Mixing
What is $lim v_n$ if $v_1=1$ and $v_{n+1}=sqrt{v_n^2+left(frac{1}{5}right)^n}$ for $nge 1$?
$begingroup$
Let ${v_n}$ be a sequence defined by $v_1=1$ and $v_{n+1}=sqrt{v_{n}^2 + left(frac{1}{5}right)^n}$ for $nge 1$. Then find $limlimits_{n rightarrow infty} v_n$.
I take $v_n = v_{n+1} = l$, now i have $l^2 = l^2 + 0$ that is $l = 1$ so $limlimits_{n rightarrow infty} v_n= 1$
Is its correct ? Any hints/solution
real-analysis
edited Jan 6 at 17:18
rtybase
10.7k21533
asked Jan 6 at 14:51
Messi fifaMessi fifa
53111
$endgroup$
add a comment |
$begingroup$
Let ${v_n}$ be a sequence defined by $v_1=1$ and $v_{n+1}=sqrt{v_{n}^2 + left(frac{1}{5}right)^n}$ for $nge 1$. Then find $limlimits_{n rightarrow infty} v_n$.
I take $v_n = v_{n+1} = l$, now i have $l^2 = l^2 + 0$ that is $l = 1$ so $limlimits_{n rightarrow infty} v_n= 1$
Is its correct ? Any hints/solution
real-analysis
edited Jan 6 at 17:18
rtybase
10.7k21533
asked Jan 6 at 14:51
Messi fifaMessi fifa
53111
$endgroup$
1
$begingroup$
No, $l^2= l^2+ 0$ does NOT give "l= 1". It is true for all l.
$endgroup$
– user247327
Jan 6 at 15:09
add a comment |
$begingroup$
Let ${v_n}$ be a sequence defined by $v_1=1$ and $v_{n+1}=sqrt{v_{n}^2 + left(frac{1}{5}right)^n}$ for $nge 1$. Then find $limlimits_{n rightarrow infty} v_n$.
I take $v_n = v_{n+1} = l$, now i have $l^2 = l^2 + 0$ that is $l = 1$ so $limlimits_{n rightarrow infty} v_n= 1$
Is its correct ? Any hints/solution
real-analysis
edited Jan 6 at 17:18
rtybase
10.7k21533
asked Jan 6 at 14:51
Messi fifaMessi fifa
53111
$endgroup$
Let ${v_n}$ be a sequence defined by $v_1=1$ and $v_{n+1}=sqrt{v_{n}^2 + left(frac{1}{5}right)^n}$ for $nge 1$. Then find $limlimits_{n rightarrow infty} v_n$.
I take $v_n = v_{n+1} = l$, now i have $l^2 = l^2 + 0$ that is $l = 1$ so $limlimits_{n rightarrow infty} v_n= 1$
Is its correct ? Any hints/solution
real-analysis
real-analysis
edited Jan 6 at 17:18
rtybase
10.7k21533
asked Jan 6 at 14:51
Messi fifaMessi fifa
53111
edited Jan 6 at 17:18
rtybase
10.7k21533
asked Jan 6 at 14:51
Messi fifaMessi fifa
53111
edited Jan 6 at 17:18
rtybase
10.7k21533
edited Jan 6 at 17:18
rtybase
10.7k21533
edited Jan 6 at 17:18
rtybase
10.7k21533
10.7k21533
asked Jan 6 at 14:51
Messi fifaMessi fifa
53111
asked Jan 6 at 14:51
Messi fifaMessi fifa
53111
asked Jan 6 at 14:51
Messi fifaMessi fifa
53111
53111
1
$begingroup$
No, $l^2= l^2+ 0$ does NOT give "l= 1". It is true for all l.
$endgroup$
– user247327
Jan 6 at 15:09
add a comment |
1
$begingroup$
No, $l^2= l^2+ 0$ does NOT give "l= 1". It is true for all l.
$endgroup$
– user247327
Jan 6 at 15:09
1
1
$begingroup$
No, $l^2= l^2+ 0$ does NOT give "l= 1". It is true for all l.
$endgroup$
– user247327
Jan 6 at 15:09
$begingroup$
No, $l^2= l^2+ 0$ does NOT give "l= 1". It is true for all l.
$endgroup$
– user247327
Jan 6 at 15:09
add a comment |
1 Answer
1
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oldest
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$begingroup$
No, it is not correct. Note that $l^2=l^2$ is true for any number.
Set $x_n:=v_n^2$. From your recursion you get $x_{n+1}=x_n+(frac{1}{5})^n$ and $x_1=1$, and hence the explicit formula
$$
x_n=x_1+sum_{j=1}^{n-1}left(frac{1}{5}right)^j=sum_{j=0}^{n-1}left(frac{1}{5}right)^j.
$$
This geometric series converges to $frac{5}{4}$, and therefore, $(v_n)$ converges to
$$
frac{sqrt{5}}{2}.
$$
answered Jan 6 at 16:36
sranthropsranthrop
7,0741925
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add a comment |
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$begingroup$
No, it is not correct. Note that $l^2=l^2$ is true for any number.
Set $x_n:=v_n^2$. From your recursion you get $x_{n+1}=x_n+(frac{1}{5})^n$ and $x_1=1$, and hence the explicit formula
$$
x_n=x_1+sum_{j=1}^{n-1}left(frac{1}{5}right)^j=sum_{j=0}^{n-1}left(frac{1}{5}right)^j.
$$
This geometric series converges to $frac{5}{4}$, and therefore, $(v_n)$ converges to
$$
frac{sqrt{5}}{2}.
$$
answered Jan 6 at 16:36
sranthropsranthrop
7,0741925
$endgroup$
add a comment |
$begingroup$
No, it is not correct. Note that $l^2=l^2$ is true for any number.
Set $x_n:=v_n^2$. From your recursion you get $x_{n+1}=x_n+(frac{1}{5})^n$ and $x_1=1$, and hence the explicit formula
$$
x_n=x_1+sum_{j=1}^{n-1}left(frac{1}{5}right)^j=sum_{j=0}^{n-1}left(frac{1}{5}right)^j.
$$
This geometric series converges to $frac{5}{4}$, and therefore, $(v_n)$ converges to
$$
frac{sqrt{5}}{2}.
$$
answered Jan 6 at 16:36
sranthropsranthrop
7,0741925
$endgroup$
add a comment |
$begingroup$
No, it is not correct. Note that $l^2=l^2$ is true for any number.
Set $x_n:=v_n^2$. From your recursion you get $x_{n+1}=x_n+(frac{1}{5})^n$ and $x_1=1$, and hence the explicit formula
$$
x_n=x_1+sum_{j=1}^{n-1}left(frac{1}{5}right)^j=sum_{j=0}^{n-1}left(frac{1}{5}right)^j.
$$
This geometric series converges to $frac{5}{4}$, and therefore, $(v_n)$ converges to
$$
frac{sqrt{5}}{2}.
$$
answered Jan 6 at 16:36
sranthropsranthrop
7,0741925
$endgroup$
No, it is not correct. Note that $l^2=l^2$ is true for any number.
Set $x_n:=v_n^2$. From your recursion you get $x_{n+1}=x_n+(frac{1}{5})^n$ and $x_1=1$, and hence the explicit formula
$$
x_n=x_1+sum_{j=1}^{n-1}left(frac{1}{5}right)^j=sum_{j=0}^{n-1}left(frac{1}{5}right)^j.
$$
This geometric series converges to $frac{5}{4}$, and therefore, $(v_n)$ converges to
$$
frac{sqrt{5}}{2}.
$$
answered Jan 6 at 16:36
sranthropsranthrop
7,0741925
answered Jan 6 at 16:36
sranthropsranthrop
7,0741925
answered Jan 6 at 16:36
sranthropsranthrop
7,0741925
answered Jan 6 at 16:36
sranthropsranthrop
7,0741925
7,0741925
add a comment |
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$begingroup$
No, $l^2= l^2+ 0$ does NOT give "l= 1". It is true for all l.
$endgroup$
– user247327
Jan 6 at 15:09