Showing sequence is Cauchy by Definition












4














Question: I have to show that sequence $(x_n)$ defined by



$x_n=frac{n+(-1)^n}{2n-1}$ , $n=1,2,3,...$.



Is Cauchy sequence using definition only.



My attempt: (I can see given sequence is Cauchy because it is convergent sequence in $mathbb{R}$)But I have to show it Cauchy by using definition of Cauchy sequence only. So I considered



$|x_n-x_m|= |frac{n+(-1)^n}{2n-1}-frac{m+(-1)^m}{2m-1}|$



$=|frac{(n+(-1)^n)(2m-1)-(m+(-1)^m)(2n-1)}{(2n-1)(2m+1)}|$



$=|frac{m-n+(-1)^n 2m+(-1)^{m+1}2n+(-1)^{n+1}+(-1)^{m+2}}{(2n-1)(2m-1)}|$



but I am unable to proceed further :-( please Help me...






real-analysis sequences-and-series cauchy-sequences







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  • 1




    Perhaps write it as $frac{n}{2n-1} + frac{(-1)^n}{2n-1}$, and then examine the difference seperately? Let me think a bit. Honestly, my original approach would be like yours, convergent therefore Cauchy.
    – TrostAft
    Dec 2 '18 at 6:49












  • Still i am unable to show, above difference is less than epsilon. please help...
    – Akash Patalwanshi
    Dec 2 '18 at 6:53






  • 1




    Well $frac{n}{2n-1} - frac{m}{2m-1}$ should be small, since they're about the same for large $n, m in mathbb{N}$. Same for $frac{(-1)^n}{2n-1} - frac{(-1)^m}{2m-1}$ should also be small.
    – TrostAft
    Dec 2 '18 at 6:56
















4














Question: I have to show that sequence $(x_n)$ defined by



$x_n=frac{n+(-1)^n}{2n-1}$ , $n=1,2,3,...$.



Is Cauchy sequence using definition only.



My attempt: (I can see given sequence is Cauchy because it is convergent sequence in $mathbb{R}$)But I have to show it Cauchy by using definition of Cauchy sequence only. So I considered



$|x_n-x_m|= |frac{n+(-1)^n}{2n-1}-frac{m+(-1)^m}{2m-1}|$



$=|frac{(n+(-1)^n)(2m-1)-(m+(-1)^m)(2n-1)}{(2n-1)(2m+1)}|$



$=|frac{m-n+(-1)^n 2m+(-1)^{m+1}2n+(-1)^{n+1}+(-1)^{m+2}}{(2n-1)(2m-1)}|$



but I am unable to proceed further :-( please Help me...






real-analysis sequences-and-series cauchy-sequences







share|cite|improve this question




















  • 1




    Perhaps write it as $frac{n}{2n-1} + frac{(-1)^n}{2n-1}$, and then examine the difference seperately? Let me think a bit. Honestly, my original approach would be like yours, convergent therefore Cauchy.
    – TrostAft
    Dec 2 '18 at 6:49












  • Still i am unable to show, above difference is less than epsilon. please help...
    – Akash Patalwanshi
    Dec 2 '18 at 6:53






  • 1




    Well $frac{n}{2n-1} - frac{m}{2m-1}$ should be small, since they're about the same for large $n, m in mathbb{N}$. Same for $frac{(-1)^n}{2n-1} - frac{(-1)^m}{2m-1}$ should also be small.
    – TrostAft
    Dec 2 '18 at 6:56














4












4








4







Question: I have to show that sequence $(x_n)$ defined by



$x_n=frac{n+(-1)^n}{2n-1}$ , $n=1,2,3,...$.



Is Cauchy sequence using definition only.



My attempt: (I can see given sequence is Cauchy because it is convergent sequence in $mathbb{R}$)But I have to show it Cauchy by using definition of Cauchy sequence only. So I considered



$|x_n-x_m|= |frac{n+(-1)^n}{2n-1}-frac{m+(-1)^m}{2m-1}|$



$=|frac{(n+(-1)^n)(2m-1)-(m+(-1)^m)(2n-1)}{(2n-1)(2m+1)}|$



$=|frac{m-n+(-1)^n 2m+(-1)^{m+1}2n+(-1)^{n+1}+(-1)^{m+2}}{(2n-1)(2m-1)}|$



but I am unable to proceed further :-( please Help me...






real-analysis sequences-and-series cauchy-sequences







share|cite|improve this question















Question: I have to show that sequence $(x_n)$ defined by



$x_n=frac{n+(-1)^n}{2n-1}$ , $n=1,2,3,...$.



Is Cauchy sequence using definition only.



My attempt: (I can see given sequence is Cauchy because it is convergent sequence in $mathbb{R}$)But I have to show it Cauchy by using definition of Cauchy sequence only. So I considered



$|x_n-x_m|= |frac{n+(-1)^n}{2n-1}-frac{m+(-1)^m}{2m-1}|$



$=|frac{(n+(-1)^n)(2m-1)-(m+(-1)^m)(2n-1)}{(2n-1)(2m+1)}|$



$=|frac{m-n+(-1)^n 2m+(-1)^{m+1}2n+(-1)^{n+1}+(-1)^{m+2}}{(2n-1)(2m-1)}|$



but I am unable to proceed further :-( please Help me...






real-analysis sequences-and-series cauchy-sequences




real-analysis sequences-and-series cauchy-sequences






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share|cite|improve this question













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edited Dec 2 '18 at 6:45

























asked Dec 2 '18 at 6:37









Akash Patalwanshi

9771816




9771816








  • 1




    Perhaps write it as $frac{n}{2n-1} + frac{(-1)^n}{2n-1}$, and then examine the difference seperately? Let me think a bit. Honestly, my original approach would be like yours, convergent therefore Cauchy.
    – TrostAft
    Dec 2 '18 at 6:49












  • Still i am unable to show, above difference is less than epsilon. please help...
    – Akash Patalwanshi
    Dec 2 '18 at 6:53






  • 1




    Well $frac{n}{2n-1} - frac{m}{2m-1}$ should be small, since they're about the same for large $n, m in mathbb{N}$. Same for $frac{(-1)^n}{2n-1} - frac{(-1)^m}{2m-1}$ should also be small.
    – TrostAft
    Dec 2 '18 at 6:56














  • 1




    Perhaps write it as $frac{n}{2n-1} + frac{(-1)^n}{2n-1}$, and then examine the difference seperately? Let me think a bit. Honestly, my original approach would be like yours, convergent therefore Cauchy.
    – TrostAft
    Dec 2 '18 at 6:49












  • Still i am unable to show, above difference is less than epsilon. please help...
    – Akash Patalwanshi
    Dec 2 '18 at 6:53






  • 1




    Well $frac{n}{2n-1} - frac{m}{2m-1}$ should be small, since they're about the same for large $n, m in mathbb{N}$. Same for $frac{(-1)^n}{2n-1} - frac{(-1)^m}{2m-1}$ should also be small.
    – TrostAft
    Dec 2 '18 at 6:56








1




1




Perhaps write it as $frac{n}{2n-1} + frac{(-1)^n}{2n-1}$, and then examine the difference seperately? Let me think a bit. Honestly, my original approach would be like yours, convergent therefore Cauchy.
– TrostAft
Dec 2 '18 at 6:49






Perhaps write it as $frac{n}{2n-1} + frac{(-1)^n}{2n-1}$, and then examine the difference seperately? Let me think a bit. Honestly, my original approach would be like yours, convergent therefore Cauchy.
– TrostAft
Dec 2 '18 at 6:49














Still i am unable to show, above difference is less than epsilon. please help...
– Akash Patalwanshi
Dec 2 '18 at 6:53




Still i am unable to show, above difference is less than epsilon. please help...
– Akash Patalwanshi
Dec 2 '18 at 6:53




1




1




Well $frac{n}{2n-1} - frac{m}{2m-1}$ should be small, since they're about the same for large $n, m in mathbb{N}$. Same for $frac{(-1)^n}{2n-1} - frac{(-1)^m}{2m-1}$ should also be small.
– TrostAft
Dec 2 '18 at 6:56




Well $frac{n}{2n-1} - frac{m}{2m-1}$ should be small, since they're about the same for large $n, m in mathbb{N}$. Same for $frac{(-1)^n}{2n-1} - frac{(-1)^m}{2m-1}$ should also be small.
– TrostAft
Dec 2 '18 at 6:56










3 Answers
3






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5














Suppose $m>n.$ $|x_m-x_n|=left|frac{m+(-1)^m}{2m-1}-frac{n+(-1)^n}{2n-1}right|leq left|frac{m}{2m-1}-frac{n}{2n-1}right|+frac{1}{2m-1}+frac{1}{2n-1}=frac{m-n}{(2m-1)(2n-1)}+frac{1}{2m-1}+frac{1}{2n-1}leq frac{3}{2n-1}<epsilon$



for sufficiently large $m,n.$






share|cite|improve this answer





















  • Sir, love your second step. But please could you elaborate about how you reach $≤frac{3}{2n-1}$ from its preceding step...
    – Akash Patalwanshi
    Dec 2 '18 at 7:04








  • 1




    $m-nleq 2m-1$ and $1/(2m-1)leq 1/(2n-1)$
    – John_Wick
    Dec 2 '18 at 7:06










  • Thank you so much sir. Really very beautiful answer :-)
    – Akash Patalwanshi
    Dec 2 '18 at 7:07



















1














Using logic, algebra and simple inequality rules, one can show that for $n ge 3$, $x_n$ is between the prior two terms, $x_{n-1}$ and $x_{n-2}$. So if $N ge 3$ and $m ge N$, the number $x_m$ lies inside the interval formed by $x_{N-1}$ and $x_{N-2}$. But then of course the distance between $x_m$ and another term $x_n$ with $n ge N$ can not be greater than the distance between $x_{N-1}$ and $x_{N-2}$.



If $N = 2k+1$, using the above reasoning, we know that $x_{N-2} lt x_{N-1}$. Also



$tag 1 x_{N-1} = frac{2k + 1}{4k-1}$



and



$tag 2 x_{N-2} = frac{2k-2}{4k-3}$



Using algebra, we can write



$tag 3 x_{N-1} - x_{N-2} = frac{8k-5}{(4k-1)(4k-3)} ; text{ with } k ge 1$



By making $text{(3)}$ arbitrarily '$varepsilon$ small', the assertion that $x_n$ is a Cauchy sequence can be backed up by finding an $N$ so that $|x_m - x_n| le varepsilon$ for $m,n ge N$.



If it is your preference, you can make '$le varepsilon$' into '$lt varepsilon$' in at least two ways. But nothing changes - the two definitions of a Cauchy sequence, one with '$le varepsilon$' and the other using '$lt varepsilon$', are equivalent.






share|cite|improve this answer































    -1














    Using the work that you already have, you can bound your final equality by letting the -1 products be 1, i.e, $|frac{m-n+(-1)^n 2m+(-1)^{m+1}2n+(-1)^{n+1}+(-1)^{m+2}}{(2n-1)(2m-1)}|$ < $|frac{m-n+2m+2n+1 + 1}{(2n-1)(2m-1)}|$. You can then expand the denominator and you'll have a cancellation.






    share|cite|improve this answer























    • didn't get your answer :-( "by letting -1 products be 1"?
      – Akash Patalwanshi
      Dec 2 '18 at 7:11










    • If we let (-1)^m and (-1)^n just be 1, then we have a bound for your last equality.
      – modest_mildew
      Dec 2 '18 at 7:25











    Your Answer





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    3 Answers
    3






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    oldest

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    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5














    Suppose $m>n.$ $|x_m-x_n|=left|frac{m+(-1)^m}{2m-1}-frac{n+(-1)^n}{2n-1}right|leq left|frac{m}{2m-1}-frac{n}{2n-1}right|+frac{1}{2m-1}+frac{1}{2n-1}=frac{m-n}{(2m-1)(2n-1)}+frac{1}{2m-1}+frac{1}{2n-1}leq frac{3}{2n-1}<epsilon$



    for sufficiently large $m,n.$






    share|cite|improve this answer





















    • Sir, love your second step. But please could you elaborate about how you reach $≤frac{3}{2n-1}$ from its preceding step...
      – Akash Patalwanshi
      Dec 2 '18 at 7:04








    • 1




      $m-nleq 2m-1$ and $1/(2m-1)leq 1/(2n-1)$
      – John_Wick
      Dec 2 '18 at 7:06










    • Thank you so much sir. Really very beautiful answer :-)
      – Akash Patalwanshi
      Dec 2 '18 at 7:07
















    5














    Suppose $m>n.$ $|x_m-x_n|=left|frac{m+(-1)^m}{2m-1}-frac{n+(-1)^n}{2n-1}right|leq left|frac{m}{2m-1}-frac{n}{2n-1}right|+frac{1}{2m-1}+frac{1}{2n-1}=frac{m-n}{(2m-1)(2n-1)}+frac{1}{2m-1}+frac{1}{2n-1}leq frac{3}{2n-1}<epsilon$



    for sufficiently large $m,n.$






    share|cite|improve this answer





















    • Sir, love your second step. But please could you elaborate about how you reach $≤frac{3}{2n-1}$ from its preceding step...
      – Akash Patalwanshi
      Dec 2 '18 at 7:04








    • 1




      $m-nleq 2m-1$ and $1/(2m-1)leq 1/(2n-1)$
      – John_Wick
      Dec 2 '18 at 7:06










    • Thank you so much sir. Really very beautiful answer :-)
      – Akash Patalwanshi
      Dec 2 '18 at 7:07














    5












    5








    5






    Suppose $m>n.$ $|x_m-x_n|=left|frac{m+(-1)^m}{2m-1}-frac{n+(-1)^n}{2n-1}right|leq left|frac{m}{2m-1}-frac{n}{2n-1}right|+frac{1}{2m-1}+frac{1}{2n-1}=frac{m-n}{(2m-1)(2n-1)}+frac{1}{2m-1}+frac{1}{2n-1}leq frac{3}{2n-1}<epsilon$



    for sufficiently large $m,n.$






    share|cite|improve this answer












    Suppose $m>n.$ $|x_m-x_n|=left|frac{m+(-1)^m}{2m-1}-frac{n+(-1)^n}{2n-1}right|leq left|frac{m}{2m-1}-frac{n}{2n-1}right|+frac{1}{2m-1}+frac{1}{2n-1}=frac{m-n}{(2m-1)(2n-1)}+frac{1}{2m-1}+frac{1}{2n-1}leq frac{3}{2n-1}<epsilon$



    for sufficiently large $m,n.$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 2 '18 at 6:57









    John_Wick

    1,366111




    1,366111












    • Sir, love your second step. But please could you elaborate about how you reach $≤frac{3}{2n-1}$ from its preceding step...
      – Akash Patalwanshi
      Dec 2 '18 at 7:04








    • 1




      $m-nleq 2m-1$ and $1/(2m-1)leq 1/(2n-1)$
      – John_Wick
      Dec 2 '18 at 7:06










    • Thank you so much sir. Really very beautiful answer :-)
      – Akash Patalwanshi
      Dec 2 '18 at 7:07


















    • Sir, love your second step. But please could you elaborate about how you reach $≤frac{3}{2n-1}$ from its preceding step...
      – Akash Patalwanshi
      Dec 2 '18 at 7:04








    • 1




      $m-nleq 2m-1$ and $1/(2m-1)leq 1/(2n-1)$
      – John_Wick
      Dec 2 '18 at 7:06










    • Thank you so much sir. Really very beautiful answer :-)
      – Akash Patalwanshi
      Dec 2 '18 at 7:07
















    Sir, love your second step. But please could you elaborate about how you reach $≤frac{3}{2n-1}$ from its preceding step...
    – Akash Patalwanshi
    Dec 2 '18 at 7:04






    Sir, love your second step. But please could you elaborate about how you reach $≤frac{3}{2n-1}$ from its preceding step...
    – Akash Patalwanshi
    Dec 2 '18 at 7:04






    1




    1




    $m-nleq 2m-1$ and $1/(2m-1)leq 1/(2n-1)$
    – John_Wick
    Dec 2 '18 at 7:06




    $m-nleq 2m-1$ and $1/(2m-1)leq 1/(2n-1)$
    – John_Wick
    Dec 2 '18 at 7:06












    Thank you so much sir. Really very beautiful answer :-)
    – Akash Patalwanshi
    Dec 2 '18 at 7:07




    Thank you so much sir. Really very beautiful answer :-)
    – Akash Patalwanshi
    Dec 2 '18 at 7:07











    1














    Using logic, algebra and simple inequality rules, one can show that for $n ge 3$, $x_n$ is between the prior two terms, $x_{n-1}$ and $x_{n-2}$. So if $N ge 3$ and $m ge N$, the number $x_m$ lies inside the interval formed by $x_{N-1}$ and $x_{N-2}$. But then of course the distance between $x_m$ and another term $x_n$ with $n ge N$ can not be greater than the distance between $x_{N-1}$ and $x_{N-2}$.



    If $N = 2k+1$, using the above reasoning, we know that $x_{N-2} lt x_{N-1}$. Also



    $tag 1 x_{N-1} = frac{2k + 1}{4k-1}$



    and



    $tag 2 x_{N-2} = frac{2k-2}{4k-3}$



    Using algebra, we can write



    $tag 3 x_{N-1} - x_{N-2} = frac{8k-5}{(4k-1)(4k-3)} ; text{ with } k ge 1$



    By making $text{(3)}$ arbitrarily '$varepsilon$ small', the assertion that $x_n$ is a Cauchy sequence can be backed up by finding an $N$ so that $|x_m - x_n| le varepsilon$ for $m,n ge N$.



    If it is your preference, you can make '$le varepsilon$' into '$lt varepsilon$' in at least two ways. But nothing changes - the two definitions of a Cauchy sequence, one with '$le varepsilon$' and the other using '$lt varepsilon$', are equivalent.






    share|cite|improve this answer




























      1














      Using logic, algebra and simple inequality rules, one can show that for $n ge 3$, $x_n$ is between the prior two terms, $x_{n-1}$ and $x_{n-2}$. So if $N ge 3$ and $m ge N$, the number $x_m$ lies inside the interval formed by $x_{N-1}$ and $x_{N-2}$. But then of course the distance between $x_m$ and another term $x_n$ with $n ge N$ can not be greater than the distance between $x_{N-1}$ and $x_{N-2}$.



      If $N = 2k+1$, using the above reasoning, we know that $x_{N-2} lt x_{N-1}$. Also



      $tag 1 x_{N-1} = frac{2k + 1}{4k-1}$



      and



      $tag 2 x_{N-2} = frac{2k-2}{4k-3}$



      Using algebra, we can write



      $tag 3 x_{N-1} - x_{N-2} = frac{8k-5}{(4k-1)(4k-3)} ; text{ with } k ge 1$



      By making $text{(3)}$ arbitrarily '$varepsilon$ small', the assertion that $x_n$ is a Cauchy sequence can be backed up by finding an $N$ so that $|x_m - x_n| le varepsilon$ for $m,n ge N$.



      If it is your preference, you can make '$le varepsilon$' into '$lt varepsilon$' in at least two ways. But nothing changes - the two definitions of a Cauchy sequence, one with '$le varepsilon$' and the other using '$lt varepsilon$', are equivalent.






      share|cite|improve this answer


























        1












        1








        1






        Using logic, algebra and simple inequality rules, one can show that for $n ge 3$, $x_n$ is between the prior two terms, $x_{n-1}$ and $x_{n-2}$. So if $N ge 3$ and $m ge N$, the number $x_m$ lies inside the interval formed by $x_{N-1}$ and $x_{N-2}$. But then of course the distance between $x_m$ and another term $x_n$ with $n ge N$ can not be greater than the distance between $x_{N-1}$ and $x_{N-2}$.



        If $N = 2k+1$, using the above reasoning, we know that $x_{N-2} lt x_{N-1}$. Also



        $tag 1 x_{N-1} = frac{2k + 1}{4k-1}$



        and



        $tag 2 x_{N-2} = frac{2k-2}{4k-3}$



        Using algebra, we can write



        $tag 3 x_{N-1} - x_{N-2} = frac{8k-5}{(4k-1)(4k-3)} ; text{ with } k ge 1$



        By making $text{(3)}$ arbitrarily '$varepsilon$ small', the assertion that $x_n$ is a Cauchy sequence can be backed up by finding an $N$ so that $|x_m - x_n| le varepsilon$ for $m,n ge N$.



        If it is your preference, you can make '$le varepsilon$' into '$lt varepsilon$' in at least two ways. But nothing changes - the two definitions of a Cauchy sequence, one with '$le varepsilon$' and the other using '$lt varepsilon$', are equivalent.






        share|cite|improve this answer














        Using logic, algebra and simple inequality rules, one can show that for $n ge 3$, $x_n$ is between the prior two terms, $x_{n-1}$ and $x_{n-2}$. So if $N ge 3$ and $m ge N$, the number $x_m$ lies inside the interval formed by $x_{N-1}$ and $x_{N-2}$. But then of course the distance between $x_m$ and another term $x_n$ with $n ge N$ can not be greater than the distance between $x_{N-1}$ and $x_{N-2}$.



        If $N = 2k+1$, using the above reasoning, we know that $x_{N-2} lt x_{N-1}$. Also



        $tag 1 x_{N-1} = frac{2k + 1}{4k-1}$



        and



        $tag 2 x_{N-2} = frac{2k-2}{4k-3}$



        Using algebra, we can write



        $tag 3 x_{N-1} - x_{N-2} = frac{8k-5}{(4k-1)(4k-3)} ; text{ with } k ge 1$



        By making $text{(3)}$ arbitrarily '$varepsilon$ small', the assertion that $x_n$ is a Cauchy sequence can be backed up by finding an $N$ so that $|x_m - x_n| le varepsilon$ for $m,n ge N$.



        If it is your preference, you can make '$le varepsilon$' into '$lt varepsilon$' in at least two ways. But nothing changes - the two definitions of a Cauchy sequence, one with '$le varepsilon$' and the other using '$lt varepsilon$', are equivalent.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 2 '18 at 15:42

























        answered Dec 2 '18 at 13:22









        CopyPasteIt

        4,0301627




        4,0301627























            -1














            Using the work that you already have, you can bound your final equality by letting the -1 products be 1, i.e, $|frac{m-n+(-1)^n 2m+(-1)^{m+1}2n+(-1)^{n+1}+(-1)^{m+2}}{(2n-1)(2m-1)}|$ < $|frac{m-n+2m+2n+1 + 1}{(2n-1)(2m-1)}|$. You can then expand the denominator and you'll have a cancellation.






            share|cite|improve this answer























            • didn't get your answer :-( "by letting -1 products be 1"?
              – Akash Patalwanshi
              Dec 2 '18 at 7:11










            • If we let (-1)^m and (-1)^n just be 1, then we have a bound for your last equality.
              – modest_mildew
              Dec 2 '18 at 7:25
















            -1














            Using the work that you already have, you can bound your final equality by letting the -1 products be 1, i.e, $|frac{m-n+(-1)^n 2m+(-1)^{m+1}2n+(-1)^{n+1}+(-1)^{m+2}}{(2n-1)(2m-1)}|$ < $|frac{m-n+2m+2n+1 + 1}{(2n-1)(2m-1)}|$. You can then expand the denominator and you'll have a cancellation.






            share|cite|improve this answer























            • didn't get your answer :-( "by letting -1 products be 1"?
              – Akash Patalwanshi
              Dec 2 '18 at 7:11










            • If we let (-1)^m and (-1)^n just be 1, then we have a bound for your last equality.
              – modest_mildew
              Dec 2 '18 at 7:25














            -1












            -1








            -1






            Using the work that you already have, you can bound your final equality by letting the -1 products be 1, i.e, $|frac{m-n+(-1)^n 2m+(-1)^{m+1}2n+(-1)^{n+1}+(-1)^{m+2}}{(2n-1)(2m-1)}|$ < $|frac{m-n+2m+2n+1 + 1}{(2n-1)(2m-1)}|$. You can then expand the denominator and you'll have a cancellation.






            share|cite|improve this answer














            Using the work that you already have, you can bound your final equality by letting the -1 products be 1, i.e, $|frac{m-n+(-1)^n 2m+(-1)^{m+1}2n+(-1)^{n+1}+(-1)^{m+2}}{(2n-1)(2m-1)}|$ < $|frac{m-n+2m+2n+1 + 1}{(2n-1)(2m-1)}|$. You can then expand the denominator and you'll have a cancellation.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 2 '18 at 7:32

























            answered Dec 2 '18 at 7:06









            modest_mildew

            112




            112












            • didn't get your answer :-( "by letting -1 products be 1"?
              – Akash Patalwanshi
              Dec 2 '18 at 7:11










            • If we let (-1)^m and (-1)^n just be 1, then we have a bound for your last equality.
              – modest_mildew
              Dec 2 '18 at 7:25


















            • didn't get your answer :-( "by letting -1 products be 1"?
              – Akash Patalwanshi
              Dec 2 '18 at 7:11










            • If we let (-1)^m and (-1)^n just be 1, then we have a bound for your last equality.
              – modest_mildew
              Dec 2 '18 at 7:25
















            didn't get your answer :-( "by letting -1 products be 1"?
            – Akash Patalwanshi
            Dec 2 '18 at 7:11




            didn't get your answer :-( "by letting -1 products be 1"?
            – Akash Patalwanshi
            Dec 2 '18 at 7:11












            If we let (-1)^m and (-1)^n just be 1, then we have a bound for your last equality.
            – modest_mildew
            Dec 2 '18 at 7:25




            If we let (-1)^m and (-1)^n just be 1, then we have a bound for your last equality.
            – modest_mildew
            Dec 2 '18 at 7:25


















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