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How to prove that the sequence defined by $a_1=0$, $a_2=1$, $a_n=frac{a_{n-1}+a_{n-2}}{2}$ converges to...
How to prove that the sequence defined by $a_1=0$, $a_2=1$, $a_n=frac{a_{n-1}+a_{n-2}}{2}$ converges to $frac23$?
If we analyse terms:
$$
0,1,frac{1}{2},frac{3}{4},frac{5}{8},cdots.
$$
I'm asked to do this using a previously proved theorem which says that
if you have two sequences ${b_n}$ and ${c_n}$ converging both to the same limit $L$, then the sequence $a_n$ defined as $b_1,c_1,b_2,c_2,b_3,dots$ converge to $L$.
In this case, $b_n$ would be $0,cfrac{1}{2},cfrac{5}{8},cfrac{21}{32},dots$ and $c_n$ would be $1,cfrac{3}{4},cfrac{11}{16},cfrac{43}{64},dots$
From here it's seems all I need to do is prove that $b_n$ and $c_n$ converge to $cfrac{2}{3}$ and then, by the theorem, $a_n$ converges to $cfrac{2}{3}$.
I need to define them because I need to prove $b_n$ and $c_n$ are monotone and bounded, in order to use the monotone convergence theorem. I've got troubles when trying to define $b_n$ and $c_n$. Can anyone help me to define them?
calculus real-analysis sequences-and-series limits
edited Nov 21 '18 at 3:40
user587192
1,782215
asked Nov 21 '18 at 2:49
Tom Arbuckle
367
add a comment |
How to prove that the sequence defined by $a_1=0$, $a_2=1$, $a_n=frac{a_{n-1}+a_{n-2}}{2}$ converges to $frac23$?
If we analyse terms:
$$
0,1,frac{1}{2},frac{3}{4},frac{5}{8},cdots.
$$
I'm asked to do this using a previously proved theorem which says that
if you have two sequences ${b_n}$ and ${c_n}$ converging both to the same limit $L$, then the sequence $a_n$ defined as $b_1,c_1,b_2,c_2,b_3,dots$ converge to $L$.
In this case, $b_n$ would be $0,cfrac{1}{2},cfrac{5}{8},cfrac{21}{32},dots$ and $c_n$ would be $1,cfrac{3}{4},cfrac{11}{16},cfrac{43}{64},dots$
From here it's seems all I need to do is prove that $b_n$ and $c_n$ converge to $cfrac{2}{3}$ and then, by the theorem, $a_n$ converges to $cfrac{2}{3}$.
I need to define them because I need to prove $b_n$ and $c_n$ are monotone and bounded, in order to use the monotone convergence theorem. I've got troubles when trying to define $b_n$ and $c_n$. Can anyone help me to define them?
calculus real-analysis sequences-and-series limits
edited Nov 21 '18 at 3:40
user587192
1,782215
asked Nov 21 '18 at 2:49
Tom Arbuckle
367
it seems you only have one choice. try to define $a_n$ in terms of $b_n$ and $c_n$ from the theorem.
– The Count
Nov 21 '18 at 2:53
But i need to define $b_n$ and $c_n$ to prove each one is monotone and bounded, in order to use the monotone convergence theorem.
– Tom Arbuckle
Nov 21 '18 at 2:57
i mean write the equation for a general $a_n$.
– The Count
Nov 21 '18 at 2:58
If you read carefully, you will see that it´s a date they´re giving me. The title says it: $a_n=frac{a_{n-1}+a_{n+2}}{2}$
– Tom Arbuckle
Nov 21 '18 at 3:02
A similar question here
– rtybase
Nov 21 '18 at 10:32
add a comment |
How to prove that the sequence defined by $a_1=0$, $a_2=1$, $a_n=frac{a_{n-1}+a_{n-2}}{2}$ converges to $frac23$?
If we analyse terms:
$$
0,1,frac{1}{2},frac{3}{4},frac{5}{8},cdots.
$$
I'm asked to do this using a previously proved theorem which says that
if you have two sequences ${b_n}$ and ${c_n}$ converging both to the same limit $L$, then the sequence $a_n$ defined as $b_1,c_1,b_2,c_2,b_3,dots$ converge to $L$.
In this case, $b_n$ would be $0,cfrac{1}{2},cfrac{5}{8},cfrac{21}{32},dots$ and $c_n$ would be $1,cfrac{3}{4},cfrac{11}{16},cfrac{43}{64},dots$
From here it's seems all I need to do is prove that $b_n$ and $c_n$ converge to $cfrac{2}{3}$ and then, by the theorem, $a_n$ converges to $cfrac{2}{3}$.
I need to define them because I need to prove $b_n$ and $c_n$ are monotone and bounded, in order to use the monotone convergence theorem. I've got troubles when trying to define $b_n$ and $c_n$. Can anyone help me to define them?
calculus real-analysis sequences-and-series limits
edited Nov 21 '18 at 3:40
user587192
1,782215
asked Nov 21 '18 at 2:49
Tom Arbuckle
367
How to prove that the sequence defined by $a_1=0$, $a_2=1$, $a_n=frac{a_{n-1}+a_{n-2}}{2}$ converges to $frac23$?
If we analyse terms:
$$
0,1,frac{1}{2},frac{3}{4},frac{5}{8},cdots.
$$
I'm asked to do this using a previously proved theorem which says that
if you have two sequences ${b_n}$ and ${c_n}$ converging both to the same limit $L$, then the sequence $a_n$ defined as $b_1,c_1,b_2,c_2,b_3,dots$ converge to $L$.
In this case, $b_n$ would be $0,cfrac{1}{2},cfrac{5}{8},cfrac{21}{32},dots$ and $c_n$ would be $1,cfrac{3}{4},cfrac{11}{16},cfrac{43}{64},dots$
From here it's seems all I need to do is prove that $b_n$ and $c_n$ converge to $cfrac{2}{3}$ and then, by the theorem, $a_n$ converges to $cfrac{2}{3}$.
I need to define them because I need to prove $b_n$ and $c_n$ are monotone and bounded, in order to use the monotone convergence theorem. I've got troubles when trying to define $b_n$ and $c_n$. Can anyone help me to define them?
calculus real-analysis sequences-and-series limits
calculus real-analysis sequences-and-series limits
edited Nov 21 '18 at 3:40
user587192
1,782215
asked Nov 21 '18 at 2:49
Tom Arbuckle
367
edited Nov 21 '18 at 3:40
user587192
1,782215
asked Nov 21 '18 at 2:49
Tom Arbuckle
367
edited Nov 21 '18 at 3:40
user587192
1,782215
edited Nov 21 '18 at 3:40
user587192
1,782215
edited Nov 21 '18 at 3:40
user587192
1,782215
1,782215
asked Nov 21 '18 at 2:49
Tom Arbuckle
367
asked Nov 21 '18 at 2:49
Tom Arbuckle
367
asked Nov 21 '18 at 2:49
Tom Arbuckle
367
367
it seems you only have one choice. try to define $a_n$ in terms of $b_n$ and $c_n$ from the theorem.
– The Count
Nov 21 '18 at 2:53
But i need to define $b_n$ and $c_n$ to prove each one is monotone and bounded, in order to use the monotone convergence theorem.
– Tom Arbuckle
Nov 21 '18 at 2:57
i mean write the equation for a general $a_n$.
– The Count
Nov 21 '18 at 2:58
If you read carefully, you will see that it´s a date they´re giving me. The title says it: $a_n=frac{a_{n-1}+a_{n+2}}{2}$
– Tom Arbuckle
Nov 21 '18 at 3:02
A similar question here
– rtybase
Nov 21 '18 at 10:32
add a comment |
it seems you only have one choice. try to define $a_n$ in terms of $b_n$ and $c_n$ from the theorem.
– The Count
Nov 21 '18 at 2:53
But i need to define $b_n$ and $c_n$ to prove each one is monotone and bounded, in order to use the monotone convergence theorem.
– Tom Arbuckle
Nov 21 '18 at 2:57
i mean write the equation for a general $a_n$.
– The Count
Nov 21 '18 at 2:58
If you read carefully, you will see that it´s a date they´re giving me. The title says it: $a_n=frac{a_{n-1}+a_{n+2}}{2}$
– Tom Arbuckle
Nov 21 '18 at 3:02
A similar question here
– rtybase
Nov 21 '18 at 10:32
it seems you only have one choice. try to define $a_n$ in terms of $b_n$ and $c_n$ from the theorem.
– The Count
Nov 21 '18 at 2:53
it seems you only have one choice. try to define $a_n$ in terms of $b_n$ and $c_n$ from the theorem.
– The Count
Nov 21 '18 at 2:53
But i need to define $b_n$ and $c_n$ to prove each one is monotone and bounded, in order to use the monotone convergence theorem.
– Tom Arbuckle
Nov 21 '18 at 2:57
But i need to define $b_n$ and $c_n$ to prove each one is monotone and bounded, in order to use the monotone convergence theorem.
– Tom Arbuckle
Nov 21 '18 at 2:57
i mean write the equation for a general $a_n$.
– The Count
Nov 21 '18 at 2:58
i mean write the equation for a general $a_n$.
– The Count
Nov 21 '18 at 2:58
If you read carefully, you will see that it´s a date they´re giving me. The title says it: $a_n=frac{a_{n-1}+a_{n+2}}{2}$
– Tom Arbuckle
Nov 21 '18 at 3:02
If you read carefully, you will see that it´s a date they´re giving me. The title says it: $a_n=frac{a_{n-1}+a_{n+2}}{2}$
– Tom Arbuckle
Nov 21 '18 at 3:02
A similar question here
– rtybase
Nov 21 '18 at 10:32
A similar question here
– rtybase
Nov 21 '18 at 10:32
add a comment |
5 Answers
5
active
oldest
votes
Here is an alternative (maybe easier) way.
By definition of the sequence one has
$$
begin{align}
a_1+a_2&=2a_3\
a_2+a_3&=2a_4\
&vdots\
a_{n-1}+a_{n}&=2a_{n+1}
end{align}
$$
Adding together these identities one gets
$$
a_{n+1}=1-frac12 a_{n},quad ngeqslant 3.tag{1}
$$
Let $b_n=a_n-frac23$. Then (1) implies that
$$
b_{n+1}=-frac12 b_ntag{2}
$$
[Note: One can easily find the form of $b_n$ by setting $a_{n+1}+b=-frac12(a_n+b)$.]
Now one only needs to show that
$$
lim_{ntoinfty}b_n=0.
$$
But (2) gives:
$$
b_{n}=q^{n-3}b_3,quad ngeqslant 3tag{3}
$$
where $|q|=frac12$.
Your $(b_n)$ and $(c_n)$ can be read from (3) if you want.
[Added later.] Yet, there is another way to do this problem using linear algebra. Noticing that
$$
a_{n+1}=frac{1}{2}a_n+frac12a_{n-1},quad
a_{n+2}=frac34a_n+frac14a_{n-1},quad nge 1,
$$
one can write $b_{n+2}=Ab_n$ where $b_n=(a_{n-1},a_n)^T$ and
$$
A=frac14begin{pmatrix}
2&2\
1&3
end{pmatrix}
=SJS^{-1}
$$
where
$$
J=begin{pmatrix}
frac14&0\
0&1
end{pmatrix},quad
S=begin{pmatrix}
-2&1\
1&1
end{pmatrix},quad
S^{-1}=frac13begin{pmatrix}
-1&1\
1&2
end{pmatrix}.
$$
Now,
$$
b_{2n+1}=A^{n}b_1,quad nge1.tag{4}
$$
But as $ntoinfty$,
$$
A^{n}=SJ^{n}S^{-1}to S
begin{pmatrix}
0&0\0&1
end{pmatrix}S^{-1}=
frac13begin{pmatrix}
1&2\1&2
end{pmatrix}.tag{5}
$$
Combining (4) and (5) one gets
$$
(a_{2n},a_{2n+1})to (frac23,frac23)quadtext{as }ntoinfty.
$$
Now you can apply the theorem you have to conclude that
$$
lim_{ntoinfty}a_n=frac23.
$$
answered Nov 21 '18 at 3:21
user587192
1,782215
Excuse me, i don't understand how do you get (1) from $a_{n-1}+a_n=2a_{n+1}$. Could you explain me more that step?
– Tom Arbuckle
Nov 22 '18 at 1:22
@TomArbuckle: Adding up all the identities above (1), one has $$ (a_1+cdots+a_{n-1})+(a_2+cdots+a_n)=2(a_3+cdots+a_{n+1}), $$ which implies that $$ a_1+2a_2+color{red}{(2a_3+cdots+2a_{n-1})}+a_n =color{red}{2(a_3+cdots+a_{n-1})}+2a_n+2a_{n+1}. $$ So $$ 0+2times 1=a_1+2a_2=a_n+a_{n+1}. $$
– user587192
Nov 22 '18 at 13:24
add a comment |
Note that $0,1,5,21,85,...$ is the sequence $frac{4^n -1}{3}$ for $n geq 0$. Also, $*, 2, 8,32,128,dots = 2^{2n-1}$. Note that the missing term $*$ will fit any pattern, so we need only worry about larger values. These observations show the first sequence converges to $2/3$.
The other sequence is harder to spot the pattern: $1,3,11,43,dots = frac{2^{2n+1}+1}{3}$. You should have no problem finding a pattern for the denominators.
answered Nov 21 '18 at 3:20
JavaMan
11k12655
add a comment |
Rewriting the recursion you obtain
- $a_n=frac{a_{n-1}+a_{n-2}}{2} Leftrightarrow a_n - frac{1}{2}a_{n-1}-frac{1}{2}a_{n-2}= 0$
This is a linear difference equation with the characteristic polynomial
$$x^2 -frac{1}{2}x-frac{1}{2} = 0 Leftrightarrow left(x + frac{1}{2} right)(x-1)= 0$$
So, the general solution is
$$acdot 1^n + b cdot left(-frac{1}{2} right)^n = a+bleft(-frac{1}{2} right)^n stackrel{a_1 =0, a_2 = 1}{Longrightarrow}frac{2}{3} + frac{4}{3}left(-frac{1}{2} right)^n stackrel{n to infty}{longrightarrow}frac{2}{3}$$
answered Nov 21 '18 at 8:58
trancelocation
9,1151521
add a comment |
Guide: Define
$$b_n := frac{b_{n-1}+c_{n-1}}{2} qquad c_n := frac{b_{n}+c_{n-1}}{2}$$
with $b_1 := a_1$ and $c_1 := a_2$ for all $n in Bbb{N}$. It's not hard to see that
$$b_n = a_{2n-1} qquad c_n = a_{2n}$$
by induction.
answered Nov 21 '18 at 3:02
GNUSupporter 8964民主女神 地下教會
12.8k72445
Could you help me and guide me to show that, for example, $b_n$ is monotone increasing?
– Tom Arbuckle
Nov 21 '18 at 3:10
@TomArbuckle That's again a straightforward exercise of induction. Since this site in called Mathematics Stack Exchange, it's an exchange of ideas, rather than a do-my-HW site.
– GNUSupporter 8964民主女神 地下教會
Nov 21 '18 at 3:17
add a comment |
We have that $$2a_n=a_{n-1}+a_{n-2}$$therefore $$2(a_n-a_{n-1})=a_{n-2}-a_{n-1}$$defining $b_n= a_n-a_{n-1}$ we obtain$$2b_n=-{b_{n-1}}$$since $b_2=a_2-a_1=1$ we have $$b_n=4(-{1over 2})^n$$which yields to $$a_n=a_1+sum_{k=2}^{n}b_k=sum_{k=2}^{n}b_k$$and we can write $$lim_{nto infty}a_n=sum_{k=2}^{infty}b_k=b_2+b_3+b_4+cdots= {1over 1-left(-{1over 2}right)}={2over 3}$$
answered Nov 30 '18 at 15:49
Mostafa Ayaz
13.7k3936
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
Here is an alternative (maybe easier) way.
By definition of the sequence one has
$$
begin{align}
a_1+a_2&=2a_3\
a_2+a_3&=2a_4\
&vdots\
a_{n-1}+a_{n}&=2a_{n+1}
end{align}
$$
Adding together these identities one gets
$$
a_{n+1}=1-frac12 a_{n},quad ngeqslant 3.tag{1}
$$
Let $b_n=a_n-frac23$. Then (1) implies that
$$
b_{n+1}=-frac12 b_ntag{2}
$$
[Note: One can easily find the form of $b_n$ by setting $a_{n+1}+b=-frac12(a_n+b)$.]
Now one only needs to show that
$$
lim_{ntoinfty}b_n=0.
$$
But (2) gives:
$$
b_{n}=q^{n-3}b_3,quad ngeqslant 3tag{3}
$$
where $|q|=frac12$.
Your $(b_n)$ and $(c_n)$ can be read from (3) if you want.
[Added later.] Yet, there is another way to do this problem using linear algebra. Noticing that
$$
a_{n+1}=frac{1}{2}a_n+frac12a_{n-1},quad
a_{n+2}=frac34a_n+frac14a_{n-1},quad nge 1,
$$
one can write $b_{n+2}=Ab_n$ where $b_n=(a_{n-1},a_n)^T$ and
$$
A=frac14begin{pmatrix}
2&2\
1&3
end{pmatrix}
=SJS^{-1}
$$
where
$$
J=begin{pmatrix}
frac14&0\
0&1
end{pmatrix},quad
S=begin{pmatrix}
-2&1\
1&1
end{pmatrix},quad
S^{-1}=frac13begin{pmatrix}
-1&1\
1&2
end{pmatrix}.
$$
Now,
$$
b_{2n+1}=A^{n}b_1,quad nge1.tag{4}
$$
But as $ntoinfty$,
$$
A^{n}=SJ^{n}S^{-1}to S
begin{pmatrix}
0&0\0&1
end{pmatrix}S^{-1}=
frac13begin{pmatrix}
1&2\1&2
end{pmatrix}.tag{5}
$$
Combining (4) and (5) one gets
$$
(a_{2n},a_{2n+1})to (frac23,frac23)quadtext{as }ntoinfty.
$$
Now you can apply the theorem you have to conclude that
$$
lim_{ntoinfty}a_n=frac23.
$$
answered Nov 21 '18 at 3:21
user587192
1,782215
Excuse me, i don't understand how do you get (1) from $a_{n-1}+a_n=2a_{n+1}$. Could you explain me more that step?
– Tom Arbuckle
Nov 22 '18 at 1:22
@TomArbuckle: Adding up all the identities above (1), one has $$ (a_1+cdots+a_{n-1})+(a_2+cdots+a_n)=2(a_3+cdots+a_{n+1}), $$ which implies that $$ a_1+2a_2+color{red}{(2a_3+cdots+2a_{n-1})}+a_n =color{red}{2(a_3+cdots+a_{n-1})}+2a_n+2a_{n+1}. $$ So $$ 0+2times 1=a_1+2a_2=a_n+a_{n+1}. $$
– user587192
Nov 22 '18 at 13:24
add a comment |
Here is an alternative (maybe easier) way.
By definition of the sequence one has
$$
begin{align}
a_1+a_2&=2a_3\
a_2+a_3&=2a_4\
&vdots\
a_{n-1}+a_{n}&=2a_{n+1}
end{align}
$$
Adding together these identities one gets
$$
a_{n+1}=1-frac12 a_{n},quad ngeqslant 3.tag{1}
$$
Let $b_n=a_n-frac23$. Then (1) implies that
$$
b_{n+1}=-frac12 b_ntag{2}
$$
[Note: One can easily find the form of $b_n$ by setting $a_{n+1}+b=-frac12(a_n+b)$.]
Now one only needs to show that
$$
lim_{ntoinfty}b_n=0.
$$
But (2) gives:
$$
b_{n}=q^{n-3}b_3,quad ngeqslant 3tag{3}
$$
where $|q|=frac12$.
Your $(b_n)$ and $(c_n)$ can be read from (3) if you want.
[Added later.] Yet, there is another way to do this problem using linear algebra. Noticing that
$$
a_{n+1}=frac{1}{2}a_n+frac12a_{n-1},quad
a_{n+2}=frac34a_n+frac14a_{n-1},quad nge 1,
$$
one can write $b_{n+2}=Ab_n$ where $b_n=(a_{n-1},a_n)^T$ and
$$
A=frac14begin{pmatrix}
2&2\
1&3
end{pmatrix}
=SJS^{-1}
$$
where
$$
J=begin{pmatrix}
frac14&0\
0&1
end{pmatrix},quad
S=begin{pmatrix}
-2&1\
1&1
end{pmatrix},quad
S^{-1}=frac13begin{pmatrix}
-1&1\
1&2
end{pmatrix}.
$$
Now,
$$
b_{2n+1}=A^{n}b_1,quad nge1.tag{4}
$$
But as $ntoinfty$,
$$
A^{n}=SJ^{n}S^{-1}to S
begin{pmatrix}
0&0\0&1
end{pmatrix}S^{-1}=
frac13begin{pmatrix}
1&2\1&2
end{pmatrix}.tag{5}
$$
Combining (4) and (5) one gets
$$
(a_{2n},a_{2n+1})to (frac23,frac23)quadtext{as }ntoinfty.
$$
Now you can apply the theorem you have to conclude that
$$
lim_{ntoinfty}a_n=frac23.
$$
answered Nov 21 '18 at 3:21
user587192
1,782215
Excuse me, i don't understand how do you get (1) from $a_{n-1}+a_n=2a_{n+1}$. Could you explain me more that step?
– Tom Arbuckle
Nov 22 '18 at 1:22
@TomArbuckle: Adding up all the identities above (1), one has $$ (a_1+cdots+a_{n-1})+(a_2+cdots+a_n)=2(a_3+cdots+a_{n+1}), $$ which implies that $$ a_1+2a_2+color{red}{(2a_3+cdots+2a_{n-1})}+a_n =color{red}{2(a_3+cdots+a_{n-1})}+2a_n+2a_{n+1}. $$ So $$ 0+2times 1=a_1+2a_2=a_n+a_{n+1}. $$
– user587192
Nov 22 '18 at 13:24
add a comment |
Here is an alternative (maybe easier) way.
By definition of the sequence one has
$$
begin{align}
a_1+a_2&=2a_3\
a_2+a_3&=2a_4\
&vdots\
a_{n-1}+a_{n}&=2a_{n+1}
end{align}
$$
Adding together these identities one gets
$$
a_{n+1}=1-frac12 a_{n},quad ngeqslant 3.tag{1}
$$
Let $b_n=a_n-frac23$. Then (1) implies that
$$
b_{n+1}=-frac12 b_ntag{2}
$$
[Note: One can easily find the form of $b_n$ by setting $a_{n+1}+b=-frac12(a_n+b)$.]
Now one only needs to show that
$$
lim_{ntoinfty}b_n=0.
$$
But (2) gives:
$$
b_{n}=q^{n-3}b_3,quad ngeqslant 3tag{3}
$$
where $|q|=frac12$.
Your $(b_n)$ and $(c_n)$ can be read from (3) if you want.
[Added later.] Yet, there is another way to do this problem using linear algebra. Noticing that
$$
a_{n+1}=frac{1}{2}a_n+frac12a_{n-1},quad
a_{n+2}=frac34a_n+frac14a_{n-1},quad nge 1,
$$
one can write $b_{n+2}=Ab_n$ where $b_n=(a_{n-1},a_n)^T$ and
$$
A=frac14begin{pmatrix}
2&2\
1&3
end{pmatrix}
=SJS^{-1}
$$
where
$$
J=begin{pmatrix}
frac14&0\
0&1
end{pmatrix},quad
S=begin{pmatrix}
-2&1\
1&1
end{pmatrix},quad
S^{-1}=frac13begin{pmatrix}
-1&1\
1&2
end{pmatrix}.
$$
Now,
$$
b_{2n+1}=A^{n}b_1,quad nge1.tag{4}
$$
But as $ntoinfty$,
$$
A^{n}=SJ^{n}S^{-1}to S
begin{pmatrix}
0&0\0&1
end{pmatrix}S^{-1}=
frac13begin{pmatrix}
1&2\1&2
end{pmatrix}.tag{5}
$$
Combining (4) and (5) one gets
$$
(a_{2n},a_{2n+1})to (frac23,frac23)quadtext{as }ntoinfty.
$$
Now you can apply the theorem you have to conclude that
$$
lim_{ntoinfty}a_n=frac23.
$$
answered Nov 21 '18 at 3:21
user587192
1,782215
Here is an alternative (maybe easier) way.
By definition of the sequence one has
$$
begin{align}
a_1+a_2&=2a_3\
a_2+a_3&=2a_4\
&vdots\
a_{n-1}+a_{n}&=2a_{n+1}
end{align}
$$
Adding together these identities one gets
$$
a_{n+1}=1-frac12 a_{n},quad ngeqslant 3.tag{1}
$$
Let $b_n=a_n-frac23$. Then (1) implies that
$$
b_{n+1}=-frac12 b_ntag{2}
$$
[Note: One can easily find the form of $b_n$ by setting $a_{n+1}+b=-frac12(a_n+b)$.]
Now one only needs to show that
$$
lim_{ntoinfty}b_n=0.
$$
But (2) gives:
$$
b_{n}=q^{n-3}b_3,quad ngeqslant 3tag{3}
$$
where $|q|=frac12$.
Your $(b_n)$ and $(c_n)$ can be read from (3) if you want.
[Added later.] Yet, there is another way to do this problem using linear algebra. Noticing that
$$
a_{n+1}=frac{1}{2}a_n+frac12a_{n-1},quad
a_{n+2}=frac34a_n+frac14a_{n-1},quad nge 1,
$$
one can write $b_{n+2}=Ab_n$ where $b_n=(a_{n-1},a_n)^T$ and
$$
A=frac14begin{pmatrix}
2&2\
1&3
end{pmatrix}
=SJS^{-1}
$$
where
$$
J=begin{pmatrix}
frac14&0\
0&1
end{pmatrix},quad
S=begin{pmatrix}
-2&1\
1&1
end{pmatrix},quad
S^{-1}=frac13begin{pmatrix}
-1&1\
1&2
end{pmatrix}.
$$
Now,
$$
b_{2n+1}=A^{n}b_1,quad nge1.tag{4}
$$
But as $ntoinfty$,
$$
A^{n}=SJ^{n}S^{-1}to S
begin{pmatrix}
0&0\0&1
end{pmatrix}S^{-1}=
frac13begin{pmatrix}
1&2\1&2
end{pmatrix}.tag{5}
$$
Combining (4) and (5) one gets
$$
(a_{2n},a_{2n+1})to (frac23,frac23)quadtext{as }ntoinfty.
$$
Now you can apply the theorem you have to conclude that
$$
lim_{ntoinfty}a_n=frac23.
$$
answered Nov 21 '18 at 3:21
user587192
1,782215
edited Nov 21 '18 at 16:13
answered Nov 21 '18 at 3:21
user587192
1,782215
answered Nov 21 '18 at 3:21
user587192
1,782215
answered Nov 21 '18 at 3:21
user587192
1,782215
1,782215
Excuse me, i don't understand how do you get (1) from $a_{n-1}+a_n=2a_{n+1}$. Could you explain me more that step?
– Tom Arbuckle
Nov 22 '18 at 1:22
@TomArbuckle: Adding up all the identities above (1), one has $$ (a_1+cdots+a_{n-1})+(a_2+cdots+a_n)=2(a_3+cdots+a_{n+1}), $$ which implies that $$ a_1+2a_2+color{red}{(2a_3+cdots+2a_{n-1})}+a_n =color{red}{2(a_3+cdots+a_{n-1})}+2a_n+2a_{n+1}. $$ So $$ 0+2times 1=a_1+2a_2=a_n+a_{n+1}. $$
– user587192
Nov 22 '18 at 13:24
add a comment |
Excuse me, i don't understand how do you get (1) from $a_{n-1}+a_n=2a_{n+1}$. Could you explain me more that step?
– Tom Arbuckle
Nov 22 '18 at 1:22
@TomArbuckle: Adding up all the identities above (1), one has $$ (a_1+cdots+a_{n-1})+(a_2+cdots+a_n)=2(a_3+cdots+a_{n+1}), $$ which implies that $$ a_1+2a_2+color{red}{(2a_3+cdots+2a_{n-1})}+a_n =color{red}{2(a_3+cdots+a_{n-1})}+2a_n+2a_{n+1}. $$ So $$ 0+2times 1=a_1+2a_2=a_n+a_{n+1}. $$
– user587192
Nov 22 '18 at 13:24
Excuse me, i don't understand how do you get (1) from $a_{n-1}+a_n=2a_{n+1}$. Could you explain me more that step?
– Tom Arbuckle
Nov 22 '18 at 1:22
Excuse me, i don't understand how do you get (1) from $a_{n-1}+a_n=2a_{n+1}$. Could you explain me more that step?
– Tom Arbuckle
Nov 22 '18 at 1:22
@TomArbuckle: Adding up all the identities above (1), one has $$ (a_1+cdots+a_{n-1})+(a_2+cdots+a_n)=2(a_3+cdots+a_{n+1}), $$ which implies that $$ a_1+2a_2+color{red}{(2a_3+cdots+2a_{n-1})}+a_n =color{red}{2(a_3+cdots+a_{n-1})}+2a_n+2a_{n+1}. $$ So $$ 0+2times 1=a_1+2a_2=a_n+a_{n+1}. $$
– user587192
Nov 22 '18 at 13:24
@TomArbuckle: Adding up all the identities above (1), one has $$ (a_1+cdots+a_{n-1})+(a_2+cdots+a_n)=2(a_3+cdots+a_{n+1}), $$ which implies that $$ a_1+2a_2+color{red}{(2a_3+cdots+2a_{n-1})}+a_n =color{red}{2(a_3+cdots+a_{n-1})}+2a_n+2a_{n+1}. $$ So $$ 0+2times 1=a_1+2a_2=a_n+a_{n+1}. $$
– user587192
Nov 22 '18 at 13:24
add a comment |
Note that $0,1,5,21,85,...$ is the sequence $frac{4^n -1}{3}$ for $n geq 0$. Also, $*, 2, 8,32,128,dots = 2^{2n-1}$. Note that the missing term $*$ will fit any pattern, so we need only worry about larger values. These observations show the first sequence converges to $2/3$.
The other sequence is harder to spot the pattern: $1,3,11,43,dots = frac{2^{2n+1}+1}{3}$. You should have no problem finding a pattern for the denominators.
answered Nov 21 '18 at 3:20
JavaMan
11k12655
add a comment |
Note that $0,1,5,21,85,...$ is the sequence $frac{4^n -1}{3}$ for $n geq 0$. Also, $*, 2, 8,32,128,dots = 2^{2n-1}$. Note that the missing term $*$ will fit any pattern, so we need only worry about larger values. These observations show the first sequence converges to $2/3$.
The other sequence is harder to spot the pattern: $1,3,11,43,dots = frac{2^{2n+1}+1}{3}$. You should have no problem finding a pattern for the denominators.
answered Nov 21 '18 at 3:20
JavaMan
11k12655
add a comment |
Note that $0,1,5,21,85,...$ is the sequence $frac{4^n -1}{3}$ for $n geq 0$. Also, $*, 2, 8,32,128,dots = 2^{2n-1}$. Note that the missing term $*$ will fit any pattern, so we need only worry about larger values. These observations show the first sequence converges to $2/3$.
The other sequence is harder to spot the pattern: $1,3,11,43,dots = frac{2^{2n+1}+1}{3}$. You should have no problem finding a pattern for the denominators.
answered Nov 21 '18 at 3:20
JavaMan
11k12655
Note that $0,1,5,21,85,...$ is the sequence $frac{4^n -1}{3}$ for $n geq 0$. Also, $*, 2, 8,32,128,dots = 2^{2n-1}$. Note that the missing term $*$ will fit any pattern, so we need only worry about larger values. These observations show the first sequence converges to $2/3$.
The other sequence is harder to spot the pattern: $1,3,11,43,dots = frac{2^{2n+1}+1}{3}$. You should have no problem finding a pattern for the denominators.
answered Nov 21 '18 at 3:20
JavaMan
11k12655
answered Nov 21 '18 at 3:20
JavaMan
11k12655
answered Nov 21 '18 at 3:20
JavaMan
11k12655
answered Nov 21 '18 at 3:20
JavaMan
11k12655
11k12655
add a comment |
add a comment |
Rewriting the recursion you obtain
- $a_n=frac{a_{n-1}+a_{n-2}}{2} Leftrightarrow a_n - frac{1}{2}a_{n-1}-frac{1}{2}a_{n-2}= 0$
This is a linear difference equation with the characteristic polynomial
$$x^2 -frac{1}{2}x-frac{1}{2} = 0 Leftrightarrow left(x + frac{1}{2} right)(x-1)= 0$$
So, the general solution is
$$acdot 1^n + b cdot left(-frac{1}{2} right)^n = a+bleft(-frac{1}{2} right)^n stackrel{a_1 =0, a_2 = 1}{Longrightarrow}frac{2}{3} + frac{4}{3}left(-frac{1}{2} right)^n stackrel{n to infty}{longrightarrow}frac{2}{3}$$
answered Nov 21 '18 at 8:58
trancelocation
9,1151521
add a comment |
Rewriting the recursion you obtain
- $a_n=frac{a_{n-1}+a_{n-2}}{2} Leftrightarrow a_n - frac{1}{2}a_{n-1}-frac{1}{2}a_{n-2}= 0$
This is a linear difference equation with the characteristic polynomial
$$x^2 -frac{1}{2}x-frac{1}{2} = 0 Leftrightarrow left(x + frac{1}{2} right)(x-1)= 0$$
So, the general solution is
$$acdot 1^n + b cdot left(-frac{1}{2} right)^n = a+bleft(-frac{1}{2} right)^n stackrel{a_1 =0, a_2 = 1}{Longrightarrow}frac{2}{3} + frac{4}{3}left(-frac{1}{2} right)^n stackrel{n to infty}{longrightarrow}frac{2}{3}$$
answered Nov 21 '18 at 8:58
trancelocation
9,1151521
add a comment |
Rewriting the recursion you obtain
- $a_n=frac{a_{n-1}+a_{n-2}}{2} Leftrightarrow a_n - frac{1}{2}a_{n-1}-frac{1}{2}a_{n-2}= 0$
This is a linear difference equation with the characteristic polynomial
$$x^2 -frac{1}{2}x-frac{1}{2} = 0 Leftrightarrow left(x + frac{1}{2} right)(x-1)= 0$$
So, the general solution is
$$acdot 1^n + b cdot left(-frac{1}{2} right)^n = a+bleft(-frac{1}{2} right)^n stackrel{a_1 =0, a_2 = 1}{Longrightarrow}frac{2}{3} + frac{4}{3}left(-frac{1}{2} right)^n stackrel{n to infty}{longrightarrow}frac{2}{3}$$
answered Nov 21 '18 at 8:58
trancelocation
9,1151521
Rewriting the recursion you obtain
- $a_n=frac{a_{n-1}+a_{n-2}}{2} Leftrightarrow a_n - frac{1}{2}a_{n-1}-frac{1}{2}a_{n-2}= 0$
This is a linear difference equation with the characteristic polynomial
$$x^2 -frac{1}{2}x-frac{1}{2} = 0 Leftrightarrow left(x + frac{1}{2} right)(x-1)= 0$$
So, the general solution is
$$acdot 1^n + b cdot left(-frac{1}{2} right)^n = a+bleft(-frac{1}{2} right)^n stackrel{a_1 =0, a_2 = 1}{Longrightarrow}frac{2}{3} + frac{4}{3}left(-frac{1}{2} right)^n stackrel{n to infty}{longrightarrow}frac{2}{3}$$
answered Nov 21 '18 at 8:58
trancelocation
9,1151521
answered Nov 21 '18 at 8:58
trancelocation
9,1151521
answered Nov 21 '18 at 8:58
trancelocation
9,1151521
answered Nov 21 '18 at 8:58
trancelocation
9,1151521
9,1151521
add a comment |
add a comment |
Guide: Define
$$b_n := frac{b_{n-1}+c_{n-1}}{2} qquad c_n := frac{b_{n}+c_{n-1}}{2}$$
with $b_1 := a_1$ and $c_1 := a_2$ for all $n in Bbb{N}$. It's not hard to see that
$$b_n = a_{2n-1} qquad c_n = a_{2n}$$
by induction.
answered Nov 21 '18 at 3:02
GNUSupporter 8964民主女神 地下教會
12.8k72445
Could you help me and guide me to show that, for example, $b_n$ is monotone increasing?
– Tom Arbuckle
Nov 21 '18 at 3:10
@TomArbuckle That's again a straightforward exercise of induction. Since this site in called Mathematics Stack Exchange, it's an exchange of ideas, rather than a do-my-HW site.
– GNUSupporter 8964民主女神 地下教會
Nov 21 '18 at 3:17
add a comment |
Guide: Define
$$b_n := frac{b_{n-1}+c_{n-1}}{2} qquad c_n := frac{b_{n}+c_{n-1}}{2}$$
with $b_1 := a_1$ and $c_1 := a_2$ for all $n in Bbb{N}$. It's not hard to see that
$$b_n = a_{2n-1} qquad c_n = a_{2n}$$
by induction.
answered Nov 21 '18 at 3:02
GNUSupporter 8964民主女神 地下教會
12.8k72445
Could you help me and guide me to show that, for example, $b_n$ is monotone increasing?
– Tom Arbuckle
Nov 21 '18 at 3:10
@TomArbuckle That's again a straightforward exercise of induction. Since this site in called Mathematics Stack Exchange, it's an exchange of ideas, rather than a do-my-HW site.
– GNUSupporter 8964民主女神 地下教會
Nov 21 '18 at 3:17
add a comment |
Guide: Define
$$b_n := frac{b_{n-1}+c_{n-1}}{2} qquad c_n := frac{b_{n}+c_{n-1}}{2}$$
with $b_1 := a_1$ and $c_1 := a_2$ for all $n in Bbb{N}$. It's not hard to see that
$$b_n = a_{2n-1} qquad c_n = a_{2n}$$
by induction.
answered Nov 21 '18 at 3:02
GNUSupporter 8964民主女神 地下教會
12.8k72445
Guide: Define
$$b_n := frac{b_{n-1}+c_{n-1}}{2} qquad c_n := frac{b_{n}+c_{n-1}}{2}$$
with $b_1 := a_1$ and $c_1 := a_2$ for all $n in Bbb{N}$. It's not hard to see that
$$b_n = a_{2n-1} qquad c_n = a_{2n}$$
by induction.
answered Nov 21 '18 at 3:02
GNUSupporter 8964民主女神 地下教會
12.8k72445
edited Nov 21 '18 at 3:11
answered Nov 21 '18 at 3:02
GNUSupporter 8964民主女神 地下教會
12.8k72445
answered Nov 21 '18 at 3:02
GNUSupporter 8964民主女神 地下教會
12.8k72445
answered Nov 21 '18 at 3:02
GNUSupporter 8964民主女神 地下教會
12.8k72445
12.8k72445
Could you help me and guide me to show that, for example, $b_n$ is monotone increasing?
– Tom Arbuckle
Nov 21 '18 at 3:10
@TomArbuckle That's again a straightforward exercise of induction. Since this site in called Mathematics Stack Exchange, it's an exchange of ideas, rather than a do-my-HW site.
– GNUSupporter 8964民主女神 地下教會
Nov 21 '18 at 3:17
add a comment |
Could you help me and guide me to show that, for example, $b_n$ is monotone increasing?
– Tom Arbuckle
Nov 21 '18 at 3:10
@TomArbuckle That's again a straightforward exercise of induction. Since this site in called Mathematics Stack Exchange, it's an exchange of ideas, rather than a do-my-HW site.
– GNUSupporter 8964民主女神 地下教會
Nov 21 '18 at 3:17
Could you help me and guide me to show that, for example, $b_n$ is monotone increasing?
– Tom Arbuckle
Nov 21 '18 at 3:10
Could you help me and guide me to show that, for example, $b_n$ is monotone increasing?
– Tom Arbuckle
Nov 21 '18 at 3:10
@TomArbuckle That's again a straightforward exercise of induction. Since this site in called Mathematics Stack Exchange, it's an exchange of ideas, rather than a do-my-HW site.
– GNUSupporter 8964民主女神 地下教會
Nov 21 '18 at 3:17
@TomArbuckle That's again a straightforward exercise of induction. Since this site in called Mathematics Stack Exchange, it's an exchange of ideas, rather than a do-my-HW site.
– GNUSupporter 8964民主女神 地下教會
Nov 21 '18 at 3:17
add a comment |
We have that $$2a_n=a_{n-1}+a_{n-2}$$therefore $$2(a_n-a_{n-1})=a_{n-2}-a_{n-1}$$defining $b_n= a_n-a_{n-1}$ we obtain$$2b_n=-{b_{n-1}}$$since $b_2=a_2-a_1=1$ we have $$b_n=4(-{1over 2})^n$$which yields to $$a_n=a_1+sum_{k=2}^{n}b_k=sum_{k=2}^{n}b_k$$and we can write $$lim_{nto infty}a_n=sum_{k=2}^{infty}b_k=b_2+b_3+b_4+cdots= {1over 1-left(-{1over 2}right)}={2over 3}$$
answered Nov 30 '18 at 15:49
Mostafa Ayaz
13.7k3936
add a comment |
We have that $$2a_n=a_{n-1}+a_{n-2}$$therefore $$2(a_n-a_{n-1})=a_{n-2}-a_{n-1}$$defining $b_n= a_n-a_{n-1}$ we obtain$$2b_n=-{b_{n-1}}$$since $b_2=a_2-a_1=1$ we have $$b_n=4(-{1over 2})^n$$which yields to $$a_n=a_1+sum_{k=2}^{n}b_k=sum_{k=2}^{n}b_k$$and we can write $$lim_{nto infty}a_n=sum_{k=2}^{infty}b_k=b_2+b_3+b_4+cdots= {1over 1-left(-{1over 2}right)}={2over 3}$$
answered Nov 30 '18 at 15:49
Mostafa Ayaz
13.7k3936
add a comment |
We have that $$2a_n=a_{n-1}+a_{n-2}$$therefore $$2(a_n-a_{n-1})=a_{n-2}-a_{n-1}$$defining $b_n= a_n-a_{n-1}$ we obtain$$2b_n=-{b_{n-1}}$$since $b_2=a_2-a_1=1$ we have $$b_n=4(-{1over 2})^n$$which yields to $$a_n=a_1+sum_{k=2}^{n}b_k=sum_{k=2}^{n}b_k$$and we can write $$lim_{nto infty}a_n=sum_{k=2}^{infty}b_k=b_2+b_3+b_4+cdots= {1over 1-left(-{1over 2}right)}={2over 3}$$
answered Nov 30 '18 at 15:49
Mostafa Ayaz
13.7k3936
We have that $$2a_n=a_{n-1}+a_{n-2}$$therefore $$2(a_n-a_{n-1})=a_{n-2}-a_{n-1}$$defining $b_n= a_n-a_{n-1}$ we obtain$$2b_n=-{b_{n-1}}$$since $b_2=a_2-a_1=1$ we have $$b_n=4(-{1over 2})^n$$which yields to $$a_n=a_1+sum_{k=2}^{n}b_k=sum_{k=2}^{n}b_k$$and we can write $$lim_{nto infty}a_n=sum_{k=2}^{infty}b_k=b_2+b_3+b_4+cdots= {1over 1-left(-{1over 2}right)}={2over 3}$$
answered Nov 30 '18 at 15:49
Mostafa Ayaz
13.7k3936
answered Nov 30 '18 at 15:49
Mostafa Ayaz
13.7k3936
answered Nov 30 '18 at 15:49
Mostafa Ayaz
13.7k3936
answered Nov 30 '18 at 15:49
Mostafa Ayaz
13.7k3936
13.7k3936
add a comment |
add a comment |
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it seems you only have one choice. try to define $a_n$ in terms of $b_n$ and $c_n$ from the theorem.
– The Count
Nov 21 '18 at 2:53
But i need to define $b_n$ and $c_n$ to prove each one is monotone and bounded, in order to use the monotone convergence theorem.
– Tom Arbuckle
Nov 21 '18 at 2:57
i mean write the equation for a general $a_n$.
– The Count
Nov 21 '18 at 2:58
If you read carefully, you will see that it´s a date they´re giving me. The title says it: $a_n=frac{a_{n-1}+a_{n+2}}{2}$
– Tom Arbuckle
Nov 21 '18 at 3:02
A similar question here
– rtybase
Nov 21 '18 at 10:32