Mixing
How does $x_{n+1}ge x_nland y_{y+1}le y_n land lim_{ntoinfty}(x_n-y_n) = 0implies lim x_n = lim y_n$ and both...
$begingroup$
Let $x_n$ and $y_n$ denote two sequences. The sequences are given such that:
$$
x_{n+1} ge x_n \
y_{n+1} le y_n \
lim_{ntoinfty}(x_n-y_n) = 0
$$
Prove that both $x_n$ and $y_n$ are convergent and:
$$
lim_{ntoinfty} x_n = lim_{ntoinfty}y_n
$$
I'm not sure how to proceed with the proof. I've started with inspecting the relations between $x_n$ and $y_n$:
$$
x_{n+1} ge x_n \
y_{n+1} le y_n \ iff
\
x_{n+1} ge x_n \
-y_{n+1} ge -y_n
$$
From which it follows that:
$$
x_{n+1} - y_{n+1} ge x_n - y_n tag 1
$$
Denote $z_n = x_n - y_n$, then by $(1)$ it follows that $z_n$ is a monotonically increasing sequence:
$$
z_{n+1} ge z_n
$$
By monotone convergence theorem it follows that if $lim z_n = 0$ and $z_{n+1} ge z_n$ then $z_n le 0$. Also $z_n$ is convergent hence bounded:
$$
m le z_n le M
$$
By the fact that $z_n < 0$ it follows that $x_n < y_n$.
The problem is i do not see how to combine those facts in order to show what is requested in the problem statement. Seems like the problem may be reduced to using MCT for $x_n$ and $y_n$ alone or use the properties of $limsup$, $liminf$ or somehow use the triangular inequality but i do not see how.
What is a proper way to achieve that?
calculus sequences-and-series limits
asked Jan 10 at 13:58
romanroman
2,17321224
$endgroup$
|
show 1 more comment
$begingroup$
Let $x_n$ and $y_n$ denote two sequences. The sequences are given such that:
$$
x_{n+1} ge x_n \
y_{n+1} le y_n \
lim_{ntoinfty}(x_n-y_n) = 0
$$
Prove that both $x_n$ and $y_n$ are convergent and:
$$
lim_{ntoinfty} x_n = lim_{ntoinfty}y_n
$$
I'm not sure how to proceed with the proof. I've started with inspecting the relations between $x_n$ and $y_n$:
$$
x_{n+1} ge x_n \
y_{n+1} le y_n \ iff
\
x_{n+1} ge x_n \
-y_{n+1} ge -y_n
$$
From which it follows that:
$$
x_{n+1} - y_{n+1} ge x_n - y_n tag 1
$$
Denote $z_n = x_n - y_n$, then by $(1)$ it follows that $z_n$ is a monotonically increasing sequence:
$$
z_{n+1} ge z_n
$$
By monotone convergence theorem it follows that if $lim z_n = 0$ and $z_{n+1} ge z_n$ then $z_n le 0$. Also $z_n$ is convergent hence bounded:
$$
m le z_n le M
$$
By the fact that $z_n < 0$ it follows that $x_n < y_n$.
The problem is i do not see how to combine those facts in order to show what is requested in the problem statement. Seems like the problem may be reduced to using MCT for $x_n$ and $y_n$ alone or use the properties of $limsup$, $liminf$ or somehow use the triangular inequality but i do not see how.
What is a proper way to achieve that?
calculus sequences-and-series limits
asked Jan 10 at 13:58
romanroman
2,17321224
$endgroup$
2
$begingroup$
Hint: Can you use $x_n<y_n$ to show that ${x_n}$ is bounded?
$endgroup$
– SmileyCraft
Jan 10 at 14:04
$begingroup$
@Uncountable $lim(x_n - y_n) = lim (n - (-n)) ne 0$ so it violates the initial conditions.
$endgroup$
– roman
Jan 10 at 14:05
$begingroup$
Yes, roman, I realised that as soon as I posted it. I apologise.
$endgroup$
– Uncountable
Jan 10 at 14:07
$begingroup$
@SmileyCraft I would appreciate if you could elaborate on this. Intuitively the statement is true however I don't see where the boundedness of $x_n$ comes from
$endgroup$
– roman
Jan 10 at 14:13
1
$begingroup$
@roman By induction $y_nleq y_1$, so $y_1$ is an upper bound of ${x_n}$.
$endgroup$
– SmileyCraft
Jan 10 at 14:15
|
show 1 more comment
$begingroup$
Let $x_n$ and $y_n$ denote two sequences. The sequences are given such that:
$$
x_{n+1} ge x_n \
y_{n+1} le y_n \
lim_{ntoinfty}(x_n-y_n) = 0
$$
Prove that both $x_n$ and $y_n$ are convergent and:
$$
lim_{ntoinfty} x_n = lim_{ntoinfty}y_n
$$
I'm not sure how to proceed with the proof. I've started with inspecting the relations between $x_n$ and $y_n$:
$$
x_{n+1} ge x_n \
y_{n+1} le y_n \ iff
\
x_{n+1} ge x_n \
-y_{n+1} ge -y_n
$$
From which it follows that:
$$
x_{n+1} - y_{n+1} ge x_n - y_n tag 1
$$
Denote $z_n = x_n - y_n$, then by $(1)$ it follows that $z_n$ is a monotonically increasing sequence:
$$
z_{n+1} ge z_n
$$
By monotone convergence theorem it follows that if $lim z_n = 0$ and $z_{n+1} ge z_n$ then $z_n le 0$. Also $z_n$ is convergent hence bounded:
$$
m le z_n le M
$$
By the fact that $z_n < 0$ it follows that $x_n < y_n$.
The problem is i do not see how to combine those facts in order to show what is requested in the problem statement. Seems like the problem may be reduced to using MCT for $x_n$ and $y_n$ alone or use the properties of $limsup$, $liminf$ or somehow use the triangular inequality but i do not see how.
What is a proper way to achieve that?
calculus sequences-and-series limits
asked Jan 10 at 13:58
romanroman
2,17321224
$endgroup$
Let $x_n$ and $y_n$ denote two sequences. The sequences are given such that:
$$
x_{n+1} ge x_n \
y_{n+1} le y_n \
lim_{ntoinfty}(x_n-y_n) = 0
$$
Prove that both $x_n$ and $y_n$ are convergent and:
$$
lim_{ntoinfty} x_n = lim_{ntoinfty}y_n
$$
I'm not sure how to proceed with the proof. I've started with inspecting the relations between $x_n$ and $y_n$:
$$
x_{n+1} ge x_n \
y_{n+1} le y_n \ iff
\
x_{n+1} ge x_n \
-y_{n+1} ge -y_n
$$
From which it follows that:
$$
x_{n+1} - y_{n+1} ge x_n - y_n tag 1
$$
Denote $z_n = x_n - y_n$, then by $(1)$ it follows that $z_n$ is a monotonically increasing sequence:
$$
z_{n+1} ge z_n
$$
By monotone convergence theorem it follows that if $lim z_n = 0$ and $z_{n+1} ge z_n$ then $z_n le 0$. Also $z_n$ is convergent hence bounded:
$$
m le z_n le M
$$
By the fact that $z_n < 0$ it follows that $x_n < y_n$.
The problem is i do not see how to combine those facts in order to show what is requested in the problem statement. Seems like the problem may be reduced to using MCT for $x_n$ and $y_n$ alone or use the properties of $limsup$, $liminf$ or somehow use the triangular inequality but i do not see how.
What is a proper way to achieve that?
calculus sequences-and-series limits
calculus sequences-and-series limits
asked Jan 10 at 13:58
romanroman
2,17321224
asked Jan 10 at 13:58
romanroman
2,17321224
asked Jan 10 at 13:58
romanroman
2,17321224
asked Jan 10 at 13:58
romanroman
2,17321224
asked Jan 10 at 13:58
romanroman
2,17321224
2,17321224
2
$begingroup$
Hint: Can you use $x_n<y_n$ to show that ${x_n}$ is bounded?
$endgroup$
– SmileyCraft
Jan 10 at 14:04
$begingroup$
@Uncountable $lim(x_n - y_n) = lim (n - (-n)) ne 0$ so it violates the initial conditions.
$endgroup$
– roman
Jan 10 at 14:05
$begingroup$
Yes, roman, I realised that as soon as I posted it. I apologise.
$endgroup$
– Uncountable
Jan 10 at 14:07
$begingroup$
@SmileyCraft I would appreciate if you could elaborate on this. Intuitively the statement is true however I don't see where the boundedness of $x_n$ comes from
$endgroup$
– roman
Jan 10 at 14:13
1
$begingroup$
@roman By induction $y_nleq y_1$, so $y_1$ is an upper bound of ${x_n}$.
$endgroup$
– SmileyCraft
Jan 10 at 14:15
|
show 1 more comment
2
$begingroup$
Hint: Can you use $x_n<y_n$ to show that ${x_n}$ is bounded?
$endgroup$
– SmileyCraft
Jan 10 at 14:04
$begingroup$
@Uncountable $lim(x_n - y_n) = lim (n - (-n)) ne 0$ so it violates the initial conditions.
$endgroup$
– roman
Jan 10 at 14:05
$begingroup$
Yes, roman, I realised that as soon as I posted it. I apologise.
$endgroup$
– Uncountable
Jan 10 at 14:07
$begingroup$
@SmileyCraft I would appreciate if you could elaborate on this. Intuitively the statement is true however I don't see where the boundedness of $x_n$ comes from
$endgroup$
– roman
Jan 10 at 14:13
1
$begingroup$
@roman By induction $y_nleq y_1$, so $y_1$ is an upper bound of ${x_n}$.
$endgroup$
– SmileyCraft
Jan 10 at 14:15
2
2
$begingroup$
Hint: Can you use $x_n<y_n$ to show that ${x_n}$ is bounded?
$endgroup$
– SmileyCraft
Jan 10 at 14:04
$begingroup$
Hint: Can you use $x_n<y_n$ to show that ${x_n}$ is bounded?
$endgroup$
– SmileyCraft
Jan 10 at 14:04
$begingroup$
@Uncountable $lim(x_n - y_n) = lim (n - (-n)) ne 0$ so it violates the initial conditions.
$endgroup$
– roman
Jan 10 at 14:05
$begingroup$
@Uncountable $lim(x_n - y_n) = lim (n - (-n)) ne 0$ so it violates the initial conditions.
$endgroup$
– roman
Jan 10 at 14:05
$begingroup$
Yes, roman, I realised that as soon as I posted it. I apologise.
$endgroup$
– Uncountable
Jan 10 at 14:07
$begingroup$
Yes, roman, I realised that as soon as I posted it. I apologise.
$endgroup$
– Uncountable
Jan 10 at 14:07
$begingroup$
@SmileyCraft I would appreciate if you could elaborate on this. Intuitively the statement is true however I don't see where the boundedness of $x_n$ comes from
$endgroup$
– roman
Jan 10 at 14:13
$begingroup$
@SmileyCraft I would appreciate if you could elaborate on this. Intuitively the statement is true however I don't see where the boundedness of $x_n$ comes from
$endgroup$
– roman
Jan 10 at 14:13
1
1
$begingroup$
@roman By induction $y_nleq y_1$, so $y_1$ is an upper bound of ${x_n}$.
$endgroup$
– SmileyCraft
Jan 10 at 14:15
$begingroup$
@roman By induction $y_nleq y_1$, so $y_1$ is an upper bound of ${x_n}$.
$endgroup$
– SmileyCraft
Jan 10 at 14:15
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
I would proceed as follows, without reference to the “nested intervals lemma” (which is essentially equivalent to what you want to prove):
As you already observed, $z_n = x_n - y_n$ is increasing, with $lim_{n to infty} z_n = 0$. It follows that $x_n le y_n$ for all $n in Bbb N$.
Then
$$
x_n le x_{n+1} le y_{n+1} le y_1 quad (n in Bbb N)
$$
so that the sequence $(x_n)$ is increasing and bounded above, and therefore convergent: $lim_{n to infty} x_n$ exists.
Similarly,
$$
y_n ge y_{n+1} ge x_{n+1} ge x_1 quad (n in Bbb N)
$$
implies that $(y_n)$ is decreasing and bounded below, so that $lim_{n to infty} y_n$ exists as well.
Finally (since both limits exits),
$$
0 = lim_{ntoinfty}(x_n-y_n) = lim_{n to infty} x_n - lim_{n to infty} y_n
$$
so that the limits are equal.
answered Jan 10 at 18:20
Martin RMartin R
28.1k33355
$endgroup$
$begingroup$
Nice approach, thank you!
$endgroup$
– roman
Jan 10 at 19:39
add a comment |
$begingroup$
Using a hint by @SmileyCraft we might show that $x_n$ is bounded and increasing hence convergent. Consider the following inequality:
$$
x_n le y_n
$$
Since $x_n$ is monotonically decreasing we have that:
$$
x_n le x_{n+1}
$$
By $y_n$ is monotonically decreasing:
$$
y_{n+1} le y_{n}
$$
Expanding that further and using the fact $x_n le y_n$ we may obtain:
$$
x_1 le x_2 le cdots le x_{n-1} le x_{n} le x_{n+1} le y_{n+1} le y_n le y_{n-1} le cdots le y_2 le y_1
$$
Which a set of nested intervals. By the Nested intervals lemma and the fact that
$$
lim_{ntoinfty}(x_n - y_n) = 0
$$
we know that there is a single point $L$ which belongs to all of the intervals. By monotonicity of $x_n$ and $y_n$ this point appears to be the limit for both sequences. So by this $x_n$ and $y_n$ converge and:
$$
lim_{ntoinfty} x_n = lim_{ntoinfty} y_n =L
$$
answered Jan 10 at 14:26
romanroman
2,17321224
$endgroup$
$begingroup$
I hope this is correct
$endgroup$
– roman
Jan 10 at 14:53
add a comment |
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2 Answers
2
active
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2 Answers
2
active
oldest
votes
active
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oldest
votes
$begingroup$
I would proceed as follows, without reference to the “nested intervals lemma” (which is essentially equivalent to what you want to prove):
As you already observed, $z_n = x_n - y_n$ is increasing, with $lim_{n to infty} z_n = 0$. It follows that $x_n le y_n$ for all $n in Bbb N$.
Then
$$
x_n le x_{n+1} le y_{n+1} le y_1 quad (n in Bbb N)
$$
so that the sequence $(x_n)$ is increasing and bounded above, and therefore convergent: $lim_{n to infty} x_n$ exists.
Similarly,
$$
y_n ge y_{n+1} ge x_{n+1} ge x_1 quad (n in Bbb N)
$$
implies that $(y_n)$ is decreasing and bounded below, so that $lim_{n to infty} y_n$ exists as well.
Finally (since both limits exits),
$$
0 = lim_{ntoinfty}(x_n-y_n) = lim_{n to infty} x_n - lim_{n to infty} y_n
$$
so that the limits are equal.
answered Jan 10 at 18:20
Martin RMartin R
28.1k33355
$endgroup$
$begingroup$
Nice approach, thank you!
$endgroup$
– roman
Jan 10 at 19:39
add a comment |
$begingroup$
I would proceed as follows, without reference to the “nested intervals lemma” (which is essentially equivalent to what you want to prove):
As you already observed, $z_n = x_n - y_n$ is increasing, with $lim_{n to infty} z_n = 0$. It follows that $x_n le y_n$ for all $n in Bbb N$.
Then
$$
x_n le x_{n+1} le y_{n+1} le y_1 quad (n in Bbb N)
$$
so that the sequence $(x_n)$ is increasing and bounded above, and therefore convergent: $lim_{n to infty} x_n$ exists.
Similarly,
$$
y_n ge y_{n+1} ge x_{n+1} ge x_1 quad (n in Bbb N)
$$
implies that $(y_n)$ is decreasing and bounded below, so that $lim_{n to infty} y_n$ exists as well.
Finally (since both limits exits),
$$
0 = lim_{ntoinfty}(x_n-y_n) = lim_{n to infty} x_n - lim_{n to infty} y_n
$$
so that the limits are equal.
answered Jan 10 at 18:20
Martin RMartin R
28.1k33355
$endgroup$
$begingroup$
Nice approach, thank you!
$endgroup$
– roman
Jan 10 at 19:39
add a comment |
$begingroup$
I would proceed as follows, without reference to the “nested intervals lemma” (which is essentially equivalent to what you want to prove):
As you already observed, $z_n = x_n - y_n$ is increasing, with $lim_{n to infty} z_n = 0$. It follows that $x_n le y_n$ for all $n in Bbb N$.
Then
$$
x_n le x_{n+1} le y_{n+1} le y_1 quad (n in Bbb N)
$$
so that the sequence $(x_n)$ is increasing and bounded above, and therefore convergent: $lim_{n to infty} x_n$ exists.
Similarly,
$$
y_n ge y_{n+1} ge x_{n+1} ge x_1 quad (n in Bbb N)
$$
implies that $(y_n)$ is decreasing and bounded below, so that $lim_{n to infty} y_n$ exists as well.
Finally (since both limits exits),
$$
0 = lim_{ntoinfty}(x_n-y_n) = lim_{n to infty} x_n - lim_{n to infty} y_n
$$
so that the limits are equal.
answered Jan 10 at 18:20
Martin RMartin R
28.1k33355
$endgroup$
I would proceed as follows, without reference to the “nested intervals lemma” (which is essentially equivalent to what you want to prove):
As you already observed, $z_n = x_n - y_n$ is increasing, with $lim_{n to infty} z_n = 0$. It follows that $x_n le y_n$ for all $n in Bbb N$.
Then
$$
x_n le x_{n+1} le y_{n+1} le y_1 quad (n in Bbb N)
$$
so that the sequence $(x_n)$ is increasing and bounded above, and therefore convergent: $lim_{n to infty} x_n$ exists.
Similarly,
$$
y_n ge y_{n+1} ge x_{n+1} ge x_1 quad (n in Bbb N)
$$
implies that $(y_n)$ is decreasing and bounded below, so that $lim_{n to infty} y_n$ exists as well.
Finally (since both limits exits),
$$
0 = lim_{ntoinfty}(x_n-y_n) = lim_{n to infty} x_n - lim_{n to infty} y_n
$$
so that the limits are equal.
answered Jan 10 at 18:20
Martin RMartin R
28.1k33355
answered Jan 10 at 18:20
Martin RMartin R
28.1k33355
answered Jan 10 at 18:20
Martin RMartin R
28.1k33355
answered Jan 10 at 18:20
Martin RMartin R
28.1k33355
28.1k33355
$begingroup$
Nice approach, thank you!
$endgroup$
– roman
Jan 10 at 19:39
add a comment |
$begingroup$
Nice approach, thank you!
$endgroup$
– roman
Jan 10 at 19:39
$begingroup$
Nice approach, thank you!
$endgroup$
– roman
Jan 10 at 19:39
$begingroup$
Nice approach, thank you!
$endgroup$
– roman
Jan 10 at 19:39
add a comment |
$begingroup$
Using a hint by @SmileyCraft we might show that $x_n$ is bounded and increasing hence convergent. Consider the following inequality:
$$
x_n le y_n
$$
Since $x_n$ is monotonically decreasing we have that:
$$
x_n le x_{n+1}
$$
By $y_n$ is monotonically decreasing:
$$
y_{n+1} le y_{n}
$$
Expanding that further and using the fact $x_n le y_n$ we may obtain:
$$
x_1 le x_2 le cdots le x_{n-1} le x_{n} le x_{n+1} le y_{n+1} le y_n le y_{n-1} le cdots le y_2 le y_1
$$
Which a set of nested intervals. By the Nested intervals lemma and the fact that
$$
lim_{ntoinfty}(x_n - y_n) = 0
$$
we know that there is a single point $L$ which belongs to all of the intervals. By monotonicity of $x_n$ and $y_n$ this point appears to be the limit for both sequences. So by this $x_n$ and $y_n$ converge and:
$$
lim_{ntoinfty} x_n = lim_{ntoinfty} y_n =L
$$
answered Jan 10 at 14:26
romanroman
2,17321224
$endgroup$
$begingroup$
I hope this is correct
$endgroup$
– roman
Jan 10 at 14:53
add a comment |
$begingroup$
Using a hint by @SmileyCraft we might show that $x_n$ is bounded and increasing hence convergent. Consider the following inequality:
$$
x_n le y_n
$$
Since $x_n$ is monotonically decreasing we have that:
$$
x_n le x_{n+1}
$$
By $y_n$ is monotonically decreasing:
$$
y_{n+1} le y_{n}
$$
Expanding that further and using the fact $x_n le y_n$ we may obtain:
$$
x_1 le x_2 le cdots le x_{n-1} le x_{n} le x_{n+1} le y_{n+1} le y_n le y_{n-1} le cdots le y_2 le y_1
$$
Which a set of nested intervals. By the Nested intervals lemma and the fact that
$$
lim_{ntoinfty}(x_n - y_n) = 0
$$
we know that there is a single point $L$ which belongs to all of the intervals. By monotonicity of $x_n$ and $y_n$ this point appears to be the limit for both sequences. So by this $x_n$ and $y_n$ converge and:
$$
lim_{ntoinfty} x_n = lim_{ntoinfty} y_n =L
$$
answered Jan 10 at 14:26
romanroman
2,17321224
$endgroup$
$begingroup$
I hope this is correct
$endgroup$
– roman
Jan 10 at 14:53
add a comment |
$begingroup$
Using a hint by @SmileyCraft we might show that $x_n$ is bounded and increasing hence convergent. Consider the following inequality:
$$
x_n le y_n
$$
Since $x_n$ is monotonically decreasing we have that:
$$
x_n le x_{n+1}
$$
By $y_n$ is monotonically decreasing:
$$
y_{n+1} le y_{n}
$$
Expanding that further and using the fact $x_n le y_n$ we may obtain:
$$
x_1 le x_2 le cdots le x_{n-1} le x_{n} le x_{n+1} le y_{n+1} le y_n le y_{n-1} le cdots le y_2 le y_1
$$
Which a set of nested intervals. By the Nested intervals lemma and the fact that
$$
lim_{ntoinfty}(x_n - y_n) = 0
$$
we know that there is a single point $L$ which belongs to all of the intervals. By monotonicity of $x_n$ and $y_n$ this point appears to be the limit for both sequences. So by this $x_n$ and $y_n$ converge and:
$$
lim_{ntoinfty} x_n = lim_{ntoinfty} y_n =L
$$
answered Jan 10 at 14:26
romanroman
2,17321224
$endgroup$
Using a hint by @SmileyCraft we might show that $x_n$ is bounded and increasing hence convergent. Consider the following inequality:
$$
x_n le y_n
$$
Since $x_n$ is monotonically decreasing we have that:
$$
x_n le x_{n+1}
$$
By $y_n$ is monotonically decreasing:
$$
y_{n+1} le y_{n}
$$
Expanding that further and using the fact $x_n le y_n$ we may obtain:
$$
x_1 le x_2 le cdots le x_{n-1} le x_{n} le x_{n+1} le y_{n+1} le y_n le y_{n-1} le cdots le y_2 le y_1
$$
Which a set of nested intervals. By the Nested intervals lemma and the fact that
$$
lim_{ntoinfty}(x_n - y_n) = 0
$$
we know that there is a single point $L$ which belongs to all of the intervals. By monotonicity of $x_n$ and $y_n$ this point appears to be the limit for both sequences. So by this $x_n$ and $y_n$ converge and:
$$
lim_{ntoinfty} x_n = lim_{ntoinfty} y_n =L
$$
answered Jan 10 at 14:26
romanroman
2,17321224
edited Jan 10 at 14:32
answered Jan 10 at 14:26
romanroman
2,17321224
answered Jan 10 at 14:26
romanroman
2,17321224
answered Jan 10 at 14:26
romanroman
2,17321224
2,17321224
$begingroup$
I hope this is correct
$endgroup$
– roman
Jan 10 at 14:53
add a comment |
$begingroup$
I hope this is correct
$endgroup$
– roman
Jan 10 at 14:53
$begingroup$
I hope this is correct
$endgroup$
– roman
Jan 10 at 14:53
$begingroup$
I hope this is correct
$endgroup$
– roman
Jan 10 at 14:53
add a comment |
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$begingroup$
Hint: Can you use $x_n<y_n$ to show that ${x_n}$ is bounded?
$endgroup$
– SmileyCraft
Jan 10 at 14:04
$begingroup$
@Uncountable $lim(x_n - y_n) = lim (n - (-n)) ne 0$ so it violates the initial conditions.
$endgroup$
– roman
Jan 10 at 14:05
$begingroup$
Yes, roman, I realised that as soon as I posted it. I apologise.
$endgroup$
– Uncountable
Jan 10 at 14:07
$begingroup$
@SmileyCraft I would appreciate if you could elaborate on this. Intuitively the statement is true however I don't see where the boundedness of $x_n$ comes from
$endgroup$
– roman
Jan 10 at 14:13
1
$begingroup$
@roman By induction $y_nleq y_1$, so $y_1$ is an upper bound of ${x_n}$.
$endgroup$
– SmileyCraft
Jan 10 at 14:15