Prove the alternating sum of a decreasing sequence converging to $0$ is Cauchy.Suppose for all positive integers $n$, $|x_n-y_n|< frac1n$ Prove that $(x_n)$ is also Cauchy.Proof check for completenessProve that $d_n$ is a Cauchy sequence in $mathbbR$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 $leftfrac2x_ny_nright$ 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 $y_n$=$x_n^frac13$, is also a Cauchy sequence.$x_n rightarrow x$ iff the modified sequence is Cauchy

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


Suppose for all positive integers $n$, $|x_n-y_n|< frac1n$ Prove that $(x_n)$ is also Cauchy.Proof check for completenessProve that $d_n$ is a Cauchy sequence in $mathbbR$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 $leftfrac2x_ny_nright$ 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 $y_n$=$x_n^frac13$, is also a Cauchy sequence.$x_n rightarrow x$ iff the modified sequence is Cauchy













2












$begingroup$


Let $(x_n)$ be a decreasing sequence with $x_n > 0$ for all $n in mathbbN$, and $(x_n) to 0$. Let $(y_n)$ be defined for all $n in mathbbN$ 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
beginalign*
|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 ?
endalign*



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
    Apr 25 at 1:17










  • $begingroup$
    @RobertShore is my answer okay?
    $endgroup$
    – Subhasis Biswas
    Apr 25 at 1:37










  • $begingroup$
    @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
    Apr 25 at 4:05







  • 1




    $begingroup$
    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
    Apr 25 at 5:54
















2












$begingroup$


Let $(x_n)$ be a decreasing sequence with $x_n > 0$ for all $n in mathbbN$, and $(x_n) to 0$. Let $(y_n)$ be defined for all $n in mathbbN$ 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
beginalign*
|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 ?
endalign*



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
    Apr 25 at 1:17










  • $begingroup$
    @RobertShore is my answer okay?
    $endgroup$
    – Subhasis Biswas
    Apr 25 at 1:37










  • $begingroup$
    @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
    Apr 25 at 4:05







  • 1




    $begingroup$
    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
    Apr 25 at 5:54














2












2








2


2



$begingroup$


Let $(x_n)$ be a decreasing sequence with $x_n > 0$ for all $n in mathbbN$, and $(x_n) to 0$. Let $(y_n)$ be defined for all $n in mathbbN$ 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
beginalign*
|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 ?
endalign*



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 $n in mathbbN$, and $(x_n) to 0$. Let $(y_n)$ be defined for all $n in mathbbN$ 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
beginalign*
|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 ?
endalign*



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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 25 at 6:53









Asaf Karagila

309k33442776




309k33442776










asked Apr 25 at 0:51









oranjioranji

616




616







  • 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
    Apr 25 at 1:17










  • $begingroup$
    @RobertShore is my answer okay?
    $endgroup$
    – Subhasis Biswas
    Apr 25 at 1:37










  • $begingroup$
    @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
    Apr 25 at 4:05







  • 1




    $begingroup$
    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
    Apr 25 at 5:54













  • 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
    Apr 25 at 1:17










  • $begingroup$
    @RobertShore is my answer okay?
    $endgroup$
    – Subhasis Biswas
    Apr 25 at 1:37










  • $begingroup$
    @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
    Apr 25 at 4:05







  • 1




    $begingroup$
    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
    Apr 25 at 5:54








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
Apr 25 at 1:17




$begingroup$
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
Apr 25 at 1:17












$begingroup$
@RobertShore is my answer okay?
$endgroup$
– Subhasis Biswas
Apr 25 at 1:37




$begingroup$
@RobertShore is my answer okay?
$endgroup$
– Subhasis Biswas
Apr 25 at 1:37












$begingroup$
@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
Apr 25 at 4:05





$begingroup$
@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
Apr 25 at 4:05





1




1




$begingroup$
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
Apr 25 at 5:54





$begingroup$
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
Apr 25 at 5:54











2 Answers
2






active

oldest

votes


















4












$begingroup$

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 $sum_k=1^inftyfrac(-1)^kk$.



What you can do is grouping the terms of the partial sums $s_n= sum_j=1^n(-1)^jx_j$ as follows:



  • Let $m = n+k, k,n in mathbbN$

Now, you can write $|s_m - s_n|$ in two different ways:



$$|s_n+k - s_n| = begincases
|x_n+1 - (x_n+2-x_n+3) - cdots - (x_n+2i-x_n+2i+1)| & k = 2i+1 \
|x_n+1 - (x_n+2-x_n+3) - cdots - (x_n+2i-2-x_n+2i-1) - x_2i| & k = 2i \
endcases
$$



$$|s_n+k - s_n| = begincases
|(x_n+1 - x_n+2) + cdots + (x_n+2i-1-x_n+2i) + x_n+2i+1| & k = 2i+1 \
|(x_n+1 - x_n+2) + cdots + (x_n+2i-1-x_n+2i) | & k = 2i \
endcases
$$



Using the fact that $x_n searrow 0$, it follows immediately that for all $k in mathbbN$ 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
    Apr 25 at 4:06






  • 1




    $begingroup$
    @SubhasisBiswas So, I did it for you :-D
    $endgroup$
    – trancelocation
    Apr 25 at 4:07


















4












$begingroup$

This is also known as the "Leibnitz's Test".



We write $s_n = x_1-x_2+x_3-...+(-1)^n+1x_n$



$s_2n+2-s_2n=u_2n+1-u_2n+2 geq0$ 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. $U = 2n+1 : n in mathbbN$ and $V = 2n : n in mathbbN$ form a partition of $mathbbN$ 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
    Apr 25 at 4:02










  • $begingroup$
    I'll edit this answer.
    $endgroup$
    – Subhasis Biswas
    Apr 25 at 4:05











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

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 $sum_k=1^inftyfrac(-1)^kk$.



What you can do is grouping the terms of the partial sums $s_n= sum_j=1^n(-1)^jx_j$ as follows:



  • Let $m = n+k, k,n in mathbbN$

Now, you can write $|s_m - s_n|$ in two different ways:



$$|s_n+k - s_n| = begincases
|x_n+1 - (x_n+2-x_n+3) - cdots - (x_n+2i-x_n+2i+1)| & k = 2i+1 \
|x_n+1 - (x_n+2-x_n+3) - cdots - (x_n+2i-2-x_n+2i-1) - x_2i| & k = 2i \
endcases
$$



$$|s_n+k - s_n| = begincases
|(x_n+1 - x_n+2) + cdots + (x_n+2i-1-x_n+2i) + x_n+2i+1| & k = 2i+1 \
|(x_n+1 - x_n+2) + cdots + (x_n+2i-1-x_n+2i) | & k = 2i \
endcases
$$



Using the fact that $x_n searrow 0$, it follows immediately that for all $k in mathbbN$ 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
    Apr 25 at 4:06






  • 1




    $begingroup$
    @SubhasisBiswas So, I did it for you :-D
    $endgroup$
    – trancelocation
    Apr 25 at 4:07















4












$begingroup$

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 $sum_k=1^inftyfrac(-1)^kk$.



What you can do is grouping the terms of the partial sums $s_n= sum_j=1^n(-1)^jx_j$ as follows:



  • Let $m = n+k, k,n in mathbbN$

Now, you can write $|s_m - s_n|$ in two different ways:



$$|s_n+k - s_n| = begincases
|x_n+1 - (x_n+2-x_n+3) - cdots - (x_n+2i-x_n+2i+1)| & k = 2i+1 \
|x_n+1 - (x_n+2-x_n+3) - cdots - (x_n+2i-2-x_n+2i-1) - x_2i| & k = 2i \
endcases
$$



$$|s_n+k - s_n| = begincases
|(x_n+1 - x_n+2) + cdots + (x_n+2i-1-x_n+2i) + x_n+2i+1| & k = 2i+1 \
|(x_n+1 - x_n+2) + cdots + (x_n+2i-1-x_n+2i) | & k = 2i \
endcases
$$



Using the fact that $x_n searrow 0$, it follows immediately that for all $k in mathbbN$ 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
    Apr 25 at 4:06






  • 1




    $begingroup$
    @SubhasisBiswas So, I did it for you :-D
    $endgroup$
    – trancelocation
    Apr 25 at 4:07













4












4








4





$begingroup$

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 $sum_k=1^inftyfrac(-1)^kk$.



What you can do is grouping the terms of the partial sums $s_n= sum_j=1^n(-1)^jx_j$ as follows:



  • Let $m = n+k, k,n in mathbbN$

Now, you can write $|s_m - s_n|$ in two different ways:



$$|s_n+k - s_n| = begincases
|x_n+1 - (x_n+2-x_n+3) - cdots - (x_n+2i-x_n+2i+1)| & k = 2i+1 \
|x_n+1 - (x_n+2-x_n+3) - cdots - (x_n+2i-2-x_n+2i-1) - x_2i| & k = 2i \
endcases
$$



$$|s_n+k - s_n| = begincases
|(x_n+1 - x_n+2) + cdots + (x_n+2i-1-x_n+2i) + x_n+2i+1| & k = 2i+1 \
|(x_n+1 - x_n+2) + cdots + (x_n+2i-1-x_n+2i) | & k = 2i \
endcases
$$



Using the fact that $x_n searrow 0$, it follows immediately that for all $k in mathbbN$ 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 $sum_k=1^inftyfrac(-1)^kk$.



What you can do is grouping the terms of the partial sums $s_n= sum_j=1^n(-1)^jx_j$ as follows:



  • Let $m = n+k, k,n in mathbbN$

Now, you can write $|s_m - s_n|$ in two different ways:



$$|s_n+k - s_n| = begincases
|x_n+1 - (x_n+2-x_n+3) - cdots - (x_n+2i-x_n+2i+1)| & k = 2i+1 \
|x_n+1 - (x_n+2-x_n+3) - cdots - (x_n+2i-2-x_n+2i-1) - x_2i| & k = 2i \
endcases
$$



$$|s_n+k - s_n| = begincases
|(x_n+1 - x_n+2) + cdots + (x_n+2i-1-x_n+2i) + x_n+2i+1| & k = 2i+1 \
|(x_n+1 - x_n+2) + cdots + (x_n+2i-1-x_n+2i) | & k = 2i \
endcases
$$



Using the fact that $x_n searrow 0$, it follows immediately that for all $k in mathbbN$ 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
$$|s_m - s_n| leq x_n+1 leq x_N_epsilon < epsilon$$







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edited Apr 25 at 4:22

























answered Apr 25 at 4:05









trancelocationtrancelocation

14.7k1929




14.7k1929











  • $begingroup$
    This is exactly what I was about to do.
    $endgroup$
    – Subhasis Biswas
    Apr 25 at 4:06






  • 1




    $begingroup$
    @SubhasisBiswas So, I did it for you :-D
    $endgroup$
    – trancelocation
    Apr 25 at 4:07
















  • $begingroup$
    This is exactly what I was about to do.
    $endgroup$
    – Subhasis Biswas
    Apr 25 at 4:06






  • 1




    $begingroup$
    @SubhasisBiswas So, I did it for you :-D
    $endgroup$
    – trancelocation
    Apr 25 at 4:07















$begingroup$
This is exactly what I was about to do.
$endgroup$
– Subhasis Biswas
Apr 25 at 4:06




$begingroup$
This is exactly what I was about to do.
$endgroup$
– Subhasis Biswas
Apr 25 at 4:06




1




1




$begingroup$
@SubhasisBiswas So, I did it for you :-D
$endgroup$
– trancelocation
Apr 25 at 4:07




$begingroup$
@SubhasisBiswas So, I did it for you :-D
$endgroup$
– trancelocation
Apr 25 at 4:07











4












$begingroup$

This is also known as the "Leibnitz's Test".



We write $s_n = x_1-x_2+x_3-...+(-1)^n+1x_n$



$s_2n+2-s_2n=u_2n+1-u_2n+2 geq0$ 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. $U = 2n+1 : n in mathbbN$ and $V = 2n : n in mathbbN$ form a partition of $mathbbN$ 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
    Apr 25 at 4:02










  • $begingroup$
    I'll edit this answer.
    $endgroup$
    – Subhasis Biswas
    Apr 25 at 4:05















4












$begingroup$

This is also known as the "Leibnitz's Test".



We write $s_n = x_1-x_2+x_3-...+(-1)^n+1x_n$



$s_2n+2-s_2n=u_2n+1-u_2n+2 geq0$ 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. $U = 2n+1 : n in mathbbN$ and $V = 2n : n in mathbbN$ form a partition of $mathbbN$ 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
    Apr 25 at 4:02










  • $begingroup$
    I'll edit this answer.
    $endgroup$
    – Subhasis Biswas
    Apr 25 at 4:05













4












4








4





$begingroup$

This is also known as the "Leibnitz's Test".



We write $s_n = x_1-x_2+x_3-...+(-1)^n+1x_n$



$s_2n+2-s_2n=u_2n+1-u_2n+2 geq0$ 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. $U = 2n+1 : n in mathbbN$ and $V = 2n : n in mathbbN$ form a partition of $mathbbN$ 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 $s_n = x_1-x_2+x_3-...+(-1)^n+1x_n$



$s_2n+2-s_2n=u_2n+1-u_2n+2 geq0$ 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. $U = 2n+1 : n in mathbbN$ and $V = 2n : n in mathbbN$ form a partition of $mathbbN$ and they both converge to the same limit.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Apr 25 at 1:30









Subhasis BiswasSubhasis Biswas

655512




655512











  • $begingroup$
    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
    Apr 25 at 4:02










  • $begingroup$
    I'll edit this answer.
    $endgroup$
    – Subhasis Biswas
    Apr 25 at 4:05
















  • $begingroup$
    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
    Apr 25 at 4:02










  • $begingroup$
    I'll edit this answer.
    $endgroup$
    – Subhasis Biswas
    Apr 25 at 4:05















$begingroup$
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
Apr 25 at 4:02




$begingroup$
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
Apr 25 at 4:02












$begingroup$
I'll edit this answer.
$endgroup$
– Subhasis Biswas
Apr 25 at 4:05




$begingroup$
I'll edit this answer.
$endgroup$
– Subhasis Biswas
Apr 25 at 4:05

















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