convergence of Cauchy sequences defined recursively
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$X=(x_n)$ defined as $X_1=1,X_2=2$ and $X_n=dfrac12left(X_n-2+X_n-1right)$.
To finds its limit, first we should prove that it is convergent, but my book has not given its solution. Instead, it just states that it is bounded between $1$ and $2$, i.e., $1< X_n < 2$ and guides to do it myself. It also states that $X_n$ is not monotone.
After some calculations, we find that mod of $X_n - X_n-1=frac12^n-1$.
Please suggest how to proceed in showing that its Cauchy hence also convergent
thanks in advance!
convergence
edited Aug 18 at 6:33
spaceisdarkgreen
28.1k21548
asked Aug 18 at 5:21
Vivek
11
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up vote
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favorite
$X=(x_n)$ defined as $X_1=1,X_2=2$ and $X_n=dfrac12left(X_n-2+X_n-1right)$.
To finds its limit, first we should prove that it is convergent, but my book has not given its solution. Instead, it just states that it is bounded between $1$ and $2$, i.e., $1< X_n < 2$ and guides to do it myself. It also states that $X_n$ is not monotone.
After some calculations, we find that mod of $X_n - X_n-1=frac12^n-1$.
Please suggest how to proceed in showing that its Cauchy hence also convergent
thanks in advance!
convergence
edited Aug 18 at 6:33
spaceisdarkgreen
28.1k21548
asked Aug 18 at 5:21
Vivek
11
2
Better than the modulus, after some calculation, we find that $X_n-X_n-1 = (-2)^-(n-2).$ And since we can sum differences up to get an exact value of $X_n,$ there's really no need to prove it's a Cauchy sequence separately.
– spaceisdarkgreen
Aug 18 at 6:42
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$X=(x_n)$ defined as $X_1=1,X_2=2$ and $X_n=dfrac12left(X_n-2+X_n-1right)$.
To finds its limit, first we should prove that it is convergent, but my book has not given its solution. Instead, it just states that it is bounded between $1$ and $2$, i.e., $1< X_n < 2$ and guides to do it myself. It also states that $X_n$ is not monotone.
After some calculations, we find that mod of $X_n - X_n-1=frac12^n-1$.
Please suggest how to proceed in showing that its Cauchy hence also convergent
thanks in advance!
convergence
edited Aug 18 at 6:33
spaceisdarkgreen
28.1k21548
asked Aug 18 at 5:21
Vivek
11
$X=(x_n)$ defined as $X_1=1,X_2=2$ and $X_n=dfrac12left(X_n-2+X_n-1right)$.
To finds its limit, first we should prove that it is convergent, but my book has not given its solution. Instead, it just states that it is bounded between $1$ and $2$, i.e., $1< X_n < 2$ and guides to do it myself. It also states that $X_n$ is not monotone.
After some calculations, we find that mod of $X_n - X_n-1=frac12^n-1$.
Please suggest how to proceed in showing that its Cauchy hence also convergent
thanks in advance!
convergence
edited Aug 18 at 6:33
spaceisdarkgreen
28.1k21548
asked Aug 18 at 5:21
Vivek
11
edited Aug 18 at 6:33
spaceisdarkgreen
28.1k21548
edited Aug 18 at 6:33
spaceisdarkgreen
28.1k21548
edited Aug 18 at 6:33
spaceisdarkgreen
28.1k21548
28.1k21548
asked Aug 18 at 5:21
Vivek
11
asked Aug 18 at 5:21
Vivek
11
asked Aug 18 at 5:21
Vivek
11
11
2
Better than the modulus, after some calculation, we find that $X_n-X_n-1 = (-2)^-(n-2).$ And since we can sum differences up to get an exact value of $X_n,$ there's really no need to prove it's a Cauchy sequence separately.
– spaceisdarkgreen
Aug 18 at 6:42
add a comment |Â
2
Better than the modulus, after some calculation, we find that $X_n-X_n-1 = (-2)^-(n-2).$ And since we can sum differences up to get an exact value of $X_n,$ there's really no need to prove it's a Cauchy sequence separately.
– spaceisdarkgreen
Aug 18 at 6:42
2
2
Better than the modulus, after some calculation, we find that $X_n-X_n-1 = (-2)^-(n-2).$ And since we can sum differences up to get an exact value of $X_n,$ there's really no need to prove it's a Cauchy sequence separately.
– spaceisdarkgreen
Aug 18 at 6:42
Better than the modulus, after some calculation, we find that $X_n-X_n-1 = (-2)^-(n-2).$ And since we can sum differences up to get an exact value of $X_n,$ there's really no need to prove it's a Cauchy sequence separately.
– spaceisdarkgreen
Aug 18 at 6:42
add a comment |Â
1 Answer
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$sum _n=1^infty |X_n-X_n-1| <infty$ implies that $X_n$ is Cauchy. To see this take $k>m$ and note that $|X_k-X_m-1|=|sum _n=m^k (X_n-X_n-1)| leqsum _n=m^k |X_n-X_n-1| $. This last quantity tends to $0$ because teh series $sum _n=1^infty |X_n-X_n-1|$ is convergent. [ In fact $sum _n=m^k |X_n-X_n-1|=S_k-S_m-1$ where $S_n$ is the n-th partial sum of the series; if $S_n to s$ then $S_k-S_m-1 to s-s=0$ as $k>m to infty$]. We have proved that $X_n$ is Cauchy.
answered Aug 18 at 12:17
Kavi Rama Murthy
22.9k2933
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
$sum _n=1^infty |X_n-X_n-1| <infty$ implies that $X_n$ is Cauchy. To see this take $k>m$ and note that $|X_k-X_m-1|=|sum _n=m^k (X_n-X_n-1)| leqsum _n=m^k |X_n-X_n-1| $. This last quantity tends to $0$ because teh series $sum _n=1^infty |X_n-X_n-1|$ is convergent. [ In fact $sum _n=m^k |X_n-X_n-1|=S_k-S_m-1$ where $S_n$ is the n-th partial sum of the series; if $S_n to s$ then $S_k-S_m-1 to s-s=0$ as $k>m to infty$]. We have proved that $X_n$ is Cauchy.
answered Aug 18 at 12:17
Kavi Rama Murthy
22.9k2933
add a comment |Â
up vote
0
down vote
$sum _n=1^infty |X_n-X_n-1| <infty$ implies that $X_n$ is Cauchy. To see this take $k>m$ and note that $|X_k-X_m-1|=|sum _n=m^k (X_n-X_n-1)| leqsum _n=m^k |X_n-X_n-1| $. This last quantity tends to $0$ because teh series $sum _n=1^infty |X_n-X_n-1|$ is convergent. [ In fact $sum _n=m^k |X_n-X_n-1|=S_k-S_m-1$ where $S_n$ is the n-th partial sum of the series; if $S_n to s$ then $S_k-S_m-1 to s-s=0$ as $k>m to infty$]. We have proved that $X_n$ is Cauchy.
answered Aug 18 at 12:17
Kavi Rama Murthy
22.9k2933
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$sum _n=1^infty |X_n-X_n-1| <infty$ implies that $X_n$ is Cauchy. To see this take $k>m$ and note that $|X_k-X_m-1|=|sum _n=m^k (X_n-X_n-1)| leqsum _n=m^k |X_n-X_n-1| $. This last quantity tends to $0$ because teh series $sum _n=1^infty |X_n-X_n-1|$ is convergent. [ In fact $sum _n=m^k |X_n-X_n-1|=S_k-S_m-1$ where $S_n$ is the n-th partial sum of the series; if $S_n to s$ then $S_k-S_m-1 to s-s=0$ as $k>m to infty$]. We have proved that $X_n$ is Cauchy.
answered Aug 18 at 12:17
Kavi Rama Murthy
22.9k2933
$sum _n=1^infty |X_n-X_n-1| <infty$ implies that $X_n$ is Cauchy. To see this take $k>m$ and note that $|X_k-X_m-1|=|sum _n=m^k (X_n-X_n-1)| leqsum _n=m^k |X_n-X_n-1| $. This last quantity tends to $0$ because teh series $sum _n=1^infty |X_n-X_n-1|$ is convergent. [ In fact $sum _n=m^k |X_n-X_n-1|=S_k-S_m-1$ where $S_n$ is the n-th partial sum of the series; if $S_n to s$ then $S_k-S_m-1 to s-s=0$ as $k>m to infty$]. We have proved that $X_n$ is Cauchy.
answered Aug 18 at 12:17
Kavi Rama Murthy
22.9k2933
answered Aug 18 at 12:17
Kavi Rama Murthy
22.9k2933
answered Aug 18 at 12:17
Kavi Rama Murthy
22.9k2933
answered Aug 18 at 12:17
Kavi Rama Murthy
22.9k2933
22.9k2933
add a comment |Â
add a comment |Â
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2
Better than the modulus, after some calculation, we find that $X_n-X_n-1 = (-2)^-(n-2).$ And since we can sum differences up to get an exact value of $X_n,$ there's really no need to prove it's a Cauchy sequence separately.
– spaceisdarkgreen
Aug 18 at 6:42