Proof verification that ${x_n} = 0,underbrace{77dots 7}_{text{n times}}$ is a Cauchy sequence.
$begingroup$
Given a sequence ${x_n}$:
$$
x_n = 0,underbrace{77dots 7}_{text n times}
$$
Prove that ${x_n}$ is a Cauchy sequence.
Recall the definition of a fundamental sequence:
$$
x_n text{is fundamental} iff forall epsilon>0 exists Nin Bbb N: forall n, m >Nimplies |x_n - x_m| < epsilon
$$
Rewrite $x_n$:
$$
x_n = {7over 10^1} + {7over 10^2} + cdots + {7over 10^n} = sum_{k=1}^n frac{7}{10^k}
$$
By geometric series sum:
$$
x_n = sum_{k=1}^n frac{7}{10^k} = frac{7}{9}left(1 - {1over 10^n}right) \
x_m = sum_{k=1}^m frac{7}{10^k} = frac{7}{9}left(1 - {1over 10^m}right) \
$$
Suppose $m > n$:
$$
begin{align}
|x_n - x_m| &= |x_m - x_n| = \
&= left|frac{7}{9}left(1 - {1over 10^m}right) - frac{7}{9}left(1 - {1over 10^n}right)right| = \
&= left|frac{7}{9}left(1 - {1over 10^m} - 1 + {1over 10^n}right)right| =
\
&= left|frac{7}{9}left({1over 10^n} - {1over 10^m}right)right| le left|frac{7}{9}{1over 10^n}right| le frac{7}{9cdot 10^N} < epsilon
end{align}
$$
This shows we've found $N$ which depends on $epsilon$ and satisfies the definition of a Cauchy sequence.
This is the first time I'm dealing with proving a sequence is fundamental, could someone please verify whether my proof is valid?
calculus sequences-and-series limits proof-verification cauchy-sequences
asked Dec 18 '18 at 13:57
romanroman
2,28421224
$endgroup$
add a comment |
$begingroup$
Given a sequence ${x_n}$:
$$
x_n = 0,underbrace{77dots 7}_{text n times}
$$
Prove that ${x_n}$ is a Cauchy sequence.
Recall the definition of a fundamental sequence:
$$
x_n text{is fundamental} iff forall epsilon>0 exists Nin Bbb N: forall n, m >Nimplies |x_n - x_m| < epsilon
$$
Rewrite $x_n$:
$$
x_n = {7over 10^1} + {7over 10^2} + cdots + {7over 10^n} = sum_{k=1}^n frac{7}{10^k}
$$
By geometric series sum:
$$
x_n = sum_{k=1}^n frac{7}{10^k} = frac{7}{9}left(1 - {1over 10^n}right) \
x_m = sum_{k=1}^m frac{7}{10^k} = frac{7}{9}left(1 - {1over 10^m}right) \
$$
Suppose $m > n$:
$$
begin{align}
|x_n - x_m| &= |x_m - x_n| = \
&= left|frac{7}{9}left(1 - {1over 10^m}right) - frac{7}{9}left(1 - {1over 10^n}right)right| = \
&= left|frac{7}{9}left(1 - {1over 10^m} - 1 + {1over 10^n}right)right| =
\
&= left|frac{7}{9}left({1over 10^n} - {1over 10^m}right)right| le left|frac{7}{9}{1over 10^n}right| le frac{7}{9cdot 10^N} < epsilon
end{align}
$$
This shows we've found $N$ which depends on $epsilon$ and satisfies the definition of a Cauchy sequence.
This is the first time I'm dealing with proving a sequence is fundamental, could someone please verify whether my proof is valid?
calculus sequences-and-series limits proof-verification cauchy-sequences
asked Dec 18 '18 at 13:57
romanroman
2,28421224
$endgroup$
$begingroup$
It is correct :)
$endgroup$
– Hendrra
Dec 18 '18 at 14:00
$begingroup$
@Hendrra, thanks for taking your time
$endgroup$
– roman
Dec 18 '18 at 14:10
add a comment |
$begingroup$
Given a sequence ${x_n}$:
$$
x_n = 0,underbrace{77dots 7}_{text n times}
$$
Prove that ${x_n}$ is a Cauchy sequence.
Recall the definition of a fundamental sequence:
$$
x_n text{is fundamental} iff forall epsilon>0 exists Nin Bbb N: forall n, m >Nimplies |x_n - x_m| < epsilon
$$
Rewrite $x_n$:
$$
x_n = {7over 10^1} + {7over 10^2} + cdots + {7over 10^n} = sum_{k=1}^n frac{7}{10^k}
$$
By geometric series sum:
$$
x_n = sum_{k=1}^n frac{7}{10^k} = frac{7}{9}left(1 - {1over 10^n}right) \
x_m = sum_{k=1}^m frac{7}{10^k} = frac{7}{9}left(1 - {1over 10^m}right) \
$$
Suppose $m > n$:
$$
begin{align}
|x_n - x_m| &= |x_m - x_n| = \
&= left|frac{7}{9}left(1 - {1over 10^m}right) - frac{7}{9}left(1 - {1over 10^n}right)right| = \
&= left|frac{7}{9}left(1 - {1over 10^m} - 1 + {1over 10^n}right)right| =
\
&= left|frac{7}{9}left({1over 10^n} - {1over 10^m}right)right| le left|frac{7}{9}{1over 10^n}right| le frac{7}{9cdot 10^N} < epsilon
end{align}
$$
This shows we've found $N$ which depends on $epsilon$ and satisfies the definition of a Cauchy sequence.
This is the first time I'm dealing with proving a sequence is fundamental, could someone please verify whether my proof is valid?
calculus sequences-and-series limits proof-verification cauchy-sequences
asked Dec 18 '18 at 13:57
romanroman
2,28421224
$endgroup$
Given a sequence ${x_n}$:
$$
x_n = 0,underbrace{77dots 7}_{text n times}
$$
Prove that ${x_n}$ is a Cauchy sequence.
Recall the definition of a fundamental sequence:
$$
x_n text{is fundamental} iff forall epsilon>0 exists Nin Bbb N: forall n, m >Nimplies |x_n - x_m| < epsilon
$$
Rewrite $x_n$:
$$
x_n = {7over 10^1} + {7over 10^2} + cdots + {7over 10^n} = sum_{k=1}^n frac{7}{10^k}
$$
By geometric series sum:
$$
x_n = sum_{k=1}^n frac{7}{10^k} = frac{7}{9}left(1 - {1over 10^n}right) \
x_m = sum_{k=1}^m frac{7}{10^k} = frac{7}{9}left(1 - {1over 10^m}right) \
$$
Suppose $m > n$:
$$
begin{align}
|x_n - x_m| &= |x_m - x_n| = \
&= left|frac{7}{9}left(1 - {1over 10^m}right) - frac{7}{9}left(1 - {1over 10^n}right)right| = \
&= left|frac{7}{9}left(1 - {1over 10^m} - 1 + {1over 10^n}right)right| =
\
&= left|frac{7}{9}left({1over 10^n} - {1over 10^m}right)right| le left|frac{7}{9}{1over 10^n}right| le frac{7}{9cdot 10^N} < epsilon
end{align}
$$
This shows we've found $N$ which depends on $epsilon$ and satisfies the definition of a Cauchy sequence.
This is the first time I'm dealing with proving a sequence is fundamental, could someone please verify whether my proof is valid?
calculus sequences-and-series limits proof-verification cauchy-sequences
calculus sequences-and-series limits proof-verification cauchy-sequences
asked Dec 18 '18 at 13:57
romanroman
2,28421224
asked Dec 18 '18 at 13:57
romanroman
2,28421224
asked Dec 18 '18 at 13:57
romanroman
2,28421224
asked Dec 18 '18 at 13:57
romanroman
2,28421224
asked Dec 18 '18 at 13:57
romanroman
2,28421224
2,28421224
$begingroup$
It is correct :)
$endgroup$
– Hendrra
Dec 18 '18 at 14:00
$begingroup$
@Hendrra, thanks for taking your time
$endgroup$
– roman
Dec 18 '18 at 14:10
add a comment |
$begingroup$
It is correct :)
$endgroup$
– Hendrra
Dec 18 '18 at 14:00
$begingroup$
@Hendrra, thanks for taking your time
$endgroup$
– roman
Dec 18 '18 at 14:10
$begingroup$
It is correct :)
$endgroup$
– Hendrra
Dec 18 '18 at 14:00
$begingroup$
It is correct :)
$endgroup$
– Hendrra
Dec 18 '18 at 14:00
$begingroup$
@Hendrra, thanks for taking your time
$endgroup$
– roman
Dec 18 '18 at 14:10
$begingroup$
@Hendrra, thanks for taking your time
$endgroup$
– roman
Dec 18 '18 at 14:10
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yes, just write it in the forward direction.
Suppose $N > log_{10}left(frac{7}{9 epsilon}right)$, then for any $m,n in mathbb{Z}$ such that $m> n > N$, then we have $|x_n-x_m|< epsilon$.
answered Dec 18 '18 at 14:04
Siong Thye GohSiong Thye Goh
102k1466118
$endgroup$
$begingroup$
thank you for the notice
$endgroup$
– roman
Dec 18 '18 at 14:10
add a comment |
$begingroup$
Since my proof in OP is correct I would like to add a generalization also. Consider the following $x_n$:
$$
x_n = a + aq + aq^2 + cdots + aq^{n-1}
$$
Using geometric series sum for $|q| < 1$:
$$
x_n = frac{a(1-q^n)}{1-q}
$$
Since $|q| < 1$ we may rewrite it as:
$$
q = frac{1}{1+r}, r in Bbb R_{>0}
$$
Then for $m > n$:
$$begin{align}
|x_m - x_n| &= left|frac{a}{1-q} left(q^n - q^mright)right|\ &= left|frac{a}{1-q} left(frac{1}{(1+r)^n} - frac{1}{(1+r)^m}right)right| \&le left|frac{a}{1-q} left(frac{1}{(1+r)^n}right)right| \ &
le frac{a}{1-q} left(frac{1}{(1+r)^N}right) < epsilonend{align}
$$
Which shows any sequence of such kind is Cauchy. Or with direct statement:
$$
frac{1-q}{a} left((1+r)^Nright) > {1over epsilon}\
(1+r)^N > frac{a}{(1-q)epsilon} \
exists N >log_{1+r}frac{a}{(1-q)epsilon}, forall m>n>N implies |x_m - x_n| < epsilon
$$
answered Dec 18 '18 at 14:36
romanroman
2,28421224
$endgroup$
add a comment |
Your Answer
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2 Answers
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2 Answers
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active
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votes
$begingroup$
Yes, just write it in the forward direction.
Suppose $N > log_{10}left(frac{7}{9 epsilon}right)$, then for any $m,n in mathbb{Z}$ such that $m> n > N$, then we have $|x_n-x_m|< epsilon$.
answered Dec 18 '18 at 14:04
Siong Thye GohSiong Thye Goh
102k1466118
$endgroup$
$begingroup$
thank you for the notice
$endgroup$
– roman
Dec 18 '18 at 14:10
add a comment |
$begingroup$
Yes, just write it in the forward direction.
Suppose $N > log_{10}left(frac{7}{9 epsilon}right)$, then for any $m,n in mathbb{Z}$ such that $m> n > N$, then we have $|x_n-x_m|< epsilon$.
answered Dec 18 '18 at 14:04
Siong Thye GohSiong Thye Goh
102k1466118
$endgroup$
$begingroup$
thank you for the notice
$endgroup$
– roman
Dec 18 '18 at 14:10
add a comment |
$begingroup$
Yes, just write it in the forward direction.
Suppose $N > log_{10}left(frac{7}{9 epsilon}right)$, then for any $m,n in mathbb{Z}$ such that $m> n > N$, then we have $|x_n-x_m|< epsilon$.
answered Dec 18 '18 at 14:04
Siong Thye GohSiong Thye Goh
102k1466118
$endgroup$
Yes, just write it in the forward direction.
Suppose $N > log_{10}left(frac{7}{9 epsilon}right)$, then for any $m,n in mathbb{Z}$ such that $m> n > N$, then we have $|x_n-x_m|< epsilon$.
answered Dec 18 '18 at 14:04
Siong Thye GohSiong Thye Goh
102k1466118
answered Dec 18 '18 at 14:04
Siong Thye GohSiong Thye Goh
102k1466118
answered Dec 18 '18 at 14:04
Siong Thye GohSiong Thye Goh
102k1466118
answered Dec 18 '18 at 14:04
Siong Thye GohSiong Thye Goh
102k1466118
102k1466118
$begingroup$
thank you for the notice
$endgroup$
– roman
Dec 18 '18 at 14:10
add a comment |
$begingroup$
thank you for the notice
$endgroup$
– roman
Dec 18 '18 at 14:10
$begingroup$
thank you for the notice
$endgroup$
– roman
Dec 18 '18 at 14:10
$begingroup$
thank you for the notice
$endgroup$
– roman
Dec 18 '18 at 14:10
add a comment |
$begingroup$
Since my proof in OP is correct I would like to add a generalization also. Consider the following $x_n$:
$$
x_n = a + aq + aq^2 + cdots + aq^{n-1}
$$
Using geometric series sum for $|q| < 1$:
$$
x_n = frac{a(1-q^n)}{1-q}
$$
Since $|q| < 1$ we may rewrite it as:
$$
q = frac{1}{1+r}, r in Bbb R_{>0}
$$
Then for $m > n$:
$$begin{align}
|x_m - x_n| &= left|frac{a}{1-q} left(q^n - q^mright)right|\ &= left|frac{a}{1-q} left(frac{1}{(1+r)^n} - frac{1}{(1+r)^m}right)right| \&le left|frac{a}{1-q} left(frac{1}{(1+r)^n}right)right| \ &
le frac{a}{1-q} left(frac{1}{(1+r)^N}right) < epsilonend{align}
$$
Which shows any sequence of such kind is Cauchy. Or with direct statement:
$$
frac{1-q}{a} left((1+r)^Nright) > {1over epsilon}\
(1+r)^N > frac{a}{(1-q)epsilon} \
exists N >log_{1+r}frac{a}{(1-q)epsilon}, forall m>n>N implies |x_m - x_n| < epsilon
$$
answered Dec 18 '18 at 14:36
romanroman
2,28421224
$endgroup$
add a comment |
$begingroup$
Since my proof in OP is correct I would like to add a generalization also. Consider the following $x_n$:
$$
x_n = a + aq + aq^2 + cdots + aq^{n-1}
$$
Using geometric series sum for $|q| < 1$:
$$
x_n = frac{a(1-q^n)}{1-q}
$$
Since $|q| < 1$ we may rewrite it as:
$$
q = frac{1}{1+r}, r in Bbb R_{>0}
$$
Then for $m > n$:
$$begin{align}
|x_m - x_n| &= left|frac{a}{1-q} left(q^n - q^mright)right|\ &= left|frac{a}{1-q} left(frac{1}{(1+r)^n} - frac{1}{(1+r)^m}right)right| \&le left|frac{a}{1-q} left(frac{1}{(1+r)^n}right)right| \ &
le frac{a}{1-q} left(frac{1}{(1+r)^N}right) < epsilonend{align}
$$
Which shows any sequence of such kind is Cauchy. Or with direct statement:
$$
frac{1-q}{a} left((1+r)^Nright) > {1over epsilon}\
(1+r)^N > frac{a}{(1-q)epsilon} \
exists N >log_{1+r}frac{a}{(1-q)epsilon}, forall m>n>N implies |x_m - x_n| < epsilon
$$
answered Dec 18 '18 at 14:36
romanroman
2,28421224
$endgroup$
add a comment |
$begingroup$
Since my proof in OP is correct I would like to add a generalization also. Consider the following $x_n$:
$$
x_n = a + aq + aq^2 + cdots + aq^{n-1}
$$
Using geometric series sum for $|q| < 1$:
$$
x_n = frac{a(1-q^n)}{1-q}
$$
Since $|q| < 1$ we may rewrite it as:
$$
q = frac{1}{1+r}, r in Bbb R_{>0}
$$
Then for $m > n$:
$$begin{align}
|x_m - x_n| &= left|frac{a}{1-q} left(q^n - q^mright)right|\ &= left|frac{a}{1-q} left(frac{1}{(1+r)^n} - frac{1}{(1+r)^m}right)right| \&le left|frac{a}{1-q} left(frac{1}{(1+r)^n}right)right| \ &
le frac{a}{1-q} left(frac{1}{(1+r)^N}right) < epsilonend{align}
$$
Which shows any sequence of such kind is Cauchy. Or with direct statement:
$$
frac{1-q}{a} left((1+r)^Nright) > {1over epsilon}\
(1+r)^N > frac{a}{(1-q)epsilon} \
exists N >log_{1+r}frac{a}{(1-q)epsilon}, forall m>n>N implies |x_m - x_n| < epsilon
$$
answered Dec 18 '18 at 14:36
romanroman
2,28421224
$endgroup$
Since my proof in OP is correct I would like to add a generalization also. Consider the following $x_n$:
$$
x_n = a + aq + aq^2 + cdots + aq^{n-1}
$$
Using geometric series sum for $|q| < 1$:
$$
x_n = frac{a(1-q^n)}{1-q}
$$
Since $|q| < 1$ we may rewrite it as:
$$
q = frac{1}{1+r}, r in Bbb R_{>0}
$$
Then for $m > n$:
$$begin{align}
|x_m - x_n| &= left|frac{a}{1-q} left(q^n - q^mright)right|\ &= left|frac{a}{1-q} left(frac{1}{(1+r)^n} - frac{1}{(1+r)^m}right)right| \&le left|frac{a}{1-q} left(frac{1}{(1+r)^n}right)right| \ &
le frac{a}{1-q} left(frac{1}{(1+r)^N}right) < epsilonend{align}
$$
Which shows any sequence of such kind is Cauchy. Or with direct statement:
$$
frac{1-q}{a} left((1+r)^Nright) > {1over epsilon}\
(1+r)^N > frac{a}{(1-q)epsilon} \
exists N >log_{1+r}frac{a}{(1-q)epsilon}, forall m>n>N implies |x_m - x_n| < epsilon
$$
answered Dec 18 '18 at 14:36
romanroman
2,28421224
edited Dec 18 '18 at 14:55
answered Dec 18 '18 at 14:36
romanroman
2,28421224
answered Dec 18 '18 at 14:36
romanroman
2,28421224
answered Dec 18 '18 at 14:36
romanroman
2,28421224
2,28421224
add a comment |
add a comment |
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$begingroup$
It is correct :)
$endgroup$
– Hendrra
Dec 18 '18 at 14:00
$begingroup$
@Hendrra, thanks for taking your time
$endgroup$
– roman
Dec 18 '18 at 14:10