Prove the alternating sum of a decreasing sequence converging to 0 is Cauchy.












3












$begingroup$


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.










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$endgroup$








  • 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
    5 hours ago










  • $begingroup$
    @RobertShore is my answer okay?
    $endgroup$
    – Subhasis Biswas
    5 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
    53 mins ago


















3












$begingroup$


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.










share|cite|improve this question









$endgroup$








  • 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
    5 hours ago










  • $begingroup$
    @RobertShore is my answer okay?
    $endgroup$
    – Subhasis Biswas
    5 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
    53 mins ago
















3












3








3


1



$begingroup$


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.










share|cite|improve this question









$endgroup$




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






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asked 5 hours ago









oranjioranji

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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
    5 hours ago










  • $begingroup$
    @RobertShore is my answer okay?
    $endgroup$
    – Subhasis Biswas
    5 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
    53 mins ago
















  • 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
    5 hours ago










  • $begingroup$
    @RobertShore is my answer okay?
    $endgroup$
    – Subhasis Biswas
    5 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
    53 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
5 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
5 hours ago












$begingroup$
@RobertShore is my answer okay?
$endgroup$
– Subhasis Biswas
5 hours ago




$begingroup$
@RobertShore is my answer okay?
$endgroup$
– Subhasis Biswas
5 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
53 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
53 mins ago












2 Answers
2






active

oldest

votes


















3












$begingroup$

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
$$|s_{m} - s_n| leq x_{n+1} leq x_{N_{epsilon}} < epsilon$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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



















4












$begingroup$

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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$endgroup$













  • $begingroup$
    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












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2 Answers
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active

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2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

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
$$|s_{m} - s_n| leq x_{n+1} leq x_{N_{epsilon}} < epsilon$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















3












$begingroup$

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
$$|s_{m} - s_n| leq x_{n+1} leq x_{N_{epsilon}} < epsilon$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














3












3








3





$begingroup$

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
$$|s_{m} - s_n| leq x_{n+1} leq x_{N_{epsilon}} < epsilon$$






share|cite|improve this answer











$endgroup$



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
$$|s_{m} - s_n| leq x_{n+1} leq x_{N_{epsilon}} < epsilon$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 hours 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


















  • $begingroup$
    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
















$begingroup$
This is exactly what I was about to do.
$endgroup$
– Subhasis Biswas
2 hours ago




$begingroup$
This is exactly what I was about to do.
$endgroup$
– Subhasis Biswas
2 hours ago




1




1




$begingroup$
@SubhasisBiswas So, I did it for you :-D
$endgroup$
– trancelocation
2 hours ago




$begingroup$
@SubhasisBiswas So, I did it for you :-D
$endgroup$
– trancelocation
2 hours ago











4












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















4












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














4












4








4





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 5 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


















  • $begingroup$
    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


















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