$lim_{n to +infty} n cdotcos^2left(frac{n pi}{3}right)$
$begingroup$
Find $lim_{n to +infty} n cdot cos^2left(frac{n pi}{3}right)$
First I have a look at $cos(frac{n pi}{3})$
What I expect is that the sequence diverges so I want to find two sub-sequences that tend to a different value, correct?
$cos$ is periodic for $2 pi$ so for the first sub-sequence let n:=6m and I get $cos(frac{6m cdot pi}{3})=cos(2m cdot pi)$ which is 1 for every m.
For the next sub-sequence it seems to me that finding a k that gets me either $cos(3pi)$ or $cos(pi)$ would be a good idea since as then the sub-sequence will always deliver 0. However I get think of any k that satisfies that. Can someone help me out here? Thanks in advance!
real-analysis limits limits-without-lhopital
edited Dec 3 '18 at 22:31
gimusi
92.9k94494
asked Dec 3 '18 at 22:20
D. JohnD. John
283
$endgroup$
add a comment |
$begingroup$
Find $lim_{n to +infty} n cdot cos^2left(frac{n pi}{3}right)$
First I have a look at $cos(frac{n pi}{3})$
What I expect is that the sequence diverges so I want to find two sub-sequences that tend to a different value, correct?
$cos$ is periodic for $2 pi$ so for the first sub-sequence let n:=6m and I get $cos(frac{6m cdot pi}{3})=cos(2m cdot pi)$ which is 1 for every m.
For the next sub-sequence it seems to me that finding a k that gets me either $cos(3pi)$ or $cos(pi)$ would be a good idea since as then the sub-sequence will always deliver 0. However I get think of any k that satisfies that. Can someone help me out here? Thanks in advance!
real-analysis limits limits-without-lhopital
edited Dec 3 '18 at 22:31
gimusi
92.9k94494
asked Dec 3 '18 at 22:20
D. JohnD. John
283
$endgroup$
add a comment |
$begingroup$
Find $lim_{n to +infty} n cdot cos^2left(frac{n pi}{3}right)$
First I have a look at $cos(frac{n pi}{3})$
What I expect is that the sequence diverges so I want to find two sub-sequences that tend to a different value, correct?
$cos$ is periodic for $2 pi$ so for the first sub-sequence let n:=6m and I get $cos(frac{6m cdot pi}{3})=cos(2m cdot pi)$ which is 1 for every m.
For the next sub-sequence it seems to me that finding a k that gets me either $cos(3pi)$ or $cos(pi)$ would be a good idea since as then the sub-sequence will always deliver 0. However I get think of any k that satisfies that. Can someone help me out here? Thanks in advance!
real-analysis limits limits-without-lhopital
edited Dec 3 '18 at 22:31
gimusi
92.9k94494
asked Dec 3 '18 at 22:20
D. JohnD. John
283
$endgroup$
Find $lim_{n to +infty} n cdot cos^2left(frac{n pi}{3}right)$
First I have a look at $cos(frac{n pi}{3})$
What I expect is that the sequence diverges so I want to find two sub-sequences that tend to a different value, correct?
$cos$ is periodic for $2 pi$ so for the first sub-sequence let n:=6m and I get $cos(frac{6m cdot pi}{3})=cos(2m cdot pi)$ which is 1 for every m.
For the next sub-sequence it seems to me that finding a k that gets me either $cos(3pi)$ or $cos(pi)$ would be a good idea since as then the sub-sequence will always deliver 0. However I get think of any k that satisfies that. Can someone help me out here? Thanks in advance!
real-analysis limits limits-without-lhopital
real-analysis limits limits-without-lhopital
edited Dec 3 '18 at 22:31
gimusi
92.9k94494
asked Dec 3 '18 at 22:20
D. JohnD. John
283
edited Dec 3 '18 at 22:31
gimusi
92.9k94494
asked Dec 3 '18 at 22:20
D. JohnD. John
283
edited Dec 3 '18 at 22:31
gimusi
92.9k94494
edited Dec 3 '18 at 22:31
gimusi
92.9k94494
edited Dec 3 '18 at 22:31
gimusi
92.9k94494
92.9k94494
asked Dec 3 '18 at 22:20
D. JohnD. John
283
asked Dec 3 '18 at 22:20
D. JohnD. John
283
asked Dec 3 '18 at 22:20
D. JohnD. John
283
283
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Unfortunately, you cannot find two subsequences convergent to different values since the general limit is $infty$. Let $$a_{3n}=3ncos^2{npi}=3n\a_{3n+1}={(3n+1)}cos^2{(npi +{pi over 3})}={3n+1over 4}\a_{3n-1}={(3n-1)}cos^2{(npi -{pi over 3})}={3n+1over 4}$$what can you conclude from here?
answered Dec 3 '18 at 22:26
Mostafa AyazMostafa Ayaz
15.4k3939
$endgroup$
$begingroup$
Hm I'm not sure actually. Do you want me to apply the squeeze theorem?
$endgroup$
– D. John
Dec 3 '18 at 22:36
$begingroup$
I just argued that since any term of the sequence (or even infinitely many terms of any arbitrary subsequence such as $a_{n^2}$) fall(s) in at least one $a_{3n+k}$ for $k=0,1,-1$ and all those subsequences tend to $infty$ so does the sequence itself.
$endgroup$
– Mostafa Ayaz
Dec 4 '18 at 8:27
add a comment |
$begingroup$
We have that $forall nin mathbb{N}$
$$cos^2left(frac{n pi}{3}right) ge frac14$$
therefore
$$ n cdot cos^2left(frac{n pi}{3}right) ge frac n 4to infty$$
answered Dec 3 '18 at 22:23
gimusigimusi
92.9k94494
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
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$begingroup$
Unfortunately, you cannot find two subsequences convergent to different values since the general limit is $infty$. Let $$a_{3n}=3ncos^2{npi}=3n\a_{3n+1}={(3n+1)}cos^2{(npi +{pi over 3})}={3n+1over 4}\a_{3n-1}={(3n-1)}cos^2{(npi -{pi over 3})}={3n+1over 4}$$what can you conclude from here?
answered Dec 3 '18 at 22:26
Mostafa AyazMostafa Ayaz
15.4k3939
$endgroup$
$begingroup$
Hm I'm not sure actually. Do you want me to apply the squeeze theorem?
$endgroup$
– D. John
Dec 3 '18 at 22:36
$begingroup$
I just argued that since any term of the sequence (or even infinitely many terms of any arbitrary subsequence such as $a_{n^2}$) fall(s) in at least one $a_{3n+k}$ for $k=0,1,-1$ and all those subsequences tend to $infty$ so does the sequence itself.
$endgroup$
– Mostafa Ayaz
Dec 4 '18 at 8:27
add a comment |
$begingroup$
Unfortunately, you cannot find two subsequences convergent to different values since the general limit is $infty$. Let $$a_{3n}=3ncos^2{npi}=3n\a_{3n+1}={(3n+1)}cos^2{(npi +{pi over 3})}={3n+1over 4}\a_{3n-1}={(3n-1)}cos^2{(npi -{pi over 3})}={3n+1over 4}$$what can you conclude from here?
answered Dec 3 '18 at 22:26
Mostafa AyazMostafa Ayaz
15.4k3939
$endgroup$
$begingroup$
Hm I'm not sure actually. Do you want me to apply the squeeze theorem?
$endgroup$
– D. John
Dec 3 '18 at 22:36
$begingroup$
I just argued that since any term of the sequence (or even infinitely many terms of any arbitrary subsequence such as $a_{n^2}$) fall(s) in at least one $a_{3n+k}$ for $k=0,1,-1$ and all those subsequences tend to $infty$ so does the sequence itself.
$endgroup$
– Mostafa Ayaz
Dec 4 '18 at 8:27
add a comment |
$begingroup$
Unfortunately, you cannot find two subsequences convergent to different values since the general limit is $infty$. Let $$a_{3n}=3ncos^2{npi}=3n\a_{3n+1}={(3n+1)}cos^2{(npi +{pi over 3})}={3n+1over 4}\a_{3n-1}={(3n-1)}cos^2{(npi -{pi over 3})}={3n+1over 4}$$what can you conclude from here?
answered Dec 3 '18 at 22:26
Mostafa AyazMostafa Ayaz
15.4k3939
$endgroup$
Unfortunately, you cannot find two subsequences convergent to different values since the general limit is $infty$. Let $$a_{3n}=3ncos^2{npi}=3n\a_{3n+1}={(3n+1)}cos^2{(npi +{pi over 3})}={3n+1over 4}\a_{3n-1}={(3n-1)}cos^2{(npi -{pi over 3})}={3n+1over 4}$$what can you conclude from here?
answered Dec 3 '18 at 22:26
Mostafa AyazMostafa Ayaz
15.4k3939
answered Dec 3 '18 at 22:26
Mostafa AyazMostafa Ayaz
15.4k3939
answered Dec 3 '18 at 22:26
Mostafa AyazMostafa Ayaz
15.4k3939
answered Dec 3 '18 at 22:26
Mostafa AyazMostafa Ayaz
15.4k3939
15.4k3939
$begingroup$
Hm I'm not sure actually. Do you want me to apply the squeeze theorem?
$endgroup$
– D. John
Dec 3 '18 at 22:36
$begingroup$
I just argued that since any term of the sequence (or even infinitely many terms of any arbitrary subsequence such as $a_{n^2}$) fall(s) in at least one $a_{3n+k}$ for $k=0,1,-1$ and all those subsequences tend to $infty$ so does the sequence itself.
$endgroup$
– Mostafa Ayaz
Dec 4 '18 at 8:27
add a comment |
$begingroup$
Hm I'm not sure actually. Do you want me to apply the squeeze theorem?
$endgroup$
– D. John
Dec 3 '18 at 22:36
$begingroup$
I just argued that since any term of the sequence (or even infinitely many terms of any arbitrary subsequence such as $a_{n^2}$) fall(s) in at least one $a_{3n+k}$ for $k=0,1,-1$ and all those subsequences tend to $infty$ so does the sequence itself.
$endgroup$
– Mostafa Ayaz
Dec 4 '18 at 8:27
$begingroup$
Hm I'm not sure actually. Do you want me to apply the squeeze theorem?
$endgroup$
– D. John
Dec 3 '18 at 22:36
$begingroup$
Hm I'm not sure actually. Do you want me to apply the squeeze theorem?
$endgroup$
– D. John
Dec 3 '18 at 22:36
$begingroup$
I just argued that since any term of the sequence (or even infinitely many terms of any arbitrary subsequence such as $a_{n^2}$) fall(s) in at least one $a_{3n+k}$ for $k=0,1,-1$ and all those subsequences tend to $infty$ so does the sequence itself.
$endgroup$
– Mostafa Ayaz
Dec 4 '18 at 8:27
$begingroup$
I just argued that since any term of the sequence (or even infinitely many terms of any arbitrary subsequence such as $a_{n^2}$) fall(s) in at least one $a_{3n+k}$ for $k=0,1,-1$ and all those subsequences tend to $infty$ so does the sequence itself.
$endgroup$
– Mostafa Ayaz
Dec 4 '18 at 8:27
add a comment |
$begingroup$
We have that $forall nin mathbb{N}$
$$cos^2left(frac{n pi}{3}right) ge frac14$$
therefore
$$ n cdot cos^2left(frac{n pi}{3}right) ge frac n 4to infty$$
answered Dec 3 '18 at 22:23
gimusigimusi
92.9k94494
$endgroup$
add a comment |
$begingroup$
We have that $forall nin mathbb{N}$
$$cos^2left(frac{n pi}{3}right) ge frac14$$
therefore
$$ n cdot cos^2left(frac{n pi}{3}right) ge frac n 4to infty$$
answered Dec 3 '18 at 22:23
gimusigimusi
92.9k94494
$endgroup$
add a comment |
$begingroup$
We have that $forall nin mathbb{N}$
$$cos^2left(frac{n pi}{3}right) ge frac14$$
therefore
$$ n cdot cos^2left(frac{n pi}{3}right) ge frac n 4to infty$$
answered Dec 3 '18 at 22:23
gimusigimusi
92.9k94494
$endgroup$
We have that $forall nin mathbb{N}$
$$cos^2left(frac{n pi}{3}right) ge frac14$$
therefore
$$ n cdot cos^2left(frac{n pi}{3}right) ge frac n 4to infty$$
answered Dec 3 '18 at 22:23
gimusigimusi
92.9k94494
edited Dec 3 '18 at 22:28
answered Dec 3 '18 at 22:23
gimusigimusi
92.9k94494
answered Dec 3 '18 at 22:23
gimusigimusi
92.9k94494
answered Dec 3 '18 at 22:23
gimusigimusi
92.9k94494
92.9k94494
add a comment |
add a comment |
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