Prove the alternating sum of a decreasing sequence converging to 0 is Cauchy. Unicorn Meta Zoo...
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Prove the alternating sum of a decreasing sequence converging to 0 is Cauchy.
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar ManaraSuppose for all positive integers $n$, $|x_n-y_n|< frac{1}{n}$ Prove that $(x_n)$ is also Cauchy.Proof check for completenessProve that $d_n$ is a Cauchy sequence in $mathbb{R}$Prove ${aX_n +bY_n}$ is a Cauchy Sequence.Prove a sequence is a Cauchy and thus convergentIf $(x_n)$ and $(y_n)$ are Cauchy sequences, then give a direct argument that $ (x_n + y_n)$ is a Cauchy sequenceIf ${x_n}$ and ${y_n}$ are Cauchy then $left{frac{2x_n}{y_n}right}$ is CauchyLet ${x_n}$ be a Cauchy sequence of rational numbers. Define a new sequence ${y_n}$ by $y_n = (x_n)(x_{n+1})$. Show that ${y_n}$ is a CS.Let ${x_n}$ be a Cauchy sequence of real numbers, prove that a new sequence ${y_n}$, with $y_n$=$x_n^frac{1}{3}$, is also a Cauchy sequence.$x_n rightarrow x$ iff the modified sequence is Cauchy
$begingroup$
Let $(x_n)$ be a decreasing sequence with $x_n > 0$ for all $n in mathbb{N}$, and $(x_n) to 0$. Let $(y_n)$ be defined for all $n in mathbb{N}$ by
$$y_n = x_0 - x_1 + x_2 - cdots + (-1)^n x_n .$$
I want to show, using the $varepsilon$ definition, that $(y_n)$ is Cauchy.
I am trying to find, given $varepsilon > 0$, a real number $N$ such that for all $m$ and $n$ with $m > n > N$, $|y_m - y_n| < varepsilon$.
I have been going backwards to try and find $N$, and have
begin{align*}
|y_m - y_n| & = left| (x_0 - x_1 + cdots pm x_m) - (x_0 - x_1 + cdots pm x_n) right| \
|y_m - y_n| & = left| x_{n + 1} - x_{n + 2} + cdots pm x_{m} right| \
|y_m - y_n| & leq | x_{n + 1} | + | x_{n + 2} | + cdots + | x_{m} | \
|y_m - y_n| & leq ?
end{align*}
I do not know how to get a solution from there, and am not sure about the process, particurlary the last step since I feel getting rid of the minuses might prevent me from finding a solution.
real-analysis cauchy-sequences
$endgroup$
add a comment |
$begingroup$
Let $(x_n)$ be a decreasing sequence with $x_n > 0$ for all $n in mathbb{N}$, and $(x_n) to 0$. Let $(y_n)$ be defined for all $n in mathbb{N}$ by
$$y_n = x_0 - x_1 + x_2 - cdots + (-1)^n x_n .$$
I want to show, using the $varepsilon$ definition, that $(y_n)$ is Cauchy.
I am trying to find, given $varepsilon > 0$, a real number $N$ such that for all $m$ and $n$ with $m > n > N$, $|y_m - y_n| < varepsilon$.
I have been going backwards to try and find $N$, and have
begin{align*}
|y_m - y_n| & = left| (x_0 - x_1 + cdots pm x_m) - (x_0 - x_1 + cdots pm x_n) right| \
|y_m - y_n| & = left| x_{n + 1} - x_{n + 2} + cdots pm x_{m} right| \
|y_m - y_n| & leq | x_{n + 1} | + | x_{n + 2} | + cdots + | x_{m} | \
|y_m - y_n| & leq ?
end{align*}
I do not know how to get a solution from there, and am not sure about the process, particurlary the last step since I feel getting rid of the minuses might prevent me from finding a solution.
real-analysis cauchy-sequences
$endgroup$
1
$begingroup$
Because the series is alternating and decreasing, I think you can prove by induction on $m$ that $|y_m-y_n| leq |y_n|$.
$endgroup$
– Robert Shore
4 hours ago
$begingroup$
@RobertShore is my answer okay?
$endgroup$
– Subhasis Biswas
4 hours ago
$begingroup$
@RobertShore yes I can definitely show that, but it brings me to the same issue with $|y_m| leq |x_0 - x_1 + cdots pm x_m|$, and I am unsure how to proceed from there.
$endgroup$
– oranji
2 hours ago
$begingroup$
I meant to say you can prove by induction that $|y_m-y_n| leq |x_n|$. Since $lim x_n=0$, choose $N$ such that $n gt N Rightarrow |x_n| lt epsilon$. Then $|y_m-y_n| leq |x_n| lt epsilon$ so ${y_n}$ is Cauchy.
$endgroup$
– Robert Shore
13 mins ago
add a comment |
$begingroup$
Let $(x_n)$ be a decreasing sequence with $x_n > 0$ for all $n in mathbb{N}$, and $(x_n) to 0$. Let $(y_n)$ be defined for all $n in mathbb{N}$ by
$$y_n = x_0 - x_1 + x_2 - cdots + (-1)^n x_n .$$
I want to show, using the $varepsilon$ definition, that $(y_n)$ is Cauchy.
I am trying to find, given $varepsilon > 0$, a real number $N$ such that for all $m$ and $n$ with $m > n > N$, $|y_m - y_n| < varepsilon$.
I have been going backwards to try and find $N$, and have
begin{align*}
|y_m - y_n| & = left| (x_0 - x_1 + cdots pm x_m) - (x_0 - x_1 + cdots pm x_n) right| \
|y_m - y_n| & = left| x_{n + 1} - x_{n + 2} + cdots pm x_{m} right| \
|y_m - y_n| & leq | x_{n + 1} | + | x_{n + 2} | + cdots + | x_{m} | \
|y_m - y_n| & leq ?
end{align*}
I do not know how to get a solution from there, and am not sure about the process, particurlary the last step since I feel getting rid of the minuses might prevent me from finding a solution.
real-analysis cauchy-sequences
$endgroup$
Let $(x_n)$ be a decreasing sequence with $x_n > 0$ for all $n in mathbb{N}$, and $(x_n) to 0$. Let $(y_n)$ be defined for all $n in mathbb{N}$ by
$$y_n = x_0 - x_1 + x_2 - cdots + (-1)^n x_n .$$
I want to show, using the $varepsilon$ definition, that $(y_n)$ is Cauchy.
I am trying to find, given $varepsilon > 0$, a real number $N$ such that for all $m$ and $n$ with $m > n > N$, $|y_m - y_n| < varepsilon$.
I have been going backwards to try and find $N$, and have
begin{align*}
|y_m - y_n| & = left| (x_0 - x_1 + cdots pm x_m) - (x_0 - x_1 + cdots pm x_n) right| \
|y_m - y_n| & = left| x_{n + 1} - x_{n + 2} + cdots pm x_{m} right| \
|y_m - y_n| & leq | x_{n + 1} | + | x_{n + 2} | + cdots + | x_{m} | \
|y_m - y_n| & leq ?
end{align*}
I do not know how to get a solution from there, and am not sure about the process, particurlary the last step since I feel getting rid of the minuses might prevent me from finding a solution.
real-analysis cauchy-sequences
real-analysis cauchy-sequences
asked 5 hours ago
oranjioranji
666
666
1
$begingroup$
Because the series is alternating and decreasing, I think you can prove by induction on $m$ that $|y_m-y_n| leq |y_n|$.
$endgroup$
– Robert Shore
4 hours ago
$begingroup$
@RobertShore is my answer okay?
$endgroup$
– Subhasis Biswas
4 hours ago
$begingroup$
@RobertShore yes I can definitely show that, but it brings me to the same issue with $|y_m| leq |x_0 - x_1 + cdots pm x_m|$, and I am unsure how to proceed from there.
$endgroup$
– oranji
2 hours ago
$begingroup$
I meant to say you can prove by induction that $|y_m-y_n| leq |x_n|$. Since $lim x_n=0$, choose $N$ such that $n gt N Rightarrow |x_n| lt epsilon$. Then $|y_m-y_n| leq |x_n| lt epsilon$ so ${y_n}$ is Cauchy.
$endgroup$
– Robert Shore
13 mins ago
add a comment |
1
$begingroup$
Because the series is alternating and decreasing, I think you can prove by induction on $m$ that $|y_m-y_n| leq |y_n|$.
$endgroup$
– Robert Shore
4 hours ago
$begingroup$
@RobertShore is my answer okay?
$endgroup$
– Subhasis Biswas
4 hours ago
$begingroup$
@RobertShore yes I can definitely show that, but it brings me to the same issue with $|y_m| leq |x_0 - x_1 + cdots pm x_m|$, and I am unsure how to proceed from there.
$endgroup$
– oranji
2 hours ago
$begingroup$
I meant to say you can prove by induction that $|y_m-y_n| leq |x_n|$. Since $lim x_n=0$, choose $N$ such that $n gt N Rightarrow |x_n| lt epsilon$. Then $|y_m-y_n| leq |x_n| lt epsilon$ so ${y_n}$ is Cauchy.
$endgroup$
– Robert Shore
13 mins ago
1
1
$begingroup$
Because the series is alternating and decreasing, I think you can prove by induction on $m$ that $|y_m-y_n| leq |y_n|$.
$endgroup$
– Robert Shore
4 hours ago
$begingroup$
Because the series is alternating and decreasing, I think you can prove by induction on $m$ that $|y_m-y_n| leq |y_n|$.
$endgroup$
– Robert Shore
4 hours ago
$begingroup$
@RobertShore is my answer okay?
$endgroup$
– Subhasis Biswas
4 hours ago
$begingroup$
@RobertShore is my answer okay?
$endgroup$
– Subhasis Biswas
4 hours ago
$begingroup$
@RobertShore yes I can definitely show that, but it brings me to the same issue with $|y_m| leq |x_0 - x_1 + cdots pm x_m|$, and I am unsure how to proceed from there.
$endgroup$
– oranji
2 hours ago
$begingroup$
@RobertShore yes I can definitely show that, but it brings me to the same issue with $|y_m| leq |x_0 - x_1 + cdots pm x_m|$, and I am unsure how to proceed from there.
$endgroup$
– oranji
2 hours ago
$begingroup$
I meant to say you can prove by induction that $|y_m-y_n| leq |x_n|$. Since $lim x_n=0$, choose $N$ such that $n gt N Rightarrow |x_n| lt epsilon$. Then $|y_m-y_n| leq |x_n| lt epsilon$ so ${y_n}$ is Cauchy.
$endgroup$
– Robert Shore
13 mins ago
$begingroup$
I meant to say you can prove by induction that $|y_m-y_n| leq |x_n|$. Since $lim x_n=0$, choose $N$ such that $n gt N Rightarrow |x_n| lt epsilon$. Then $|y_m-y_n| leq |x_n| lt epsilon$ so ${y_n}$ is Cauchy.
$endgroup$
– Robert Shore
13 mins ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
To see that the sequence of partial sums is Cauchy, you cannot use the triangle inequality directly as you did. A famous counter example here is $sum_{k=1}^{infty}frac{(-1)^k}{k}$.
What you can do is grouping the terms of the partial sums $s_n= sum_{j=1}^n(-1)^jx_j$ as follows:
- Let $m = n+k, k,n in mathbb{N}$
Now, you can write $|s_{m} - s_n|$ in two different ways:
$$|s_{n+k} - s_n| = begin{cases}
|x_{n+1} - (x_{n+2}-x_{n+3}) - cdots - (x_{n+2i}-x_{n+2i+1})| & k = 2i+1 \
|x_{n+1} - (x_{n+2}-x_{n+3}) - cdots - (x_{n+2i-2}-x_{n+2i-1}) - x_{2i}| & k = 2i \
end{cases}
$$
$$|s_{n+k} - s_n| = begin{cases}
|(x_{n+1} - x_{n+2}) + cdots + (x_{n+2i-1}-x_{n+2i}) + x_{n+2i+1}| & k = 2i+1 \
|(x_{n+1} - x_{n+2}) + cdots + (x_{n+2i-1}-x_{n+2i}) | & k = 2i \
end{cases}
$$
Using the fact that $x_n searrow 0$, it follows immediately that for all $k in mathbb{N}$ holds
$$|s_{n+k} - s_n| leq x_{n+1}$$
Hence, for $epsilon > 0$ choose $N_{epsilon}$ such that $x_{N_{epsilon}} < epsilon$. Then, for all $m> n > N_{epsilon}$ you have
$$|s_{m} - s_n| leq x_{n+1} leq x_{N_{epsilon}} < epsilon$$
$endgroup$
$begingroup$
This is exactly what I was about to do.
$endgroup$
– Subhasis Biswas
2 hours ago
1
$begingroup$
@SubhasisBiswas So, I did it for you :-D
$endgroup$
– trancelocation
2 hours ago
add a comment |
$begingroup$
This is also known as the "Leibnitz's Test".
We write $s_n = x_1-x_2+x_3-...+(-1)^{n+1}x_n$
$s_{2n+2}-s_{2n}=u_{2n+1}-u_{2n+2} geq0$ for all $n$.
$s_{2n+1}-s_{2n-1}=-u_{2n}+u_{2n+1} leq 0$
$s_{2n} =u_1 -(u_2-u_3)-(u_4-u_5)...-u_{2n} leq u_1$, i.e. a monotone increasing sequence bounded above.
$s_{2n+1} =(u_1 -u_2)+(u_3-u_4)+...+u_{2n+1} geq u_1-u_2$, i.e. a monotone decreasing sequence bounded below.
Hence, both are convergent subsequences of $(s_n)$. But, we have $lim (s_{2n+1}-s_{2n})=u_{2n+1}=0$, therefore, they converge to the same limit.
Hence, $(s_n)$ converges, i.e. it is Cauchy.
Note: We conclude that $(s_n)$ converges because the indices of the two subsequences $(s_{2n})$ and $(s_{2n+1})$ i.e. $U ={ 2n+1 : n in mathbb{N}}$ and $V ={ 2n : n in mathbb{N}}$ form a partition of $mathbb{N}$ and they both converge to the same limit.
$endgroup$
$begingroup$
I want to use the $varepsilon$ definition of a Cauchy sequence, and not the fact that all convergent sequences are Cauchy, which is why I cannot use this solution.
$endgroup$
– oranji
2 hours ago
$begingroup$
I'll edit this answer.
$endgroup$
– Subhasis Biswas
2 hours ago
add a comment |
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2 Answers
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2 Answers
2
active
oldest
votes
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votes
$begingroup$
To see that the sequence of partial sums is Cauchy, you cannot use the triangle inequality directly as you did. A famous counter example here is $sum_{k=1}^{infty}frac{(-1)^k}{k}$.
What you can do is grouping the terms of the partial sums $s_n= sum_{j=1}^n(-1)^jx_j$ as follows:
- Let $m = n+k, k,n in mathbb{N}$
Now, you can write $|s_{m} - s_n|$ in two different ways:
$$|s_{n+k} - s_n| = begin{cases}
|x_{n+1} - (x_{n+2}-x_{n+3}) - cdots - (x_{n+2i}-x_{n+2i+1})| & k = 2i+1 \
|x_{n+1} - (x_{n+2}-x_{n+3}) - cdots - (x_{n+2i-2}-x_{n+2i-1}) - x_{2i}| & k = 2i \
end{cases}
$$
$$|s_{n+k} - s_n| = begin{cases}
|(x_{n+1} - x_{n+2}) + cdots + (x_{n+2i-1}-x_{n+2i}) + x_{n+2i+1}| & k = 2i+1 \
|(x_{n+1} - x_{n+2}) + cdots + (x_{n+2i-1}-x_{n+2i}) | & k = 2i \
end{cases}
$$
Using the fact that $x_n searrow 0$, it follows immediately that for all $k in mathbb{N}$ holds
$$|s_{n+k} - s_n| leq x_{n+1}$$
Hence, for $epsilon > 0$ choose $N_{epsilon}$ such that $x_{N_{epsilon}} < epsilon$. Then, for all $m> n > N_{epsilon}$ you have
$$|s_{m} - s_n| leq x_{n+1} leq x_{N_{epsilon}} < epsilon$$
$endgroup$
$begingroup$
This is exactly what I was about to do.
$endgroup$
– Subhasis Biswas
2 hours ago
1
$begingroup$
@SubhasisBiswas So, I did it for you :-D
$endgroup$
– trancelocation
2 hours ago
add a comment |
$begingroup$
To see that the sequence of partial sums is Cauchy, you cannot use the triangle inequality directly as you did. A famous counter example here is $sum_{k=1}^{infty}frac{(-1)^k}{k}$.
What you can do is grouping the terms of the partial sums $s_n= sum_{j=1}^n(-1)^jx_j$ as follows:
- Let $m = n+k, k,n in mathbb{N}$
Now, you can write $|s_{m} - s_n|$ in two different ways:
$$|s_{n+k} - s_n| = begin{cases}
|x_{n+1} - (x_{n+2}-x_{n+3}) - cdots - (x_{n+2i}-x_{n+2i+1})| & k = 2i+1 \
|x_{n+1} - (x_{n+2}-x_{n+3}) - cdots - (x_{n+2i-2}-x_{n+2i-1}) - x_{2i}| & k = 2i \
end{cases}
$$
$$|s_{n+k} - s_n| = begin{cases}
|(x_{n+1} - x_{n+2}) + cdots + (x_{n+2i-1}-x_{n+2i}) + x_{n+2i+1}| & k = 2i+1 \
|(x_{n+1} - x_{n+2}) + cdots + (x_{n+2i-1}-x_{n+2i}) | & k = 2i \
end{cases}
$$
Using the fact that $x_n searrow 0$, it follows immediately that for all $k in mathbb{N}$ holds
$$|s_{n+k} - s_n| leq x_{n+1}$$
Hence, for $epsilon > 0$ choose $N_{epsilon}$ such that $x_{N_{epsilon}} < epsilon$. Then, for all $m> n > N_{epsilon}$ you have
$$|s_{m} - s_n| leq x_{n+1} leq x_{N_{epsilon}} < epsilon$$
$endgroup$
$begingroup$
This is exactly what I was about to do.
$endgroup$
– Subhasis Biswas
2 hours ago
1
$begingroup$
@SubhasisBiswas So, I did it for you :-D
$endgroup$
– trancelocation
2 hours ago
add a comment |
$begingroup$
To see that the sequence of partial sums is Cauchy, you cannot use the triangle inequality directly as you did. A famous counter example here is $sum_{k=1}^{infty}frac{(-1)^k}{k}$.
What you can do is grouping the terms of the partial sums $s_n= sum_{j=1}^n(-1)^jx_j$ as follows:
- Let $m = n+k, k,n in mathbb{N}$
Now, you can write $|s_{m} - s_n|$ in two different ways:
$$|s_{n+k} - s_n| = begin{cases}
|x_{n+1} - (x_{n+2}-x_{n+3}) - cdots - (x_{n+2i}-x_{n+2i+1})| & k = 2i+1 \
|x_{n+1} - (x_{n+2}-x_{n+3}) - cdots - (x_{n+2i-2}-x_{n+2i-1}) - x_{2i}| & k = 2i \
end{cases}
$$
$$|s_{n+k} - s_n| = begin{cases}
|(x_{n+1} - x_{n+2}) + cdots + (x_{n+2i-1}-x_{n+2i}) + x_{n+2i+1}| & k = 2i+1 \
|(x_{n+1} - x_{n+2}) + cdots + (x_{n+2i-1}-x_{n+2i}) | & k = 2i \
end{cases}
$$
Using the fact that $x_n searrow 0$, it follows immediately that for all $k in mathbb{N}$ holds
$$|s_{n+k} - s_n| leq x_{n+1}$$
Hence, for $epsilon > 0$ choose $N_{epsilon}$ such that $x_{N_{epsilon}} < epsilon$. Then, for all $m> n > N_{epsilon}$ you have
$$|s_{m} - s_n| leq x_{n+1} leq x_{N_{epsilon}} < epsilon$$
$endgroup$
To see that the sequence of partial sums is Cauchy, you cannot use the triangle inequality directly as you did. A famous counter example here is $sum_{k=1}^{infty}frac{(-1)^k}{k}$.
What you can do is grouping the terms of the partial sums $s_n= sum_{j=1}^n(-1)^jx_j$ as follows:
- Let $m = n+k, k,n in mathbb{N}$
Now, you can write $|s_{m} - s_n|$ in two different ways:
$$|s_{n+k} - s_n| = begin{cases}
|x_{n+1} - (x_{n+2}-x_{n+3}) - cdots - (x_{n+2i}-x_{n+2i+1})| & k = 2i+1 \
|x_{n+1} - (x_{n+2}-x_{n+3}) - cdots - (x_{n+2i-2}-x_{n+2i-1}) - x_{2i}| & k = 2i \
end{cases}
$$
$$|s_{n+k} - s_n| = begin{cases}
|(x_{n+1} - x_{n+2}) + cdots + (x_{n+2i-1}-x_{n+2i}) + x_{n+2i+1}| & k = 2i+1 \
|(x_{n+1} - x_{n+2}) + cdots + (x_{n+2i-1}-x_{n+2i}) | & k = 2i \
end{cases}
$$
Using the fact that $x_n searrow 0$, it follows immediately that for all $k in mathbb{N}$ holds
$$|s_{n+k} - s_n| leq x_{n+1}$$
Hence, for $epsilon > 0$ choose $N_{epsilon}$ such that $x_{N_{epsilon}} < epsilon$. Then, for all $m> n > N_{epsilon}$ you have
$$|s_{m} - s_n| leq x_{n+1} leq x_{N_{epsilon}} < epsilon$$
edited 1 hour ago
answered 2 hours ago
trancelocationtrancelocation
14.6k1929
14.6k1929
$begingroup$
This is exactly what I was about to do.
$endgroup$
– Subhasis Biswas
2 hours ago
1
$begingroup$
@SubhasisBiswas So, I did it for you :-D
$endgroup$
– trancelocation
2 hours ago
add a comment |
$begingroup$
This is exactly what I was about to do.
$endgroup$
– Subhasis Biswas
2 hours ago
1
$begingroup$
@SubhasisBiswas So, I did it for you :-D
$endgroup$
– trancelocation
2 hours ago
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This is exactly what I was about to do.
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– Subhasis Biswas
2 hours ago
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This is exactly what I was about to do.
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– Subhasis Biswas
2 hours ago
1
1
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@SubhasisBiswas So, I did it for you :-D
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– trancelocation
2 hours ago
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@SubhasisBiswas So, I did it for you :-D
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– trancelocation
2 hours ago
add a comment |
$begingroup$
This is also known as the "Leibnitz's Test".
We write $s_n = x_1-x_2+x_3-...+(-1)^{n+1}x_n$
$s_{2n+2}-s_{2n}=u_{2n+1}-u_{2n+2} geq0$ for all $n$.
$s_{2n+1}-s_{2n-1}=-u_{2n}+u_{2n+1} leq 0$
$s_{2n} =u_1 -(u_2-u_3)-(u_4-u_5)...-u_{2n} leq u_1$, i.e. a monotone increasing sequence bounded above.
$s_{2n+1} =(u_1 -u_2)+(u_3-u_4)+...+u_{2n+1} geq u_1-u_2$, i.e. a monotone decreasing sequence bounded below.
Hence, both are convergent subsequences of $(s_n)$. But, we have $lim (s_{2n+1}-s_{2n})=u_{2n+1}=0$, therefore, they converge to the same limit.
Hence, $(s_n)$ converges, i.e. it is Cauchy.
Note: We conclude that $(s_n)$ converges because the indices of the two subsequences $(s_{2n})$ and $(s_{2n+1})$ i.e. $U ={ 2n+1 : n in mathbb{N}}$ and $V ={ 2n : n in mathbb{N}}$ form a partition of $mathbb{N}$ and they both converge to the same limit.
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I want to use the $varepsilon$ definition of a Cauchy sequence, and not the fact that all convergent sequences are Cauchy, which is why I cannot use this solution.
$endgroup$
– oranji
2 hours ago
$begingroup$
I'll edit this answer.
$endgroup$
– Subhasis Biswas
2 hours ago
add a comment |
$begingroup$
This is also known as the "Leibnitz's Test".
We write $s_n = x_1-x_2+x_3-...+(-1)^{n+1}x_n$
$s_{2n+2}-s_{2n}=u_{2n+1}-u_{2n+2} geq0$ for all $n$.
$s_{2n+1}-s_{2n-1}=-u_{2n}+u_{2n+1} leq 0$
$s_{2n} =u_1 -(u_2-u_3)-(u_4-u_5)...-u_{2n} leq u_1$, i.e. a monotone increasing sequence bounded above.
$s_{2n+1} =(u_1 -u_2)+(u_3-u_4)+...+u_{2n+1} geq u_1-u_2$, i.e. a monotone decreasing sequence bounded below.
Hence, both are convergent subsequences of $(s_n)$. But, we have $lim (s_{2n+1}-s_{2n})=u_{2n+1}=0$, therefore, they converge to the same limit.
Hence, $(s_n)$ converges, i.e. it is Cauchy.
Note: We conclude that $(s_n)$ converges because the indices of the two subsequences $(s_{2n})$ and $(s_{2n+1})$ i.e. $U ={ 2n+1 : n in mathbb{N}}$ and $V ={ 2n : n in mathbb{N}}$ form a partition of $mathbb{N}$ and they both converge to the same limit.
$endgroup$
$begingroup$
I want to use the $varepsilon$ definition of a Cauchy sequence, and not the fact that all convergent sequences are Cauchy, which is why I cannot use this solution.
$endgroup$
– oranji
2 hours ago
$begingroup$
I'll edit this answer.
$endgroup$
– Subhasis Biswas
2 hours ago
add a comment |
$begingroup$
This is also known as the "Leibnitz's Test".
We write $s_n = x_1-x_2+x_3-...+(-1)^{n+1}x_n$
$s_{2n+2}-s_{2n}=u_{2n+1}-u_{2n+2} geq0$ for all $n$.
$s_{2n+1}-s_{2n-1}=-u_{2n}+u_{2n+1} leq 0$
$s_{2n} =u_1 -(u_2-u_3)-(u_4-u_5)...-u_{2n} leq u_1$, i.e. a monotone increasing sequence bounded above.
$s_{2n+1} =(u_1 -u_2)+(u_3-u_4)+...+u_{2n+1} geq u_1-u_2$, i.e. a monotone decreasing sequence bounded below.
Hence, both are convergent subsequences of $(s_n)$. But, we have $lim (s_{2n+1}-s_{2n})=u_{2n+1}=0$, therefore, they converge to the same limit.
Hence, $(s_n)$ converges, i.e. it is Cauchy.
Note: We conclude that $(s_n)$ converges because the indices of the two subsequences $(s_{2n})$ and $(s_{2n+1})$ i.e. $U ={ 2n+1 : n in mathbb{N}}$ and $V ={ 2n : n in mathbb{N}}$ form a partition of $mathbb{N}$ and they both converge to the same limit.
$endgroup$
This is also known as the "Leibnitz's Test".
We write $s_n = x_1-x_2+x_3-...+(-1)^{n+1}x_n$
$s_{2n+2}-s_{2n}=u_{2n+1}-u_{2n+2} geq0$ for all $n$.
$s_{2n+1}-s_{2n-1}=-u_{2n}+u_{2n+1} leq 0$
$s_{2n} =u_1 -(u_2-u_3)-(u_4-u_5)...-u_{2n} leq u_1$, i.e. a monotone increasing sequence bounded above.
$s_{2n+1} =(u_1 -u_2)+(u_3-u_4)+...+u_{2n+1} geq u_1-u_2$, i.e. a monotone decreasing sequence bounded below.
Hence, both are convergent subsequences of $(s_n)$. But, we have $lim (s_{2n+1}-s_{2n})=u_{2n+1}=0$, therefore, they converge to the same limit.
Hence, $(s_n)$ converges, i.e. it is Cauchy.
Note: We conclude that $(s_n)$ converges because the indices of the two subsequences $(s_{2n})$ and $(s_{2n+1})$ i.e. $U ={ 2n+1 : n in mathbb{N}}$ and $V ={ 2n : n in mathbb{N}}$ form a partition of $mathbb{N}$ and they both converge to the same limit.
answered 4 hours ago
Subhasis BiswasSubhasis Biswas
628512
628512
$begingroup$
I want to use the $varepsilon$ definition of a Cauchy sequence, and not the fact that all convergent sequences are Cauchy, which is why I cannot use this solution.
$endgroup$
– oranji
2 hours ago
$begingroup$
I'll edit this answer.
$endgroup$
– Subhasis Biswas
2 hours ago
add a comment |
$begingroup$
I want to use the $varepsilon$ definition of a Cauchy sequence, and not the fact that all convergent sequences are Cauchy, which is why I cannot use this solution.
$endgroup$
– oranji
2 hours ago
$begingroup$
I'll edit this answer.
$endgroup$
– Subhasis Biswas
2 hours ago
$begingroup$
I want to use the $varepsilon$ definition of a Cauchy sequence, and not the fact that all convergent sequences are Cauchy, which is why I cannot use this solution.
$endgroup$
– oranji
2 hours ago
$begingroup$
I want to use the $varepsilon$ definition of a Cauchy sequence, and not the fact that all convergent sequences are Cauchy, which is why I cannot use this solution.
$endgroup$
– oranji
2 hours ago
$begingroup$
I'll edit this answer.
$endgroup$
– Subhasis Biswas
2 hours ago
$begingroup$
I'll edit this answer.
$endgroup$
– Subhasis Biswas
2 hours ago
add a comment |
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1
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Because the series is alternating and decreasing, I think you can prove by induction on $m$ that $|y_m-y_n| leq |y_n|$.
$endgroup$
– Robert Shore
4 hours ago
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@RobertShore is my answer okay?
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– Subhasis Biswas
4 hours ago
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@RobertShore yes I can definitely show that, but it brings me to the same issue with $|y_m| leq |x_0 - x_1 + cdots pm x_m|$, and I am unsure how to proceed from there.
$endgroup$
– oranji
2 hours ago
$begingroup$
I meant to say you can prove by induction that $|y_m-y_n| leq |x_n|$. Since $lim x_n=0$, choose $N$ such that $n gt N Rightarrow |x_n| lt epsilon$. Then $|y_m-y_n| leq |x_n| lt epsilon$ so ${y_n}$ is Cauchy.
$endgroup$
– Robert Shore
13 mins ago