Showing that a piece-wise defined sequence converges to $0$ via $epsilon$-definition?












1












$begingroup$


I wish to show if $S_n = begin{cases}
2e^{-n}, & mbox{if n is even } \
-frac{3}{n}, & mbox{if n is odd }
end{cases}$ then we have $S_n to 0$ as $n to infty$.



Does it suffice to consider the the odd/even terms of the sequence separately? e.g.



If $n$ is even, take $ N > log(2/epsilon)$. Then $n > N implies n > log(2/3) implies |2e^{-n}|< epsilon$



And if $n$ is odd, then take $N > -frac{3}{epsilon}$. Then $n> N implies n > -frac{3}{epsilon} implies |-frac{3}{n}| < epsilon$.






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You need to be more careful. $N$ is supposed to be an (positive) integer and neither of your choices are integers and one of them is even negative!
    $endgroup$
    – LoveTooNap29
    Mar 5 '18 at 17:51










  • $begingroup$
    Your first $N$ is not a function of $epsilon$ so something is wrong.
    $endgroup$
    – Jair Taylor
    Mar 5 '18 at 17:53










  • $begingroup$
    In general, the trick in this kind of situation is to find $N_1$ for the first situation and $N_2$ for the second situation and then set $N = max(N_1,N_2)$. Then if $n > N$ then $n > N_1$ and $n > N_2$ so everything works.
    $endgroup$
    – Jair Taylor
    Mar 5 '18 at 17:54










  • $begingroup$
    But there has to be just one N for each given $epsilon$.
    $endgroup$
    – Jair Taylor
    Mar 5 '18 at 17:55










  • $begingroup$
    Thanks for the reply @LoveTooNap29 Instead of saying "take $N= ldots $" would it be correct to say "take $N > ldots $"?
    $endgroup$
    – Ryan
    Mar 5 '18 at 17:56


















1












$begingroup$


I wish to show if $S_n = begin{cases}
2e^{-n}, & mbox{if n is even } \
-frac{3}{n}, & mbox{if n is odd }
end{cases}$ then we have $S_n to 0$ as $n to infty$.



Does it suffice to consider the the odd/even terms of the sequence separately? e.g.



If $n$ is even, take $ N > log(2/epsilon)$. Then $n > N implies n > log(2/3) implies |2e^{-n}|< epsilon$



And if $n$ is odd, then take $N > -frac{3}{epsilon}$. Then $n> N implies n > -frac{3}{epsilon} implies |-frac{3}{n}| < epsilon$.






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You need to be more careful. $N$ is supposed to be an (positive) integer and neither of your choices are integers and one of them is even negative!
    $endgroup$
    – LoveTooNap29
    Mar 5 '18 at 17:51










  • $begingroup$
    Your first $N$ is not a function of $epsilon$ so something is wrong.
    $endgroup$
    – Jair Taylor
    Mar 5 '18 at 17:53










  • $begingroup$
    In general, the trick in this kind of situation is to find $N_1$ for the first situation and $N_2$ for the second situation and then set $N = max(N_1,N_2)$. Then if $n > N$ then $n > N_1$ and $n > N_2$ so everything works.
    $endgroup$
    – Jair Taylor
    Mar 5 '18 at 17:54










  • $begingroup$
    But there has to be just one N for each given $epsilon$.
    $endgroup$
    – Jair Taylor
    Mar 5 '18 at 17:55










  • $begingroup$
    Thanks for the reply @LoveTooNap29 Instead of saying "take $N= ldots $" would it be correct to say "take $N > ldots $"?
    $endgroup$
    – Ryan
    Mar 5 '18 at 17:56
















1












1








1





$begingroup$


I wish to show if $S_n = begin{cases}
2e^{-n}, & mbox{if n is even } \
-frac{3}{n}, & mbox{if n is odd }
end{cases}$ then we have $S_n to 0$ as $n to infty$.



Does it suffice to consider the the odd/even terms of the sequence separately? e.g.



If $n$ is even, take $ N > log(2/epsilon)$. Then $n > N implies n > log(2/3) implies |2e^{-n}|< epsilon$



And if $n$ is odd, then take $N > -frac{3}{epsilon}$. Then $n> N implies n > -frac{3}{epsilon} implies |-frac{3}{n}| < epsilon$.






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$




I wish to show if $S_n = begin{cases}
2e^{-n}, & mbox{if n is even } \
-frac{3}{n}, & mbox{if n is odd }
end{cases}$ then we have $S_n to 0$ as $n to infty$.



Does it suffice to consider the the odd/even terms of the sequence separately? e.g.



If $n$ is even, take $ N > log(2/epsilon)$. Then $n > N implies n > log(2/3) implies |2e^{-n}|< epsilon$



And if $n$ is odd, then take $N > -frac{3}{epsilon}$. Then $n> N implies n > -frac{3}{epsilon} implies |-frac{3}{n}| < epsilon$.






real-analysis sequences-and-series




real-analysis sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 5 '18 at 17:54







Ryan

















asked Mar 5 '18 at 17:41









RyanRyan

399211




399211








  • 1




    $begingroup$
    You need to be more careful. $N$ is supposed to be an (positive) integer and neither of your choices are integers and one of them is even negative!
    $endgroup$
    – LoveTooNap29
    Mar 5 '18 at 17:51










  • $begingroup$
    Your first $N$ is not a function of $epsilon$ so something is wrong.
    $endgroup$
    – Jair Taylor
    Mar 5 '18 at 17:53










  • $begingroup$
    In general, the trick in this kind of situation is to find $N_1$ for the first situation and $N_2$ for the second situation and then set $N = max(N_1,N_2)$. Then if $n > N$ then $n > N_1$ and $n > N_2$ so everything works.
    $endgroup$
    – Jair Taylor
    Mar 5 '18 at 17:54










  • $begingroup$
    But there has to be just one N for each given $epsilon$.
    $endgroup$
    – Jair Taylor
    Mar 5 '18 at 17:55










  • $begingroup$
    Thanks for the reply @LoveTooNap29 Instead of saying "take $N= ldots $" would it be correct to say "take $N > ldots $"?
    $endgroup$
    – Ryan
    Mar 5 '18 at 17:56
















  • 1




    $begingroup$
    You need to be more careful. $N$ is supposed to be an (positive) integer and neither of your choices are integers and one of them is even negative!
    $endgroup$
    – LoveTooNap29
    Mar 5 '18 at 17:51










  • $begingroup$
    Your first $N$ is not a function of $epsilon$ so something is wrong.
    $endgroup$
    – Jair Taylor
    Mar 5 '18 at 17:53










  • $begingroup$
    In general, the trick in this kind of situation is to find $N_1$ for the first situation and $N_2$ for the second situation and then set $N = max(N_1,N_2)$. Then if $n > N$ then $n > N_1$ and $n > N_2$ so everything works.
    $endgroup$
    – Jair Taylor
    Mar 5 '18 at 17:54










  • $begingroup$
    But there has to be just one N for each given $epsilon$.
    $endgroup$
    – Jair Taylor
    Mar 5 '18 at 17:55










  • $begingroup$
    Thanks for the reply @LoveTooNap29 Instead of saying "take $N= ldots $" would it be correct to say "take $N > ldots $"?
    $endgroup$
    – Ryan
    Mar 5 '18 at 17:56










1




1




$begingroup$
You need to be more careful. $N$ is supposed to be an (positive) integer and neither of your choices are integers and one of them is even negative!
$endgroup$
– LoveTooNap29
Mar 5 '18 at 17:51




$begingroup$
You need to be more careful. $N$ is supposed to be an (positive) integer and neither of your choices are integers and one of them is even negative!
$endgroup$
– LoveTooNap29
Mar 5 '18 at 17:51












$begingroup$
Your first $N$ is not a function of $epsilon$ so something is wrong.
$endgroup$
– Jair Taylor
Mar 5 '18 at 17:53




$begingroup$
Your first $N$ is not a function of $epsilon$ so something is wrong.
$endgroup$
– Jair Taylor
Mar 5 '18 at 17:53












$begingroup$
In general, the trick in this kind of situation is to find $N_1$ for the first situation and $N_2$ for the second situation and then set $N = max(N_1,N_2)$. Then if $n > N$ then $n > N_1$ and $n > N_2$ so everything works.
$endgroup$
– Jair Taylor
Mar 5 '18 at 17:54




$begingroup$
In general, the trick in this kind of situation is to find $N_1$ for the first situation and $N_2$ for the second situation and then set $N = max(N_1,N_2)$. Then if $n > N$ then $n > N_1$ and $n > N_2$ so everything works.
$endgroup$
– Jair Taylor
Mar 5 '18 at 17:54












$begingroup$
But there has to be just one N for each given $epsilon$.
$endgroup$
– Jair Taylor
Mar 5 '18 at 17:55




$begingroup$
But there has to be just one N for each given $epsilon$.
$endgroup$
– Jair Taylor
Mar 5 '18 at 17:55












$begingroup$
Thanks for the reply @LoveTooNap29 Instead of saying "take $N= ldots $" would it be correct to say "take $N > ldots $"?
$endgroup$
– Ryan
Mar 5 '18 at 17:56






$begingroup$
Thanks for the reply @LoveTooNap29 Instead of saying "take $N= ldots $" would it be correct to say "take $N > ldots $"?
$endgroup$
– Ryan
Mar 5 '18 at 17:56












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That's right. Also notice that for $n$ being odd, we must have $$N>{3over epsilon}$$not $N>-{3over epsilon}$. The general $N$ for both even and odd terms then would become$$N>maxleft{{3over epsilon} , log{2over epsilon} right}$$






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    $begingroup$

    That's right. Also notice that for $n$ being odd, we must have $$N>{3over epsilon}$$not $N>-{3over epsilon}$. The general $N$ for both even and odd terms then would become$$N>maxleft{{3over epsilon} , log{2over epsilon} right}$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      That's right. Also notice that for $n$ being odd, we must have $$N>{3over epsilon}$$not $N>-{3over epsilon}$. The general $N$ for both even and odd terms then would become$$N>maxleft{{3over epsilon} , log{2over epsilon} right}$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        That's right. Also notice that for $n$ being odd, we must have $$N>{3over epsilon}$$not $N>-{3over epsilon}$. The general $N$ for both even and odd terms then would become$$N>maxleft{{3over epsilon} , log{2over epsilon} right}$$






        share|cite|improve this answer









        $endgroup$



        That's right. Also notice that for $n$ being odd, we must have $$N>{3over epsilon}$$not $N>-{3over epsilon}$. The general $N$ for both even and odd terms then would become$$N>maxleft{{3over epsilon} , log{2over epsilon} right}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 1 '18 at 14:10









        Mostafa AyazMostafa Ayaz

        15.6k3939




        15.6k3939






























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