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Let $x_n$ be sequence of real numbers such that $lim_ntoinfty(x_n+1-x_n)=c$
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Let $x_n$ be sequence of real numbers such that $lim_ntoinfty(x_n+1-x_n)=c$ ,where c is a positive real number .Then the sequence $dfracx_nn$
A) is not bounded
B) is bounded bu no convergent
C) converges to c
D) converges to 0
how to define $x_n$?
if $x_n=n$ then $(x_n+1-x_n) =1$ and $lim_ntoinfty(x_n+1-x_n)=1=c$ and $(lim_ntoinftydfracx_nn)=1=c$ and
if $x_n=dfrac1n$ then $(x_n+1-x_n) =(dfrac1n^2-dfrac1(n+1)^2)$ and $lim_ntoinfty(x_n+1-x_n)=0=c$ and $(lim_ntoinftydfracx_nn)=0=c$
i.e. $lim_ntoinfty(x_n)=c$
real-analysis
asked Jul 30 at 15:22
sejy
293
add a comment |Â
up vote
0
down vote
favorite
Let $x_n$ be sequence of real numbers such that $lim_ntoinfty(x_n+1-x_n)=c$ ,where c is a positive real number .Then the sequence $dfracx_nn$
A) is not bounded
B) is bounded bu no convergent
C) converges to c
D) converges to 0
how to define $x_n$?
if $x_n=n$ then $(x_n+1-x_n) =1$ and $lim_ntoinfty(x_n+1-x_n)=1=c$ and $(lim_ntoinftydfracx_nn)=1=c$ and
if $x_n=dfrac1n$ then $(x_n+1-x_n) =(dfrac1n^2-dfrac1(n+1)^2)$ and $lim_ntoinfty(x_n+1-x_n)=0=c$ and $(lim_ntoinftydfracx_nn)=0=c$
i.e. $lim_ntoinfty(x_n)=c$
real-analysis
asked Jul 30 at 15:22
sejy
293
4
This is a special case of the Cesaro-Stolz lemma (en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem) The sequence converges to $c$. Since it is a special case of the lemma, it should hopefully be easier to prove than the full lemma.
– Dylan
Jul 30 at 15:27
You can reduce to the special case of $c=0$ by considering the sequence $y_n=x_n-nc$ instead. As for "how to define $x_n$?", you can't, really. You're supposed to show that something is true for all sequences $x_n$ fulfilling some property. That means that defining $x_n$ to be one of them won't help at all to prove what you're asked.
– Arthur
Jul 30 at 15:33
Also, since it is multiple choice, you don't actually need to prove that C is always true, you just need to find examples where A, B, and D are not true. You've already done that, and so C is the correct answer.
– Dylan
Jul 30 at 15:34
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $x_n$ be sequence of real numbers such that $lim_ntoinfty(x_n+1-x_n)=c$ ,where c is a positive real number .Then the sequence $dfracx_nn$
A) is not bounded
B) is bounded bu no convergent
C) converges to c
D) converges to 0
how to define $x_n$?
if $x_n=n$ then $(x_n+1-x_n) =1$ and $lim_ntoinfty(x_n+1-x_n)=1=c$ and $(lim_ntoinftydfracx_nn)=1=c$ and
if $x_n=dfrac1n$ then $(x_n+1-x_n) =(dfrac1n^2-dfrac1(n+1)^2)$ and $lim_ntoinfty(x_n+1-x_n)=0=c$ and $(lim_ntoinftydfracx_nn)=0=c$
i.e. $lim_ntoinfty(x_n)=c$
real-analysis
asked Jul 30 at 15:22
sejy
293
Let $x_n$ be sequence of real numbers such that $lim_ntoinfty(x_n+1-x_n)=c$ ,where c is a positive real number .Then the sequence $dfracx_nn$
A) is not bounded
B) is bounded bu no convergent
C) converges to c
D) converges to 0
how to define $x_n$?
if $x_n=n$ then $(x_n+1-x_n) =1$ and $lim_ntoinfty(x_n+1-x_n)=1=c$ and $(lim_ntoinftydfracx_nn)=1=c$ and
if $x_n=dfrac1n$ then $(x_n+1-x_n) =(dfrac1n^2-dfrac1(n+1)^2)$ and $lim_ntoinfty(x_n+1-x_n)=0=c$ and $(lim_ntoinftydfracx_nn)=0=c$
i.e. $lim_ntoinfty(x_n)=c$
real-analysis
asked Jul 30 at 15:22
sejy
293
asked Jul 30 at 15:22
sejy
293
asked Jul 30 at 15:22
sejy
293
asked Jul 30 at 15:22
sejy
293
293
4
This is a special case of the Cesaro-Stolz lemma (en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem) The sequence converges to $c$. Since it is a special case of the lemma, it should hopefully be easier to prove than the full lemma.
– Dylan
Jul 30 at 15:27
You can reduce to the special case of $c=0$ by considering the sequence $y_n=x_n-nc$ instead. As for "how to define $x_n$?", you can't, really. You're supposed to show that something is true for all sequences $x_n$ fulfilling some property. That means that defining $x_n$ to be one of them won't help at all to prove what you're asked.
– Arthur
Jul 30 at 15:33
Also, since it is multiple choice, you don't actually need to prove that C is always true, you just need to find examples where A, B, and D are not true. You've already done that, and so C is the correct answer.
– Dylan
Jul 30 at 15:34
add a comment |Â
4
This is a special case of the Cesaro-Stolz lemma (en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem) The sequence converges to $c$. Since it is a special case of the lemma, it should hopefully be easier to prove than the full lemma.
– Dylan
Jul 30 at 15:27
You can reduce to the special case of $c=0$ by considering the sequence $y_n=x_n-nc$ instead. As for "how to define $x_n$?", you can't, really. You're supposed to show that something is true for all sequences $x_n$ fulfilling some property. That means that defining $x_n$ to be one of them won't help at all to prove what you're asked.
– Arthur
Jul 30 at 15:33
Also, since it is multiple choice, you don't actually need to prove that C is always true, you just need to find examples where A, B, and D are not true. You've already done that, and so C is the correct answer.
– Dylan
Jul 30 at 15:34
4
4
This is a special case of the Cesaro-Stolz lemma (en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem) The sequence converges to $c$. Since it is a special case of the lemma, it should hopefully be easier to prove than the full lemma.
– Dylan
Jul 30 at 15:27
This is a special case of the Cesaro-Stolz lemma (en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem) The sequence converges to $c$. Since it is a special case of the lemma, it should hopefully be easier to prove than the full lemma.
– Dylan
Jul 30 at 15:27
You can reduce to the special case of $c=0$ by considering the sequence $y_n=x_n-nc$ instead. As for "how to define $x_n$?", you can't, really. You're supposed to show that something is true for all sequences $x_n$ fulfilling some property. That means that defining $x_n$ to be one of them won't help at all to prove what you're asked.
– Arthur
Jul 30 at 15:33
You can reduce to the special case of $c=0$ by considering the sequence $y_n=x_n-nc$ instead. As for "how to define $x_n$?", you can't, really. You're supposed to show that something is true for all sequences $x_n$ fulfilling some property. That means that defining $x_n$ to be one of them won't help at all to prove what you're asked.
– Arthur
Jul 30 at 15:33
Also, since it is multiple choice, you don't actually need to prove that C is always true, you just need to find examples where A, B, and D are not true. You've already done that, and so C is the correct answer.
– Dylan
Jul 30 at 15:34
Also, since it is multiple choice, you don't actually need to prove that C is always true, you just need to find examples where A, B, and D are not true. You've already done that, and so C is the correct answer.
– Dylan
Jul 30 at 15:34
add a comment |Â
1 Answer
1
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oldest
votes
up vote
2
down vote
Since
$$
lim_n to infty (x_n + 1 - x_n) = c,
$$
we know that for every $varepsilon > 0$ there exists $N > 0$ such that
$$
c - varepsilon < x_n + 1 - x_n < c + varepsilon
$$
for all $n geq N$.
We thus have that for every $m > N$ that
$$
x_N + (m - N)(c - epsilon) < x_m < x_N + (m - N)(c + varepsilon)
$$
or equivalently
$$
fracx_Nm + left(1 - fracNmright)(c - varepsilon) < fracx_mm < fracx_Nm + left(1 - fracNmright)(c + varepsilon).
$$
We have that
$$
lim_m to infty fracx_Nm + left(1 - fracNmright)(c - varepsilon) = c - varepsilon
$$
and so there exists $M_1 > 0$ such that
$$
c - 2varepsilon < fracx_Nm + left(1 - fracNmright)(c - varepsilon)
$$
for all $m geq M_1$. Similarly, there exists $M_2 > 0$ such that
$$
fracx_Nm + left(1 - fracNmright)(c + varepsilon) < c + 2varepsilon
$$
for all $m geq M_2$. Thus we have that
$$
c - 2varepsilon < fracx_mm < c + 2varepsilon
$$
for all $m > M = max(M_1, M_2, N)$.
Since such an $M$ exists for every $varepsilon > 0$, we have that
$$
lim_m to infty fracx_mm = c.
$$
answered Jul 30 at 15:44
Dylan
5,67231134
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Since
$$
lim_n to infty (x_n + 1 - x_n) = c,
$$
we know that for every $varepsilon > 0$ there exists $N > 0$ such that
$$
c - varepsilon < x_n + 1 - x_n < c + varepsilon
$$
for all $n geq N$.
We thus have that for every $m > N$ that
$$
x_N + (m - N)(c - epsilon) < x_m < x_N + (m - N)(c + varepsilon)
$$
or equivalently
$$
fracx_Nm + left(1 - fracNmright)(c - varepsilon) < fracx_mm < fracx_Nm + left(1 - fracNmright)(c + varepsilon).
$$
We have that
$$
lim_m to infty fracx_Nm + left(1 - fracNmright)(c - varepsilon) = c - varepsilon
$$
and so there exists $M_1 > 0$ such that
$$
c - 2varepsilon < fracx_Nm + left(1 - fracNmright)(c - varepsilon)
$$
for all $m geq M_1$. Similarly, there exists $M_2 > 0$ such that
$$
fracx_Nm + left(1 - fracNmright)(c + varepsilon) < c + 2varepsilon
$$
for all $m geq M_2$. Thus we have that
$$
c - 2varepsilon < fracx_mm < c + 2varepsilon
$$
for all $m > M = max(M_1, M_2, N)$.
Since such an $M$ exists for every $varepsilon > 0$, we have that
$$
lim_m to infty fracx_mm = c.
$$
answered Jul 30 at 15:44
Dylan
5,67231134
add a comment |Â
up vote
2
down vote
Since
$$
lim_n to infty (x_n + 1 - x_n) = c,
$$
we know that for every $varepsilon > 0$ there exists $N > 0$ such that
$$
c - varepsilon < x_n + 1 - x_n < c + varepsilon
$$
for all $n geq N$.
We thus have that for every $m > N$ that
$$
x_N + (m - N)(c - epsilon) < x_m < x_N + (m - N)(c + varepsilon)
$$
or equivalently
$$
fracx_Nm + left(1 - fracNmright)(c - varepsilon) < fracx_mm < fracx_Nm + left(1 - fracNmright)(c + varepsilon).
$$
We have that
$$
lim_m to infty fracx_Nm + left(1 - fracNmright)(c - varepsilon) = c - varepsilon
$$
and so there exists $M_1 > 0$ such that
$$
c - 2varepsilon < fracx_Nm + left(1 - fracNmright)(c - varepsilon)
$$
for all $m geq M_1$. Similarly, there exists $M_2 > 0$ such that
$$
fracx_Nm + left(1 - fracNmright)(c + varepsilon) < c + 2varepsilon
$$
for all $m geq M_2$. Thus we have that
$$
c - 2varepsilon < fracx_mm < c + 2varepsilon
$$
for all $m > M = max(M_1, M_2, N)$.
Since such an $M$ exists for every $varepsilon > 0$, we have that
$$
lim_m to infty fracx_mm = c.
$$
answered Jul 30 at 15:44
Dylan
5,67231134
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Since
$$
lim_n to infty (x_n + 1 - x_n) = c,
$$
we know that for every $varepsilon > 0$ there exists $N > 0$ such that
$$
c - varepsilon < x_n + 1 - x_n < c + varepsilon
$$
for all $n geq N$.
We thus have that for every $m > N$ that
$$
x_N + (m - N)(c - epsilon) < x_m < x_N + (m - N)(c + varepsilon)
$$
or equivalently
$$
fracx_Nm + left(1 - fracNmright)(c - varepsilon) < fracx_mm < fracx_Nm + left(1 - fracNmright)(c + varepsilon).
$$
We have that
$$
lim_m to infty fracx_Nm + left(1 - fracNmright)(c - varepsilon) = c - varepsilon
$$
and so there exists $M_1 > 0$ such that
$$
c - 2varepsilon < fracx_Nm + left(1 - fracNmright)(c - varepsilon)
$$
for all $m geq M_1$. Similarly, there exists $M_2 > 0$ such that
$$
fracx_Nm + left(1 - fracNmright)(c + varepsilon) < c + 2varepsilon
$$
for all $m geq M_2$. Thus we have that
$$
c - 2varepsilon < fracx_mm < c + 2varepsilon
$$
for all $m > M = max(M_1, M_2, N)$.
Since such an $M$ exists for every $varepsilon > 0$, we have that
$$
lim_m to infty fracx_mm = c.
$$
answered Jul 30 at 15:44
Dylan
5,67231134
Since
$$
lim_n to infty (x_n + 1 - x_n) = c,
$$
we know that for every $varepsilon > 0$ there exists $N > 0$ such that
$$
c - varepsilon < x_n + 1 - x_n < c + varepsilon
$$
for all $n geq N$.
We thus have that for every $m > N$ that
$$
x_N + (m - N)(c - epsilon) < x_m < x_N + (m - N)(c + varepsilon)
$$
or equivalently
$$
fracx_Nm + left(1 - fracNmright)(c - varepsilon) < fracx_mm < fracx_Nm + left(1 - fracNmright)(c + varepsilon).
$$
We have that
$$
lim_m to infty fracx_Nm + left(1 - fracNmright)(c - varepsilon) = c - varepsilon
$$
and so there exists $M_1 > 0$ such that
$$
c - 2varepsilon < fracx_Nm + left(1 - fracNmright)(c - varepsilon)
$$
for all $m geq M_1$. Similarly, there exists $M_2 > 0$ such that
$$
fracx_Nm + left(1 - fracNmright)(c + varepsilon) < c + 2varepsilon
$$
for all $m geq M_2$. Thus we have that
$$
c - 2varepsilon < fracx_mm < c + 2varepsilon
$$
for all $m > M = max(M_1, M_2, N)$.
Since such an $M$ exists for every $varepsilon > 0$, we have that
$$
lim_m to infty fracx_mm = c.
$$
answered Jul 30 at 15:44
Dylan
5,67231134
edited Jul 30 at 15:51
answered Jul 30 at 15:44
Dylan
5,67231134
answered Jul 30 at 15:44
Dylan
5,67231134
answered Jul 30 at 15:44
Dylan
5,67231134
5,67231134
add a comment |Â
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4
This is a special case of the Cesaro-Stolz lemma (en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem) The sequence converges to $c$. Since it is a special case of the lemma, it should hopefully be easier to prove than the full lemma.
– Dylan
Jul 30 at 15:27
You can reduce to the special case of $c=0$ by considering the sequence $y_n=x_n-nc$ instead. As for "how to define $x_n$?", you can't, really. You're supposed to show that something is true for all sequences $x_n$ fulfilling some property. That means that defining $x_n$ to be one of them won't help at all to prove what you're asked.
– Arthur
Jul 30 at 15:33
Also, since it is multiple choice, you don't actually need to prove that C is always true, you just need to find examples where A, B, and D are not true. You've already done that, and so C is the correct answer.
– Dylan
Jul 30 at 15:34