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How to prove that recursive sequence converges and find it's limit
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For sequence given by $a_1=1$ i $a_n+1= frac1a_1+a_2+ldots+a_n$ I have to prove that it converges to some number and find this number.
I tried toshow that it's monotonic by calculating
$$
fraca_n+1a_n = frac1a_n(a_1+a_2+ldots+a_n)
$$
but I cannot say anything about the denominator. How can I try to find it's limit?
sequences-and-series limits convergence
asked 1 hour ago
avan1235
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up vote
2
down vote
favorite
For sequence given by $a_1=1$ i $a_n+1= frac1a_1+a_2+ldots+a_n$ I have to prove that it converges to some number and find this number.
I tried toshow that it's monotonic by calculating
$$
fraca_n+1a_n = frac1a_n(a_1+a_2+ldots+a_n)
$$
but I cannot say anything about the denominator. How can I try to find it's limit?
sequences-and-series limits convergence
asked 1 hour ago
avan1235
183
New contributor
avan1235 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
For sequence given by $a_1=1$ i $a_n+1= frac1a_1+a_2+ldots+a_n$ I have to prove that it converges to some number and find this number.
I tried toshow that it's monotonic by calculating
$$
fraca_n+1a_n = frac1a_n(a_1+a_2+ldots+a_n)
$$
but I cannot say anything about the denominator. How can I try to find it's limit?
sequences-and-series limits convergence
asked 1 hour ago
avan1235
183
New contributor
avan1235 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
For sequence given by $a_1=1$ i $a_n+1= frac1a_1+a_2+ldots+a_n$ I have to prove that it converges to some number and find this number.
I tried toshow that it's monotonic by calculating
$$
fraca_n+1a_n = frac1a_n(a_1+a_2+ldots+a_n)
$$
but I cannot say anything about the denominator. How can I try to find it's limit?
sequences-and-series limits convergence
sequences-and-series limits convergence
asked 1 hour ago
avan1235
183
New contributor
avan1235 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
avan1235
183
New contributor
avan1235 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
avan1235
183
New contributor
avan1235 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
avan1235
183
asked 1 hour ago
avan1235
183
183
New contributor
avan1235 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
avan1235 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
avan1235 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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3 Answers
3
active
oldest
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up vote
2
down vote
accepted
Let $s_n = sumlimits_k=1^n a_k$. We can rewrite the recurrence relation as
$$s_n+1 - s_n = a_n+1 = frac1s_n quadimplies s_n+1 = s_n + frac1s_n$$
This implies
$$s_n+1^2 = s_n^2 + 2 + frac1s_n^2 ge s_n^2 + 2$$
So for all $n > 1$, we have
$$s_n^2 = s_1^2 + sum_k=1^n-1 (s_k+1^2 - s_k^2) ge 1 + sum_k=1^n-1 2 = 2n - 1$$
Since all $a_n$ is clearly positive, we have $displaystyle;0 < a_n = frac1s_n-1 le frac1sqrt2n-3$.
By squeezing, $a_n$ converges to $0$ as $ntoinfty$.
answered 51 mins ago
achille hui
92.1k5127248
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up vote
2
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Note that $$a_n+1 = frac1left(a_1 + ldots + a_n-1right) + a_n = frac1frac1a_n + a_n =fraca_n1 + a_n^2$$ for $n geq 1$. If you can prove that is is convergent, then, calling the limit $alpha$, we must have $$alpha = fracalpha1+alpha^2,$$ and solving this gives $alpha = 0$.
As for proving convergence, the recurrence relation given in the form above should help you to verify that the sequence is indeed monotone decreasing, and further you should be able to check that the sequence is bounded below by zero. Then apply monotone convergence theorem.
answered 1 hour ago
Sam Streeter
1,245217
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up vote
0
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Prove by induction that $a_ngeq0$ for all $n$ (it's pretty straightforward).
You'll have then $a_n+1/a_n = 1/(a_nsum_i=1^n a_i) < 1$ for all $n$ (since $a_1=1$).
Thus $a_ngeq0$ is decreasing, converging to some limit $xgeq0$.
Such limit must satisfy $x=1/sum_n=1^infty a_n$.
If $x>0$, we would have convergence of $sum_n=1^inftya_n$, implying $a_nto 0$, a contradiction. Therefore, $x=limlimits_ntoinfty a_n = 0$.
answered 51 mins ago
rldias
2,8431522
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Let $s_n = sumlimits_k=1^n a_k$. We can rewrite the recurrence relation as
$$s_n+1 - s_n = a_n+1 = frac1s_n quadimplies s_n+1 = s_n + frac1s_n$$
This implies
$$s_n+1^2 = s_n^2 + 2 + frac1s_n^2 ge s_n^2 + 2$$
So for all $n > 1$, we have
$$s_n^2 = s_1^2 + sum_k=1^n-1 (s_k+1^2 - s_k^2) ge 1 + sum_k=1^n-1 2 = 2n - 1$$
Since all $a_n$ is clearly positive, we have $displaystyle;0 < a_n = frac1s_n-1 le frac1sqrt2n-3$.
By squeezing, $a_n$ converges to $0$ as $ntoinfty$.
answered 51 mins ago
achille hui
92.1k5127248
add a comment |Â
up vote
2
down vote
accepted
Let $s_n = sumlimits_k=1^n a_k$. We can rewrite the recurrence relation as
$$s_n+1 - s_n = a_n+1 = frac1s_n quadimplies s_n+1 = s_n + frac1s_n$$
This implies
$$s_n+1^2 = s_n^2 + 2 + frac1s_n^2 ge s_n^2 + 2$$
So for all $n > 1$, we have
$$s_n^2 = s_1^2 + sum_k=1^n-1 (s_k+1^2 - s_k^2) ge 1 + sum_k=1^n-1 2 = 2n - 1$$
Since all $a_n$ is clearly positive, we have $displaystyle;0 < a_n = frac1s_n-1 le frac1sqrt2n-3$.
By squeezing, $a_n$ converges to $0$ as $ntoinfty$.
answered 51 mins ago
achille hui
92.1k5127248
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Let $s_n = sumlimits_k=1^n a_k$. We can rewrite the recurrence relation as
$$s_n+1 - s_n = a_n+1 = frac1s_n quadimplies s_n+1 = s_n + frac1s_n$$
This implies
$$s_n+1^2 = s_n^2 + 2 + frac1s_n^2 ge s_n^2 + 2$$
So for all $n > 1$, we have
$$s_n^2 = s_1^2 + sum_k=1^n-1 (s_k+1^2 - s_k^2) ge 1 + sum_k=1^n-1 2 = 2n - 1$$
Since all $a_n$ is clearly positive, we have $displaystyle;0 < a_n = frac1s_n-1 le frac1sqrt2n-3$.
By squeezing, $a_n$ converges to $0$ as $ntoinfty$.
answered 51 mins ago
achille hui
92.1k5127248
Let $s_n = sumlimits_k=1^n a_k$. We can rewrite the recurrence relation as
$$s_n+1 - s_n = a_n+1 = frac1s_n quadimplies s_n+1 = s_n + frac1s_n$$
This implies
$$s_n+1^2 = s_n^2 + 2 + frac1s_n^2 ge s_n^2 + 2$$
So for all $n > 1$, we have
$$s_n^2 = s_1^2 + sum_k=1^n-1 (s_k+1^2 - s_k^2) ge 1 + sum_k=1^n-1 2 = 2n - 1$$
Since all $a_n$ is clearly positive, we have $displaystyle;0 < a_n = frac1s_n-1 le frac1sqrt2n-3$.
By squeezing, $a_n$ converges to $0$ as $ntoinfty$.
answered 51 mins ago
achille hui
92.1k5127248
answered 51 mins ago
achille hui
92.1k5127248
answered 51 mins ago
achille hui
92.1k5127248
answered 51 mins ago
achille hui
92.1k5127248
92.1k5127248
add a comment |Â
add a comment |Â
up vote
2
down vote
Note that $$a_n+1 = frac1left(a_1 + ldots + a_n-1right) + a_n = frac1frac1a_n + a_n =fraca_n1 + a_n^2$$ for $n geq 1$. If you can prove that is is convergent, then, calling the limit $alpha$, we must have $$alpha = fracalpha1+alpha^2,$$ and solving this gives $alpha = 0$.
As for proving convergence, the recurrence relation given in the form above should help you to verify that the sequence is indeed monotone decreasing, and further you should be able to check that the sequence is bounded below by zero. Then apply monotone convergence theorem.
answered 1 hour ago
Sam Streeter
1,245217
add a comment |Â
up vote
2
down vote
Note that $$a_n+1 = frac1left(a_1 + ldots + a_n-1right) + a_n = frac1frac1a_n + a_n =fraca_n1 + a_n^2$$ for $n geq 1$. If you can prove that is is convergent, then, calling the limit $alpha$, we must have $$alpha = fracalpha1+alpha^2,$$ and solving this gives $alpha = 0$.
As for proving convergence, the recurrence relation given in the form above should help you to verify that the sequence is indeed monotone decreasing, and further you should be able to check that the sequence is bounded below by zero. Then apply monotone convergence theorem.
answered 1 hour ago
Sam Streeter
1,245217
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Note that $$a_n+1 = frac1left(a_1 + ldots + a_n-1right) + a_n = frac1frac1a_n + a_n =fraca_n1 + a_n^2$$ for $n geq 1$. If you can prove that is is convergent, then, calling the limit $alpha$, we must have $$alpha = fracalpha1+alpha^2,$$ and solving this gives $alpha = 0$.
As for proving convergence, the recurrence relation given in the form above should help you to verify that the sequence is indeed monotone decreasing, and further you should be able to check that the sequence is bounded below by zero. Then apply monotone convergence theorem.
answered 1 hour ago
Sam Streeter
1,245217
Note that $$a_n+1 = frac1left(a_1 + ldots + a_n-1right) + a_n = frac1frac1a_n + a_n =fraca_n1 + a_n^2$$ for $n geq 1$. If you can prove that is is convergent, then, calling the limit $alpha$, we must have $$alpha = fracalpha1+alpha^2,$$ and solving this gives $alpha = 0$.
As for proving convergence, the recurrence relation given in the form above should help you to verify that the sequence is indeed monotone decreasing, and further you should be able to check that the sequence is bounded below by zero. Then apply monotone convergence theorem.
answered 1 hour ago
Sam Streeter
1,245217
answered 1 hour ago
Sam Streeter
1,245217
answered 1 hour ago
Sam Streeter
1,245217
answered 1 hour ago
Sam Streeter
1,245217
1,245217
add a comment |Â
add a comment |Â
up vote
0
down vote
Prove by induction that $a_ngeq0$ for all $n$ (it's pretty straightforward).
You'll have then $a_n+1/a_n = 1/(a_nsum_i=1^n a_i) < 1$ for all $n$ (since $a_1=1$).
Thus $a_ngeq0$ is decreasing, converging to some limit $xgeq0$.
Such limit must satisfy $x=1/sum_n=1^infty a_n$.
If $x>0$, we would have convergence of $sum_n=1^inftya_n$, implying $a_nto 0$, a contradiction. Therefore, $x=limlimits_ntoinfty a_n = 0$.
answered 51 mins ago
rldias
2,8431522
add a comment |Â
up vote
0
down vote
Prove by induction that $a_ngeq0$ for all $n$ (it's pretty straightforward).
You'll have then $a_n+1/a_n = 1/(a_nsum_i=1^n a_i) < 1$ for all $n$ (since $a_1=1$).
Thus $a_ngeq0$ is decreasing, converging to some limit $xgeq0$.
Such limit must satisfy $x=1/sum_n=1^infty a_n$.
If $x>0$, we would have convergence of $sum_n=1^inftya_n$, implying $a_nto 0$, a contradiction. Therefore, $x=limlimits_ntoinfty a_n = 0$.
answered 51 mins ago
rldias
2,8431522
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Prove by induction that $a_ngeq0$ for all $n$ (it's pretty straightforward).
You'll have then $a_n+1/a_n = 1/(a_nsum_i=1^n a_i) < 1$ for all $n$ (since $a_1=1$).
Thus $a_ngeq0$ is decreasing, converging to some limit $xgeq0$.
Such limit must satisfy $x=1/sum_n=1^infty a_n$.
If $x>0$, we would have convergence of $sum_n=1^inftya_n$, implying $a_nto 0$, a contradiction. Therefore, $x=limlimits_ntoinfty a_n = 0$.
answered 51 mins ago
rldias
2,8431522
Prove by induction that $a_ngeq0$ for all $n$ (it's pretty straightforward).
You'll have then $a_n+1/a_n = 1/(a_nsum_i=1^n a_i) < 1$ for all $n$ (since $a_1=1$).
Thus $a_ngeq0$ is decreasing, converging to some limit $xgeq0$.
Such limit must satisfy $x=1/sum_n=1^infty a_n$.
If $x>0$, we would have convergence of $sum_n=1^inftya_n$, implying $a_nto 0$, a contradiction. Therefore, $x=limlimits_ntoinfty a_n = 0$.
answered 51 mins ago
rldias
2,8431522
answered 51 mins ago
rldias
2,8431522
answered 51 mins ago
rldias
2,8431522
answered 51 mins ago
rldias
2,8431522
2,8431522
add a comment |Â
add a comment |Â
avan1235 is a new contributor. Be nice, and check out our Code of Conduct.
avan1235 is a new contributor. Be nice, and check out our Code of Conduct.
avan1235 is a new contributor. Be nice, and check out our Code of Conduct.
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