Showing that a piece-wise defined sequence converges to $0$ via $epsilon$-definition?
$begingroup$
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
asked Mar 5 '18 at 17:41
RyanRyan
399211
$endgroup$
|
show 3 more comments
$begingroup$
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
asked Mar 5 '18 at 17:41
RyanRyan
399211
$endgroup$
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
|
show 3 more comments
$begingroup$
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
asked Mar 5 '18 at 17:41
RyanRyan
399211
$endgroup$
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
asked Mar 5 '18 at 17:41
RyanRyan
399211
asked Mar 5 '18 at 17:41
RyanRyan
399211
edited Mar 5 '18 at 17:54
Ryan
asked Mar 5 '18 at 17:41
RyanRyan
399211
asked Mar 5 '18 at 17:41
RyanRyan
399211
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
|
show 3 more comments
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
|
show 3 more comments
1 Answer
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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$$N>maxleft{{3over epsilon} , log{2over epsilon} right}$$
answered Dec 1 '18 at 14:10
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
add a comment |
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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$$N>maxleft{{3over epsilon} , log{2over epsilon} right}$$
answered Dec 1 '18 at 14:10
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
add a comment |
$begingroup$
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$$N>maxleft{{3over epsilon} , log{2over epsilon} right}$$
answered Dec 1 '18 at 14:10
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
add a comment |
$begingroup$
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$$N>maxleft{{3over epsilon} , log{2over epsilon} right}$$
answered Dec 1 '18 at 14:10
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
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$$N>maxleft{{3over epsilon} , log{2over epsilon} right}$$
answered Dec 1 '18 at 14:10
Mostafa AyazMostafa Ayaz
15.6k3939
answered Dec 1 '18 at 14:10
Mostafa AyazMostafa Ayaz
15.6k3939
answered Dec 1 '18 at 14:10
Mostafa AyazMostafa Ayaz
15.6k3939
answered Dec 1 '18 at 14:10
Mostafa AyazMostafa Ayaz
15.6k3939
15.6k3939
add a comment |
add a comment |
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$begingroup$
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