant frac{2-d}{1+frac{3}{n^2}}\
&leqslant 2-d+0.1text{ for }n text{ large enough}
end{aligned}end{equation}
This means that for $n$ large enough we are multiplying by a number less than 0.63, and hence will give zero.
answered Jan 29 at 14:34
Alec B-GAlec B-G
52019
$endgroup$
add a comment |
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$begingroup$
For $n$ large enough $arctan(n)in (pi/2-0.1,pi/2)$, let $d=pi/2-0.1$, then $arctan(n)>d$ for $n$ large enough. Some algebra gives
begin{equation}begin{aligned}
2-frac{arctan(n) n^2-3}{3+n^2}&leqslant frac{2-d}{1+frac{3}{n^2}}\
&leqslant 2-d+0.1text{ for }n text{ large enough}
end{aligned}end{equation}
This means that for $n$ large enough we are multiplying by a number less than 0.63, and hence will give zero.
answered Jan 29 at 14:34
Alec B-GAlec B-G
52019
$endgroup$
add a comment |
$begingroup$
For $n$ large enough $arctan(n)in (pi/2-0.1,pi/2)$, let $d=pi/2-0.1$, then $arctan(n)>d$ for $n$ large enough. Some algebra gives
begin{equation}begin{aligned}
2-frac{arctan(n) n^2-3}{3+n^2}&leqslant frac{2-d}{1+frac{3}{n^2}}\
&leqslant 2-d+0.1text{ for }n text{ large enough}
end{aligned}end{equation}
This means that for $n$ large enough we are multiplying by a number less than 0.63, and hence will give zero.
answered Jan 29 at 14:34
Alec B-GAlec B-G
52019
$endgroup$
add a comment |
$begingroup$
For $n$ large enough $arctan(n)in (pi/2-0.1,pi/2)$, let $d=pi/2-0.1$, then $arctan(n)>d$ for $n$ large enough. Some algebra gives
begin{equation}begin{aligned}
2-frac{arctan(n) n^2-3}{3+n^2}&leqslant frac{2-d}{1+frac{3}{n^2}}\
&leqslant 2-d+0.1text{ for }n text{ large enough}
end{aligned}end{equation}
This means that for $n$ large enough we are multiplying by a number less than 0.63, and hence will give zero.
answered Jan 29 at 14:34
Alec B-GAlec B-G
52019
$endgroup$
For $n$ large enough $arctan(n)in (pi/2-0.1,pi/2)$, let $d=pi/2-0.1$, then $arctan(n)>d$ for $n$ large enough. Some algebra gives
begin{equation}begin{aligned}
2-frac{arctan(n) n^2-3}{3+n^2}&leqslant frac{2-d}{1+frac{3}{n^2}}\
&leqslant 2-d+0.1text{ for }n text{ large enough}
end{aligned}end{equation}
This means that for $n$ large enough we are multiplying by a number less than 0.63, and hence will give zero.
answered Jan 29 at 14:34
Alec B-GAlec B-G
52019
answered Jan 29 at 14:34
Alec B-GAlec B-G
52019
answered Jan 29 at 14:34
Alec B-GAlec B-G
52019
answered Jan 29 at 14:34
Alec B-GAlec B-G
52019
52019
add a comment |
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$begingroup$
That the factors are eventually smaller than $1$ is not enough and only shows that $x_n$ is decreasing. However, the limit of the factors is $2-frac pi2$, so $<1$, and that implies $x_nto 0$.
$endgroup$
– Hagen von Eitzen
Jan 29 at 14:33
$begingroup$
Yes, but from the fact that $x_n$ is decreasing and bounded below, by zero, it follows that it converges ? And if limit of $x_n$ is $a$, from $a = a timesleft(2-frac{arctg(n) times n^2 - 3}{3 + n^2}right)$, it follows that $a$ is zero. Is this reasoning ok ?
$endgroup$
– user639107
Jan 29 at 14:38