Request for assistance on finishing the proof that $lim_{ntoinfty}n x_n = 0$ given some initial conditions.
$begingroup$
Let ${x_n}$ denote a sequence such that:
$$
exists N in Bbb N: forall n > N implies 0 < x_{n+1} < x_n
$$
And the sequence ${y_n}$ is convergent, where ${y_n}$ is given by:
$$
y_n = sum_{k=1}^n x_n
$$
Prove that:
$$
lim_{ntoinfty} n x_n =0
$$
Below is what I've done so far.
Start with the first fact. We are given that a sequence $x_n$ is monotonically decreasing starting from some $N$. That by Weierstrass theorem means that the sequence is convergent to some $x_0 > 0$, namely:
$$
exists x_0 inBbb R_{>0}:lim_{ntoinfty} x_n = x_0 tag1
$$
Also it is bounded:
$$
exists m, M in Bbb R, forall n in Bbb N:m le x_n le M tag2
$$
We are also given that the sum is convergent, hence:
$$
exists L in Bbb R:lim_{ntoinfty} y_n = lim_{ntoinfty}sum_{k=1}^n x_k = L tag3
$$
Combining both facts above we may fix any $epsilon > 0$ and find $N$ such that:
$$
forall epsilon > 0 exists N in Bbb N: forall m, n > N implies
begin{cases}
|x_n| < epsilon \
|y_n - y_m| < epsilon
end{cases}
$$
By convergence we also have that ${y_n}$ is a Cauchy sequence.
Since both sequences converge we as well know that for ${x_n}$:
$$
lim_{ntoinfty}sup x_n = lim_{ntoinfty}inf x_n = x_0tag4
$$
And for $y_n$:
$$
lim_{ntoinfty}sup y_n = lim_{ntoinfty}inf y_n = L tag5
$$
I've been playing around with those properties for a long time yet, but I couldn't find a way to combine them in order to show that:
$$
lim_{ntoinfty}nx_n = 0
$$
Could someone please assist me on that? I would prefer a hint rather than a complete proof. Thank you!
calculus sequences-and-series limits
asked Dec 27 '18 at 12:56
romanroman
2,36421224
$endgroup$
|
show 2 more comments
$begingroup$
Let ${x_n}$ denote a sequence such that:
$$
exists N in Bbb N: forall n > N implies 0 < x_{n+1} < x_n
$$
And the sequence ${y_n}$ is convergent, where ${y_n}$ is given by:
$$
y_n = sum_{k=1}^n x_n
$$
Prove that:
$$
lim_{ntoinfty} n x_n =0
$$
Below is what I've done so far.
Start with the first fact. We are given that a sequence $x_n$ is monotonically decreasing starting from some $N$. That by Weierstrass theorem means that the sequence is convergent to some $x_0 > 0$, namely:
$$
exists x_0 inBbb R_{>0}:lim_{ntoinfty} x_n = x_0 tag1
$$
Also it is bounded:
$$
exists m, M in Bbb R, forall n in Bbb N:m le x_n le M tag2
$$
We are also given that the sum is convergent, hence:
$$
exists L in Bbb R:lim_{ntoinfty} y_n = lim_{ntoinfty}sum_{k=1}^n x_k = L tag3
$$
Combining both facts above we may fix any $epsilon > 0$ and find $N$ such that:
$$
forall epsilon > 0 exists N in Bbb N: forall m, n > N implies
begin{cases}
|x_n| < epsilon \
|y_n - y_m| < epsilon
end{cases}
$$
By convergence we also have that ${y_n}$ is a Cauchy sequence.
Since both sequences converge we as well know that for ${x_n}$:
$$
lim_{ntoinfty}sup x_n = lim_{ntoinfty}inf x_n = x_0tag4
$$
And for $y_n$:
$$
lim_{ntoinfty}sup y_n = lim_{ntoinfty}inf y_n = L tag5
$$
I've been playing around with those properties for a long time yet, but I couldn't find a way to combine them in order to show that:
$$
lim_{ntoinfty}nx_n = 0
$$
Could someone please assist me on that? I would prefer a hint rather than a complete proof. Thank you!
calculus sequences-and-series limits
asked Dec 27 '18 at 12:56
romanroman
2,36421224
$endgroup$
$begingroup$
A monotonically decreasing sequence doesn't imply it tends to $0$. However, you could use the fact that $y_n$ convereges and hence $x_n$ must converge to $0$. Unless I'm missing something about Weierstrass Theorem. Which one is it? About 8 on wikipedia disambiguation page
$endgroup$
– Anvit
Dec 27 '18 at 13:06
$begingroup$
$n < 2^n$ so Cauchy Condensation might work
$endgroup$
– Anvit
Dec 27 '18 at 13:09
$begingroup$
@Anvit By this theorem I mean that any monotonic sequence ${x_n}$ has a finite limit in case it's bounded.
$endgroup$
– roman
Dec 27 '18 at 13:11
1
$begingroup$
No, that stament alone doesn't work, define $z_n = 1 + x_n$. $z_n$ also satifies the above properties but converges to 1 (if $x_n$ converegs to $0$)
$endgroup$
– Anvit
Dec 27 '18 at 13:14
1
$begingroup$
@Anvit is right when he says that $0 lt x_{n+1} lt x_n $ doesn't imply that $lim_{n to infty} x_n = 0$. $x_n = 1 + frac{1}{n}$ is a counter exemple. On the other hand, the fact that $y_n$ is convergent is a sufficient condition for $x_n$ convergence to $0$
$endgroup$
– F.Carette
Dec 27 '18 at 13:15
|
show 2 more comments
$begingroup$
Let ${x_n}$ denote a sequence such that:
$$
exists N in Bbb N: forall n > N implies 0 < x_{n+1} < x_n
$$
And the sequence ${y_n}$ is convergent, where ${y_n}$ is given by:
$$
y_n = sum_{k=1}^n x_n
$$
Prove that:
$$
lim_{ntoinfty} n x_n =0
$$
Below is what I've done so far.
Start with the first fact. We are given that a sequence $x_n$ is monotonically decreasing starting from some $N$. That by Weierstrass theorem means that the sequence is convergent to some $x_0 > 0$, namely:
$$
exists x_0 inBbb R_{>0}:lim_{ntoinfty} x_n = x_0 tag1
$$
Also it is bounded:
$$
exists m, M in Bbb R, forall n in Bbb N:m le x_n le M tag2
$$
We are also given that the sum is convergent, hence:
$$
exists L in Bbb R:lim_{ntoinfty} y_n = lim_{ntoinfty}sum_{k=1}^n x_k = L tag3
$$
Combining both facts above we may fix any $epsilon > 0$ and find $N$ such that:
$$
forall epsilon > 0 exists N in Bbb N: forall m, n > N implies
begin{cases}
|x_n| < epsilon \
|y_n - y_m| < epsilon
end{cases}
$$
By convergence we also have that ${y_n}$ is a Cauchy sequence.
Since both sequences converge we as well know that for ${x_n}$:
$$
lim_{ntoinfty}sup x_n = lim_{ntoinfty}inf x_n = x_0tag4
$$
And for $y_n$:
$$
lim_{ntoinfty}sup y_n = lim_{ntoinfty}inf y_n = L tag5
$$
I've been playing around with those properties for a long time yet, but I couldn't find a way to combine them in order to show that:
$$
lim_{ntoinfty}nx_n = 0
$$
Could someone please assist me on that? I would prefer a hint rather than a complete proof. Thank you!
calculus sequences-and-series limits
asked Dec 27 '18 at 12:56
romanroman
2,36421224
$endgroup$
Let ${x_n}$ denote a sequence such that:
$$
exists N in Bbb N: forall n > N implies 0 < x_{n+1} < x_n
$$
And the sequence ${y_n}$ is convergent, where ${y_n}$ is given by:
$$
y_n = sum_{k=1}^n x_n
$$
Prove that:
$$
lim_{ntoinfty} n x_n =0
$$
Below is what I've done so far.
Start with the first fact. We are given that a sequence $x_n$ is monotonically decreasing starting from some $N$. That by Weierstrass theorem means that the sequence is convergent to some $x_0 > 0$, namely:
$$
exists x_0 inBbb R_{>0}:lim_{ntoinfty} x_n = x_0 tag1
$$
Also it is bounded:
$$
exists m, M in Bbb R, forall n in Bbb N:m le x_n le M tag2
$$
We are also given that the sum is convergent, hence:
$$
exists L in Bbb R:lim_{ntoinfty} y_n = lim_{ntoinfty}sum_{k=1}^n x_k = L tag3
$$
Combining both facts above we may fix any $epsilon > 0$ and find $N$ such that:
$$
forall epsilon > 0 exists N in Bbb N: forall m, n > N implies
begin{cases}
|x_n| < epsilon \
|y_n - y_m| < epsilon
end{cases}
$$
By convergence we also have that ${y_n}$ is a Cauchy sequence.
Since both sequences converge we as well know that for ${x_n}$:
$$
lim_{ntoinfty}sup x_n = lim_{ntoinfty}inf x_n = x_0tag4
$$
And for $y_n$:
$$
lim_{ntoinfty}sup y_n = lim_{ntoinfty}inf y_n = L tag5
$$
I've been playing around with those properties for a long time yet, but I couldn't find a way to combine them in order to show that:
$$
lim_{ntoinfty}nx_n = 0
$$
Could someone please assist me on that? I would prefer a hint rather than a complete proof. Thank you!
calculus sequences-and-series limits
calculus sequences-and-series limits
asked Dec 27 '18 at 12:56
romanroman
2,36421224
asked Dec 27 '18 at 12:56
romanroman
2,36421224
edited Dec 27 '18 at 13:17
roman
asked Dec 27 '18 at 12:56
romanroman
2,36421224
asked Dec 27 '18 at 12:56
romanroman
2,36421224
asked Dec 27 '18 at 12:56
romanroman
2,36421224
2,36421224
$begingroup$
A monotonically decreasing sequence doesn't imply it tends to $0$. However, you could use the fact that $y_n$ convereges and hence $x_n$ must converge to $0$. Unless I'm missing something about Weierstrass Theorem. Which one is it? About 8 on wikipedia disambiguation page
$endgroup$
– Anvit
Dec 27 '18 at 13:06
$begingroup$
$n < 2^n$ so Cauchy Condensation might work
$endgroup$
– Anvit
Dec 27 '18 at 13:09
$begingroup$
@Anvit By this theorem I mean that any monotonic sequence ${x_n}$ has a finite limit in case it's bounded.
$endgroup$
– roman
Dec 27 '18 at 13:11
1
$begingroup$
No, that stament alone doesn't work, define $z_n = 1 + x_n$. $z_n$ also satifies the above properties but converges to 1 (if $x_n$ converegs to $0$)
$endgroup$
– Anvit
Dec 27 '18 at 13:14
1
$begingroup$
@Anvit is right when he says that $0 lt x_{n+1} lt x_n $ doesn't imply that $lim_{n to infty} x_n = 0$. $x_n = 1 + frac{1}{n}$ is a counter exemple. On the other hand, the fact that $y_n$ is convergent is a sufficient condition for $x_n$ convergence to $0$
$endgroup$
– F.Carette
Dec 27 '18 at 13:15
|
show 2 more comments
$begingroup$
A monotonically decreasing sequence doesn't imply it tends to $0$. However, you could use the fact that $y_n$ convereges and hence $x_n$ must converge to $0$. Unless I'm missing something about Weierstrass Theorem. Which one is it? About 8 on wikipedia disambiguation page
$endgroup$
– Anvit
Dec 27 '18 at 13:06
$begingroup$
$n < 2^n$ so Cauchy Condensation might work
$endgroup$
– Anvit
Dec 27 '18 at 13:09
$begingroup$
@Anvit By this theorem I mean that any monotonic sequence ${x_n}$ has a finite limit in case it's bounded.
$endgroup$
– roman
Dec 27 '18 at 13:11
1
$begingroup$
No, that stament alone doesn't work, define $z_n = 1 + x_n$. $z_n$ also satifies the above properties but converges to 1 (if $x_n$ converegs to $0$)
$endgroup$
– Anvit
Dec 27 '18 at 13:14
1
$begingroup$
@Anvit is right when he says that $0 lt x_{n+1} lt x_n $ doesn't imply that $lim_{n to infty} x_n = 0$. $x_n = 1 + frac{1}{n}$ is a counter exemple. On the other hand, the fact that $y_n$ is convergent is a sufficient condition for $x_n$ convergence to $0$
$endgroup$
– F.Carette
Dec 27 '18 at 13:15
$begingroup$
A monotonically decreasing sequence doesn't imply it tends to $0$. However, you could use the fact that $y_n$ convereges and hence $x_n$ must converge to $0$. Unless I'm missing something about Weierstrass Theorem. Which one is it? About 8 on wikipedia disambiguation page
$endgroup$
– Anvit
Dec 27 '18 at 13:06
$begingroup$
A monotonically decreasing sequence doesn't imply it tends to $0$. However, you could use the fact that $y_n$ convereges and hence $x_n$ must converge to $0$. Unless I'm missing something about Weierstrass Theorem. Which one is it? About 8 on wikipedia disambiguation page
$endgroup$
– Anvit
Dec 27 '18 at 13:06
$begingroup$
$n < 2^n$ so Cauchy Condensation might work
$endgroup$
– Anvit
Dec 27 '18 at 13:09
$begingroup$
$n < 2^n$ so Cauchy Condensation might work
$endgroup$
– Anvit
Dec 27 '18 at 13:09
$begingroup$
@Anvit By this theorem I mean that any monotonic sequence ${x_n}$ has a finite limit in case it's bounded.
$endgroup$
– roman
Dec 27 '18 at 13:11
$begingroup$
@Anvit By this theorem I mean that any monotonic sequence ${x_n}$ has a finite limit in case it's bounded.
$endgroup$
– roman
Dec 27 '18 at 13:11
1
1
$begingroup$
No, that stament alone doesn't work, define $z_n = 1 + x_n$. $z_n$ also satifies the above properties but converges to 1 (if $x_n$ converegs to $0$)
$endgroup$
– Anvit
Dec 27 '18 at 13:14
$begingroup$
No, that stament alone doesn't work, define $z_n = 1 + x_n$. $z_n$ also satifies the above properties but converges to 1 (if $x_n$ converegs to $0$)
$endgroup$
– Anvit
Dec 27 '18 at 13:14
1
1
$begingroup$
@Anvit is right when he says that $0 lt x_{n+1} lt x_n $ doesn't imply that $lim_{n to infty} x_n = 0$. $x_n = 1 + frac{1}{n}$ is a counter exemple. On the other hand, the fact that $y_n$ is convergent is a sufficient condition for $x_n$ convergence to $0$
$endgroup$
– F.Carette
Dec 27 '18 at 13:15
$begingroup$
@Anvit is right when he says that $0 lt x_{n+1} lt x_n $ doesn't imply that $lim_{n to infty} x_n = 0$. $x_n = 1 + frac{1}{n}$ is a counter exemple. On the other hand, the fact that $y_n$ is convergent is a sufficient condition for $x_n$ convergence to $0$
$endgroup$
– F.Carette
Dec 27 '18 at 13:15
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Hint: $lim_{ntoinf} (y_{2n}-y_n) = 0$
Comment if you require more hints
Also, as I said, you can also use Cauchy Condensation test as an Alternative solution
answered Dec 27 '18 at 13:18
AnvitAnvit
1,733419
$endgroup$
$begingroup$
Thank you, I've added an answer to this question, hopefully correct.
$endgroup$
– roman
Dec 27 '18 at 14:21
add a comment |
$begingroup$
My try to use hint by @Anvit
Lemma:
$$
exists L in Bbb R: lim_{ntoinfty} y_n= lim_{ntoinfty}sum_{k=1}^n x_n = L implies lim_{ntoinfty} x_n = 0
$$
Proof. Since ${y_n}$ converges it must satisfy Cauchy Criterion. Consider the following:
$$
forall epsilon > 0 exists N in Bbb N: forall n, m > N implies |y_n -y_m| < epsilon
$$
Take $m=n-1$, then we have that:
$$
|x_n| = |y_n - y_{n-1}| < epsilon
$$
Which would mean:
$$
forall epsilon > 0 exists N inBbb N: forall n > N implies |x_n| < epsilon stackrel{text{def}}{iff} lim_{ntoinfty} x_n = 0
$$
So by Lemma we obtain that $x_n$ is convergent to $0$:
$$
lim_{ntoinfty}x_n = 0
$$
As desired $Box$.
Consider ${y_n}$ and $m = 2n > n$, then by Cauchy Criterion:
$$
begin{align}
|y_n - y_m| &= |y_m - y_n| \
&= |y_{2n} - y_n| \
&= left| sum_{k=n+1}^{2n}x_k right| \
&= sum_{k=n+1}^{2n}x_k < epsilon
end{align}
$$
We know that $x_n > 0$ and is monotonically decreasing towards $0$ starting from some $N$ and thus:
$$
underbrace{x_{2n} + x_{2n} + cdots + x_{2n}}_{n text{times}} < x_{n+1} + x_{n+2} + cdots + x_{2n-1} + x_{2n} = sum_{k=n+1}^{2n}x_k < epsilon \
nx_{2n} < sum_{k=n+1}^{2n}x_k < epsilon \
2ncdot x_{2n} < 2sum_{k=n+1}^{2n}x_k < 2epsilon
$$
We know that:
$$
lim_{ntoinfty} 2sum_{k=n+1}^{2n}x_k = 0
$$
Applying squeeze theorem to that:
$$
0 le lim_{ntoinfty} 2ncdot x_{2n} le lim_{ntoinfty} 2sum_{k=n+1}^{2n}x_k
$$
Using the fact that $x_n$ is monotonically decreasing we may conclude:
$$
0 le lim_{ntoinfty} 2ncdot x_{2n} le 0 iff \
lim_{ntoinfty} 2ncdot x_{2n} = 0 iff \
lim_{ntoinfty} ncdot x_{n} = 0
$$
answered Dec 27 '18 at 14:20
romanroman
2,36421224
$endgroup$
$begingroup$
Little problem I see is that your last iff statement isn't correct. Consider a sequence whose even numbers tend to zero but odds dont
$endgroup$
– Anvit
Jan 7 at 16:01
$begingroup$
@Anvit But then the sequence is not going to be monotone, is it? which violates the initial conditions, or am i missing something?
$endgroup$
– roman
Jan 7 at 16:09
1
$begingroup$
Yep, you're right. However you should mention this else it can lead to confusion. (And possible loss of marks :P )
$endgroup$
– Anvit
Jan 7 at 16:11
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: $lim_{ntoinf} (y_{2n}-y_n) = 0$
Comment if you require more hints
Also, as I said, you can also use Cauchy Condensation test as an Alternative solution
answered Dec 27 '18 at 13:18
AnvitAnvit
1,733419
$endgroup$
$begingroup$
Thank you, I've added an answer to this question, hopefully correct.
$endgroup$
– roman
Dec 27 '18 at 14:21
add a comment |
$begingroup$
Hint: $lim_{ntoinf} (y_{2n}-y_n) = 0$
Comment if you require more hints
Also, as I said, you can also use Cauchy Condensation test as an Alternative solution
answered Dec 27 '18 at 13:18
AnvitAnvit
1,733419
$endgroup$
$begingroup$
Thank you, I've added an answer to this question, hopefully correct.
$endgroup$
– roman
Dec 27 '18 at 14:21
add a comment |
$begingroup$
Hint: $lim_{ntoinf} (y_{2n}-y_n) = 0$
Comment if you require more hints
Also, as I said, you can also use Cauchy Condensation test as an Alternative solution
answered Dec 27 '18 at 13:18
AnvitAnvit
1,733419
$endgroup$
Hint: $lim_{ntoinf} (y_{2n}-y_n) = 0$
Comment if you require more hints
Also, as I said, you can also use Cauchy Condensation test as an Alternative solution
answered Dec 27 '18 at 13:18
AnvitAnvit
1,733419
answered Dec 27 '18 at 13:18
AnvitAnvit
1,733419
answered Dec 27 '18 at 13:18
AnvitAnvit
1,733419
answered Dec 27 '18 at 13:18
AnvitAnvit
1,733419
1,733419
$begingroup$
Thank you, I've added an answer to this question, hopefully correct.
$endgroup$
– roman
Dec 27 '18 at 14:21
add a comment |
$begingroup$
Thank you, I've added an answer to this question, hopefully correct.
$endgroup$
– roman
Dec 27 '18 at 14:21
$begingroup$
Thank you, I've added an answer to this question, hopefully correct.
$endgroup$
– roman
Dec 27 '18 at 14:21
$begingroup$
Thank you, I've added an answer to this question, hopefully correct.
$endgroup$
– roman
Dec 27 '18 at 14:21
add a comment |
$begingroup$
My try to use hint by @Anvit
Lemma:
$$
exists L in Bbb R: lim_{ntoinfty} y_n= lim_{ntoinfty}sum_{k=1}^n x_n = L implies lim_{ntoinfty} x_n = 0
$$
Proof. Since ${y_n}$ converges it must satisfy Cauchy Criterion. Consider the following:
$$
forall epsilon > 0 exists N in Bbb N: forall n, m > N implies |y_n -y_m| < epsilon
$$
Take $m=n-1$, then we have that:
$$
|x_n| = |y_n - y_{n-1}| < epsilon
$$
Which would mean:
$$
forall epsilon > 0 exists N inBbb N: forall n > N implies |x_n| < epsilon stackrel{text{def}}{iff} lim_{ntoinfty} x_n = 0
$$
So by Lemma we obtain that $x_n$ is convergent to $0$:
$$
lim_{ntoinfty}x_n = 0
$$
As desired $Box$.
Consider ${y_n}$ and $m = 2n > n$, then by Cauchy Criterion:
$$
begin{align}
|y_n - y_m| &= |y_m - y_n| \
&= |y_{2n} - y_n| \
&= left| sum_{k=n+1}^{2n}x_k right| \
&= sum_{k=n+1}^{2n}x_k < epsilon
end{align}
$$
We know that $x_n > 0$ and is monotonically decreasing towards $0$ starting from some $N$ and thus:
$$
underbrace{x_{2n} + x_{2n} + cdots + x_{2n}}_{n text{times}} < x_{n+1} + x_{n+2} + cdots + x_{2n-1} + x_{2n} = sum_{k=n+1}^{2n}x_k < epsilon \
nx_{2n} < sum_{k=n+1}^{2n}x_k < epsilon \
2ncdot x_{2n} < 2sum_{k=n+1}^{2n}x_k < 2epsilon
$$
We know that:
$$
lim_{ntoinfty} 2sum_{k=n+1}^{2n}x_k = 0
$$
Applying squeeze theorem to that:
$$
0 le lim_{ntoinfty} 2ncdot x_{2n} le lim_{ntoinfty} 2sum_{k=n+1}^{2n}x_k
$$
Using the fact that $x_n$ is monotonically decreasing we may conclude:
$$
0 le lim_{ntoinfty} 2ncdot x_{2n} le 0 iff \
lim_{ntoinfty} 2ncdot x_{2n} = 0 iff \
lim_{ntoinfty} ncdot x_{n} = 0
$$
answered Dec 27 '18 at 14:20
romanroman
2,36421224
$endgroup$
$begingroup$
Little problem I see is that your last iff statement isn't correct. Consider a sequence whose even numbers tend to zero but odds dont
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– Anvit
Jan 7 at 16:01
$begingroup$
@Anvit But then the sequence is not going to be monotone, is it? which violates the initial conditions, or am i missing something?
$endgroup$
– roman
Jan 7 at 16:09
1
$begingroup$
Yep, you're right. However you should mention this else it can lead to confusion. (And possible loss of marks :P )
$endgroup$
– Anvit
Jan 7 at 16:11
add a comment |
$begingroup$
My try to use hint by @Anvit
Lemma:
$$
exists L in Bbb R: lim_{ntoinfty} y_n= lim_{ntoinfty}sum_{k=1}^n x_n = L implies lim_{ntoinfty} x_n = 0
$$
Proof. Since ${y_n}$ converges it must satisfy Cauchy Criterion. Consider the following:
$$
forall epsilon > 0 exists N in Bbb N: forall n, m > N implies |y_n -y_m| < epsilon
$$
Take $m=n-1$, then we have that:
$$
|x_n| = |y_n - y_{n-1}| < epsilon
$$
Which would mean:
$$
forall epsilon > 0 exists N inBbb N: forall n > N implies |x_n| < epsilon stackrel{text{def}}{iff} lim_{ntoinfty} x_n = 0
$$
So by Lemma we obtain that $x_n$ is convergent to $0$:
$$
lim_{ntoinfty}x_n = 0
$$
As desired $Box$.
Consider ${y_n}$ and $m = 2n > n$, then by Cauchy Criterion:
$$
begin{align}
|y_n - y_m| &= |y_m - y_n| \
&= |y_{2n} - y_n| \
&= left| sum_{k=n+1}^{2n}x_k right| \
&= sum_{k=n+1}^{2n}x_k < epsilon
end{align}
$$
We know that $x_n > 0$ and is monotonically decreasing towards $0$ starting from some $N$ and thus:
$$
underbrace{x_{2n} + x_{2n} + cdots + x_{2n}}_{n text{times}} < x_{n+1} + x_{n+2} + cdots + x_{2n-1} + x_{2n} = sum_{k=n+1}^{2n}x_k < epsilon \
nx_{2n} < sum_{k=n+1}^{2n}x_k < epsilon \
2ncdot x_{2n} < 2sum_{k=n+1}^{2n}x_k < 2epsilon
$$
We know that:
$$
lim_{ntoinfty} 2sum_{k=n+1}^{2n}x_k = 0
$$
Applying squeeze theorem to that:
$$
0 le lim_{ntoinfty} 2ncdot x_{2n} le lim_{ntoinfty} 2sum_{k=n+1}^{2n}x_k
$$
Using the fact that $x_n$ is monotonically decreasing we may conclude:
$$
0 le lim_{ntoinfty} 2ncdot x_{2n} le 0 iff \
lim_{ntoinfty} 2ncdot x_{2n} = 0 iff \
lim_{ntoinfty} ncdot x_{n} = 0
$$
answered Dec 27 '18 at 14:20
romanroman
2,36421224
$endgroup$
$begingroup$
Little problem I see is that your last iff statement isn't correct. Consider a sequence whose even numbers tend to zero but odds dont
$endgroup$
– Anvit
Jan 7 at 16:01
$begingroup$
@Anvit But then the sequence is not going to be monotone, is it? which violates the initial conditions, or am i missing something?
$endgroup$
– roman
Jan 7 at 16:09
1
$begingroup$
Yep, you're right. However you should mention this else it can lead to confusion. (And possible loss of marks :P )
$endgroup$
– Anvit
Jan 7 at 16:11
add a comment |
$begingroup$
My try to use hint by @Anvit
Lemma:
$$
exists L in Bbb R: lim_{ntoinfty} y_n= lim_{ntoinfty}sum_{k=1}^n x_n = L implies lim_{ntoinfty} x_n = 0
$$
Proof. Since ${y_n}$ converges it must satisfy Cauchy Criterion. Consider the following:
$$
forall epsilon > 0 exists N in Bbb N: forall n, m > N implies |y_n -y_m| < epsilon
$$
Take $m=n-1$, then we have that:
$$
|x_n| = |y_n - y_{n-1}| < epsilon
$$
Which would mean:
$$
forall epsilon > 0 exists N inBbb N: forall n > N implies |x_n| < epsilon stackrel{text{def}}{iff} lim_{ntoinfty} x_n = 0
$$
So by Lemma we obtain that $x_n$ is convergent to $0$:
$$
lim_{ntoinfty}x_n = 0
$$
As desired $Box$.
Consider ${y_n}$ and $m = 2n > n$, then by Cauchy Criterion:
$$
begin{align}
|y_n - y_m| &= |y_m - y_n| \
&= |y_{2n} - y_n| \
&= left| sum_{k=n+1}^{2n}x_k right| \
&= sum_{k=n+1}^{2n}x_k < epsilon
end{align}
$$
We know that $x_n > 0$ and is monotonically decreasing towards $0$ starting from some $N$ and thus:
$$
underbrace{x_{2n} + x_{2n} + cdots + x_{2n}}_{n text{times}} < x_{n+1} + x_{n+2} + cdots + x_{2n-1} + x_{2n} = sum_{k=n+1}^{2n}x_k < epsilon \
nx_{2n} < sum_{k=n+1}^{2n}x_k < epsilon \
2ncdot x_{2n} < 2sum_{k=n+1}^{2n}x_k < 2epsilon
$$
We know that:
$$
lim_{ntoinfty} 2sum_{k=n+1}^{2n}x_k = 0
$$
Applying squeeze theorem to that:
$$
0 le lim_{ntoinfty} 2ncdot x_{2n} le lim_{ntoinfty} 2sum_{k=n+1}^{2n}x_k
$$
Using the fact that $x_n$ is monotonically decreasing we may conclude:
$$
0 le lim_{ntoinfty} 2ncdot x_{2n} le 0 iff \
lim_{ntoinfty} 2ncdot x_{2n} = 0 iff \
lim_{ntoinfty} ncdot x_{n} = 0
$$
answered Dec 27 '18 at 14:20
romanroman
2,36421224
$endgroup$
My try to use hint by @Anvit
Lemma:
$$
exists L in Bbb R: lim_{ntoinfty} y_n= lim_{ntoinfty}sum_{k=1}^n x_n = L implies lim_{ntoinfty} x_n = 0
$$
Proof. Since ${y_n}$ converges it must satisfy Cauchy Criterion. Consider the following:
$$
forall epsilon > 0 exists N in Bbb N: forall n, m > N implies |y_n -y_m| < epsilon
$$
Take $m=n-1$, then we have that:
$$
|x_n| = |y_n - y_{n-1}| < epsilon
$$
Which would mean:
$$
forall epsilon > 0 exists N inBbb N: forall n > N implies |x_n| < epsilon stackrel{text{def}}{iff} lim_{ntoinfty} x_n = 0
$$
So by Lemma we obtain that $x_n$ is convergent to $0$:
$$
lim_{ntoinfty}x_n = 0
$$
As desired $Box$.
Consider ${y_n}$ and $m = 2n > n$, then by Cauchy Criterion:
$$
begin{align}
|y_n - y_m| &= |y_m - y_n| \
&= |y_{2n} - y_n| \
&= left| sum_{k=n+1}^{2n}x_k right| \
&= sum_{k=n+1}^{2n}x_k < epsilon
end{align}
$$
We know that $x_n > 0$ and is monotonically decreasing towards $0$ starting from some $N$ and thus:
$$
underbrace{x_{2n} + x_{2n} + cdots + x_{2n}}_{n text{times}} < x_{n+1} + x_{n+2} + cdots + x_{2n-1} + x_{2n} = sum_{k=n+1}^{2n}x_k < epsilon \
nx_{2n} < sum_{k=n+1}^{2n}x_k < epsilon \
2ncdot x_{2n} < 2sum_{k=n+1}^{2n}x_k < 2epsilon
$$
We know that:
$$
lim_{ntoinfty} 2sum_{k=n+1}^{2n}x_k = 0
$$
Applying squeeze theorem to that:
$$
0 le lim_{ntoinfty} 2ncdot x_{2n} le lim_{ntoinfty} 2sum_{k=n+1}^{2n}x_k
$$
Using the fact that $x_n$ is monotonically decreasing we may conclude:
$$
0 le lim_{ntoinfty} 2ncdot x_{2n} le 0 iff \
lim_{ntoinfty} 2ncdot x_{2n} = 0 iff \
lim_{ntoinfty} ncdot x_{n} = 0
$$
answered Dec 27 '18 at 14:20
romanroman
2,36421224
edited Jan 7 at 16:14
answered Dec 27 '18 at 14:20
romanroman
2,36421224
answered Dec 27 '18 at 14:20
romanroman
2,36421224
answered Dec 27 '18 at 14:20
romanroman
2,36421224
2,36421224
$begingroup$
Little problem I see is that your last iff statement isn't correct. Consider a sequence whose even numbers tend to zero but odds dont
$endgroup$
– Anvit
Jan 7 at 16:01
$begingroup$
@Anvit But then the sequence is not going to be monotone, is it? which violates the initial conditions, or am i missing something?
$endgroup$
– roman
Jan 7 at 16:09
1
$begingroup$
Yep, you're right. However you should mention this else it can lead to confusion. (And possible loss of marks :P )
$endgroup$
– Anvit
Jan 7 at 16:11
add a comment |
$begingroup$
Little problem I see is that your last iff statement isn't correct. Consider a sequence whose even numbers tend to zero but odds dont
$endgroup$
– Anvit
Jan 7 at 16:01
$begingroup$
@Anvit But then the sequence is not going to be monotone, is it? which violates the initial conditions, or am i missing something?
$endgroup$
– roman
Jan 7 at 16:09
1
$begingroup$
Yep, you're right. However you should mention this else it can lead to confusion. (And possible loss of marks :P )
$endgroup$
– Anvit
Jan 7 at 16:11
$begingroup$
Little problem I see is that your last iff statement isn't correct. Consider a sequence whose even numbers tend to zero but odds dont
$endgroup$
– Anvit
Jan 7 at 16:01
$begingroup$
Little problem I see is that your last iff statement isn't correct. Consider a sequence whose even numbers tend to zero but odds dont
$endgroup$
– Anvit
Jan 7 at 16:01
$begingroup$
@Anvit But then the sequence is not going to be monotone, is it? which violates the initial conditions, or am i missing something?
$endgroup$
– roman
Jan 7 at 16:09
$begingroup$
@Anvit But then the sequence is not going to be monotone, is it? which violates the initial conditions, or am i missing something?
$endgroup$
– roman
Jan 7 at 16:09
1
1
$begingroup$
Yep, you're right. However you should mention this else it can lead to confusion. (And possible loss of marks :P )
$endgroup$
– Anvit
Jan 7 at 16:11
$begingroup$
Yep, you're right. However you should mention this else it can lead to confusion. (And possible loss of marks :P )
$endgroup$
– Anvit
Jan 7 at 16:11
add a comment |
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$begingroup$
A monotonically decreasing sequence doesn't imply it tends to $0$. However, you could use the fact that $y_n$ convereges and hence $x_n$ must converge to $0$. Unless I'm missing something about Weierstrass Theorem. Which one is it? About 8 on wikipedia disambiguation page
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– Anvit
Dec 27 '18 at 13:06
$begingroup$
$n < 2^n$ so Cauchy Condensation might work
$endgroup$
– Anvit
Dec 27 '18 at 13:09
$begingroup$
@Anvit By this theorem I mean that any monotonic sequence ${x_n}$ has a finite limit in case it's bounded.
$endgroup$
– roman
Dec 27 '18 at 13:11
1
$begingroup$
No, that stament alone doesn't work, define $z_n = 1 + x_n$. $z_n$ also satifies the above properties but converges to 1 (if $x_n$ converegs to $0$)
$endgroup$
– Anvit
Dec 27 '18 at 13:14
1
$begingroup$
@Anvit is right when he says that $0 lt x_{n+1} lt x_n $ doesn't imply that $lim_{n to infty} x_n = 0$. $x_n = 1 + frac{1}{n}$ is a counter exemple. On the other hand, the fact that $y_n$ is convergent is a sufficient condition for $x_n$ convergence to $0$
$endgroup$
– F.Carette
Dec 27 '18 at 13:15