Mixing
find the sequence limit 1
$begingroup$
I have the following sequence $(a_{n})$, $a_{1}=1$
$$a_{n+1}=begin{cases}
a_{n}+frac{1}{2} & text{ if } n is even \
frac{a_{n}}{3} & text{ if } n is odd
end{cases}$$
I need to find $$lim_{nrightarrow infty }a_{2n+1}$$
I tried something but I didn't get too far.I rewrite the sequence:$a_{1}=1$, $$a_{n+1}=begin{cases}
a_{n}+r & text{ if } n is even \
q cdot a_{n} & text{ if } n is odd
end{cases}$$
where $q,rin (0,1)$ but I don't know how to write $a_{2n+1}$ with $q$ and $r$
sequences-and-series limits
edited Feb 1 at 17:15
xbh
6,3201522
asked Feb 1 at 17:09
DaniVajaDaniVaja
977
$endgroup$
add a comment |
$begingroup$
I have the following sequence $(a_{n})$, $a_{1}=1$
$$a_{n+1}=begin{cases}
a_{n}+frac{1}{2} & text{ if } n is even \
frac{a_{n}}{3} & text{ if } n is odd
end{cases}$$
I need to find $$lim_{nrightarrow infty }a_{2n+1}$$
I tried something but I didn't get too far.I rewrite the sequence:$a_{1}=1$, $$a_{n+1}=begin{cases}
a_{n}+r & text{ if } n is even \
q cdot a_{n} & text{ if } n is odd
end{cases}$$
where $q,rin (0,1)$ but I don't know how to write $a_{2n+1}$ with $q$ and $r$
sequences-and-series limits
edited Feb 1 at 17:15
xbh
6,3201522
asked Feb 1 at 17:09
DaniVajaDaniVaja
977
$endgroup$
add a comment |
$begingroup$
I have the following sequence $(a_{n})$, $a_{1}=1$
$$a_{n+1}=begin{cases}
a_{n}+frac{1}{2} & text{ if } n is even \
frac{a_{n}}{3} & text{ if } n is odd
end{cases}$$
I need to find $$lim_{nrightarrow infty }a_{2n+1}$$
I tried something but I didn't get too far.I rewrite the sequence:$a_{1}=1$, $$a_{n+1}=begin{cases}
a_{n}+r & text{ if } n is even \
q cdot a_{n} & text{ if } n is odd
end{cases}$$
where $q,rin (0,1)$ but I don't know how to write $a_{2n+1}$ with $q$ and $r$
sequences-and-series limits
edited Feb 1 at 17:15
xbh
6,3201522
asked Feb 1 at 17:09
DaniVajaDaniVaja
977
$endgroup$
I have the following sequence $(a_{n})$, $a_{1}=1$
$$a_{n+1}=begin{cases}
a_{n}+frac{1}{2} & text{ if } n is even \
frac{a_{n}}{3} & text{ if } n is odd
end{cases}$$
I need to find $$lim_{nrightarrow infty }a_{2n+1}$$
I tried something but I didn't get too far.I rewrite the sequence:$a_{1}=1$, $$a_{n+1}=begin{cases}
a_{n}+r & text{ if } n is even \
q cdot a_{n} & text{ if } n is odd
end{cases}$$
where $q,rin (0,1)$ but I don't know how to write $a_{2n+1}$ with $q$ and $r$
sequences-and-series limits
sequences-and-series limits
edited Feb 1 at 17:15
xbh
6,3201522
asked Feb 1 at 17:09
DaniVajaDaniVaja
977
edited Feb 1 at 17:15
xbh
6,3201522
asked Feb 1 at 17:09
DaniVajaDaniVaja
977
edited Feb 1 at 17:15
xbh
6,3201522
edited Feb 1 at 17:15
xbh
6,3201522
edited Feb 1 at 17:15
xbh
6,3201522
6,3201522
asked Feb 1 at 17:09
DaniVajaDaniVaja
977
asked Feb 1 at 17:09
DaniVajaDaniVaja
977
asked Feb 1 at 17:09
DaniVajaDaniVaja
977
977
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Combine two steps in one: from one term with odd index, you have to divide by three to get the next term (with even index), then add 1/2 to get the next term with odd index.
This gives $b_n$, the series of terms with odd indices, i.e. $b_n = a_{2n+1}$, with $b_1 =1$,$b_{n+1} = frac13 b_n + frac12$. So
$$
b_{n+1} = frac12 + frac13 ( frac12 + frac13 ( cdots (frac12 + frac13 b_1 ))cdots)\
=frac12 (1 + frac13 + (frac13)^2+cdots +(frac13)^{n-1} ) + (frac13)^{n} b_1\
= frac12 frac{1 -(frac13)^{n}}{2/3} + (frac13)^{n}
$$
so the limit is $frac34$.
answered Feb 1 at 17:25
AndreasAndreas
8,4411137
$endgroup$
$begingroup$
Thank you very much for your help!Why $b_{1}=1$ ? It shouldn't be $b_{1}=a_{3}=5/6$ ?
$endgroup$
– DaniVaja
Feb 1 at 17:33
$begingroup$
Yes you are right. I started erroneously with the index $n=0$. In any case, it makes no difference for the limit $n to infty$.
$endgroup$
– Andreas
Feb 1 at 17:35
$begingroup$
It took one day to understand the solution :).Thanks a lot!
$endgroup$
– DaniVaja
Feb 2 at 19:10
$begingroup$
you're welcome!
$endgroup$
– Andreas
Feb 2 at 22:08
add a comment |
$begingroup$
We have, for each $nin Bbb N,$
$$a_{2n+1}-a_{2n}=frac 12$$
thus
$(a_n)$ is not Cauchy.
What can you conclude ?
answered Feb 1 at 17:27
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
$begingroup$
What does this say about $a_{2n} - a_{2n-1}$?
$endgroup$
– Andreas
Feb 1 at 17:33
$begingroup$
It is $frac{-2}{3}a_{2n-1}$.
$endgroup$
– hamam_Abdallah
Feb 1 at 17:37
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Combine two steps in one: from one term with odd index, you have to divide by three to get the next term (with even index), then add 1/2 to get the next term with odd index.
This gives $b_n$, the series of terms with odd indices, i.e. $b_n = a_{2n+1}$, with $b_1 =1$,$b_{n+1} = frac13 b_n + frac12$. So
$$
b_{n+1} = frac12 + frac13 ( frac12 + frac13 ( cdots (frac12 + frac13 b_1 ))cdots)\
=frac12 (1 + frac13 + (frac13)^2+cdots +(frac13)^{n-1} ) + (frac13)^{n} b_1\
= frac12 frac{1 -(frac13)^{n}}{2/3} + (frac13)^{n}
$$
so the limit is $frac34$.
answered Feb 1 at 17:25
AndreasAndreas
8,4411137
$endgroup$
$begingroup$
Thank you very much for your help!Why $b_{1}=1$ ? It shouldn't be $b_{1}=a_{3}=5/6$ ?
$endgroup$
– DaniVaja
Feb 1 at 17:33
$begingroup$
Yes you are right. I started erroneously with the index $n=0$. In any case, it makes no difference for the limit $n to infty$.
$endgroup$
– Andreas
Feb 1 at 17:35
$begingroup$
It took one day to understand the solution :).Thanks a lot!
$endgroup$
– DaniVaja
Feb 2 at 19:10
$begingroup$
you're welcome!
$endgroup$
– Andreas
Feb 2 at 22:08
add a comment |
$begingroup$
Combine two steps in one: from one term with odd index, you have to divide by three to get the next term (with even index), then add 1/2 to get the next term with odd index.
This gives $b_n$, the series of terms with odd indices, i.e. $b_n = a_{2n+1}$, with $b_1 =1$,$b_{n+1} = frac13 b_n + frac12$. So
$$
b_{n+1} = frac12 + frac13 ( frac12 + frac13 ( cdots (frac12 + frac13 b_1 ))cdots)\
=frac12 (1 + frac13 + (frac13)^2+cdots +(frac13)^{n-1} ) + (frac13)^{n} b_1\
= frac12 frac{1 -(frac13)^{n}}{2/3} + (frac13)^{n}
$$
so the limit is $frac34$.
answered Feb 1 at 17:25
AndreasAndreas
8,4411137
$endgroup$
$begingroup$
Thank you very much for your help!Why $b_{1}=1$ ? It shouldn't be $b_{1}=a_{3}=5/6$ ?
$endgroup$
– DaniVaja
Feb 1 at 17:33
$begingroup$
Yes you are right. I started erroneously with the index $n=0$. In any case, it makes no difference for the limit $n to infty$.
$endgroup$
– Andreas
Feb 1 at 17:35
$begingroup$
It took one day to understand the solution :).Thanks a lot!
$endgroup$
– DaniVaja
Feb 2 at 19:10
$begingroup$
you're welcome!
$endgroup$
– Andreas
Feb 2 at 22:08
add a comment |
$begingroup$
Combine two steps in one: from one term with odd index, you have to divide by three to get the next term (with even index), then add 1/2 to get the next term with odd index.
This gives $b_n$, the series of terms with odd indices, i.e. $b_n = a_{2n+1}$, with $b_1 =1$,$b_{n+1} = frac13 b_n + frac12$. So
$$
b_{n+1} = frac12 + frac13 ( frac12 + frac13 ( cdots (frac12 + frac13 b_1 ))cdots)\
=frac12 (1 + frac13 + (frac13)^2+cdots +(frac13)^{n-1} ) + (frac13)^{n} b_1\
= frac12 frac{1 -(frac13)^{n}}{2/3} + (frac13)^{n}
$$
so the limit is $frac34$.
answered Feb 1 at 17:25
AndreasAndreas
8,4411137
$endgroup$
Combine two steps in one: from one term with odd index, you have to divide by three to get the next term (with even index), then add 1/2 to get the next term with odd index.
This gives $b_n$, the series of terms with odd indices, i.e. $b_n = a_{2n+1}$, with $b_1 =1$,$b_{n+1} = frac13 b_n + frac12$. So
$$
b_{n+1} = frac12 + frac13 ( frac12 + frac13 ( cdots (frac12 + frac13 b_1 ))cdots)\
=frac12 (1 + frac13 + (frac13)^2+cdots +(frac13)^{n-1} ) + (frac13)^{n} b_1\
= frac12 frac{1 -(frac13)^{n}}{2/3} + (frac13)^{n}
$$
so the limit is $frac34$.
answered Feb 1 at 17:25
AndreasAndreas
8,4411137
answered Feb 1 at 17:25
AndreasAndreas
8,4411137
answered Feb 1 at 17:25
AndreasAndreas
8,4411137
answered Feb 1 at 17:25
AndreasAndreas
8,4411137
8,4411137
$begingroup$
Thank you very much for your help!Why $b_{1}=1$ ? It shouldn't be $b_{1}=a_{3}=5/6$ ?
$endgroup$
– DaniVaja
Feb 1 at 17:33
$begingroup$
Yes you are right. I started erroneously with the index $n=0$. In any case, it makes no difference for the limit $n to infty$.
$endgroup$
– Andreas
Feb 1 at 17:35
$begingroup$
It took one day to understand the solution :).Thanks a lot!
$endgroup$
– DaniVaja
Feb 2 at 19:10
$begingroup$
you're welcome!
$endgroup$
– Andreas
Feb 2 at 22:08
add a comment |
$begingroup$
Thank you very much for your help!Why $b_{1}=1$ ? It shouldn't be $b_{1}=a_{3}=5/6$ ?
$endgroup$
– DaniVaja
Feb 1 at 17:33
$begingroup$
Yes you are right. I started erroneously with the index $n=0$. In any case, it makes no difference for the limit $n to infty$.
$endgroup$
– Andreas
Feb 1 at 17:35
$begingroup$
It took one day to understand the solution :).Thanks a lot!
$endgroup$
– DaniVaja
Feb 2 at 19:10
$begingroup$
you're welcome!
$endgroup$
– Andreas
Feb 2 at 22:08
$begingroup$
Thank you very much for your help!Why $b_{1}=1$ ? It shouldn't be $b_{1}=a_{3}=5/6$ ?
$endgroup$
– DaniVaja
Feb 1 at 17:33
$begingroup$
Thank you very much for your help!Why $b_{1}=1$ ? It shouldn't be $b_{1}=a_{3}=5/6$ ?
$endgroup$
– DaniVaja
Feb 1 at 17:33
$begingroup$
Yes you are right. I started erroneously with the index $n=0$. In any case, it makes no difference for the limit $n to infty$.
$endgroup$
– Andreas
Feb 1 at 17:35
$begingroup$
Yes you are right. I started erroneously with the index $n=0$. In any case, it makes no difference for the limit $n to infty$.
$endgroup$
– Andreas
Feb 1 at 17:35
$begingroup$
It took one day to understand the solution :).Thanks a lot!
$endgroup$
– DaniVaja
Feb 2 at 19:10
$begingroup$
It took one day to understand the solution :).Thanks a lot!
$endgroup$
– DaniVaja
Feb 2 at 19:10
$begingroup$
you're welcome!
$endgroup$
– Andreas
Feb 2 at 22:08
$begingroup$
you're welcome!
$endgroup$
– Andreas
Feb 2 at 22:08
add a comment |
$begingroup$
We have, for each $nin Bbb N,$
$$a_{2n+1}-a_{2n}=frac 12$$
thus
$(a_n)$ is not Cauchy.
What can you conclude ?
answered Feb 1 at 17:27
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
$begingroup$
What does this say about $a_{2n} - a_{2n-1}$?
$endgroup$
– Andreas
Feb 1 at 17:33
$begingroup$
It is $frac{-2}{3}a_{2n-1}$.
$endgroup$
– hamam_Abdallah
Feb 1 at 17:37
add a comment |
$begingroup$
We have, for each $nin Bbb N,$
$$a_{2n+1}-a_{2n}=frac 12$$
thus
$(a_n)$ is not Cauchy.
What can you conclude ?
answered Feb 1 at 17:27
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
$begingroup$
What does this say about $a_{2n} - a_{2n-1}$?
$endgroup$
– Andreas
Feb 1 at 17:33
$begingroup$
It is $frac{-2}{3}a_{2n-1}$.
$endgroup$
– hamam_Abdallah
Feb 1 at 17:37
add a comment |
$begingroup$
We have, for each $nin Bbb N,$
$$a_{2n+1}-a_{2n}=frac 12$$
thus
$(a_n)$ is not Cauchy.
What can you conclude ?
answered Feb 1 at 17:27
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
We have, for each $nin Bbb N,$
$$a_{2n+1}-a_{2n}=frac 12$$
thus
$(a_n)$ is not Cauchy.
What can you conclude ?
answered Feb 1 at 17:27
hamam_Abdallahhamam_Abdallah
38.1k21634
answered Feb 1 at 17:27
hamam_Abdallahhamam_Abdallah
38.1k21634
answered Feb 1 at 17:27
hamam_Abdallahhamam_Abdallah
38.1k21634
answered Feb 1 at 17:27
hamam_Abdallahhamam_Abdallah
38.1k21634
38.1k21634
$begingroup$
What does this say about $a_{2n} - a_{2n-1}$?
$endgroup$
– Andreas
Feb 1 at 17:33
$begingroup$
It is $frac{-2}{3}a_{2n-1}$.
$endgroup$
– hamam_Abdallah
Feb 1 at 17:37
add a comment |
$begingroup$
What does this say about $a_{2n} - a_{2n-1}$?
$endgroup$
– Andreas
Feb 1 at 17:33
$begingroup$
It is $frac{-2}{3}a_{2n-1}$.
$endgroup$
– hamam_Abdallah
Feb 1 at 17:37
$begingroup$
What does this say about $a_{2n} - a_{2n-1}$?
$endgroup$
– Andreas
Feb 1 at 17:33
$begingroup$
What does this say about $a_{2n} - a_{2n-1}$?
$endgroup$
– Andreas
Feb 1 at 17:33
$begingroup$
It is $frac{-2}{3}a_{2n-1}$.
$endgroup$
– hamam_Abdallah
Feb 1 at 17:37
$begingroup$
It is $frac{-2}{3}a_{2n-1}$.
$endgroup$
– hamam_Abdallah
Feb 1 at 17:37
add a comment |
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