I want to check for the convergence/divergence of $cos(1/n)$?
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2
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for checking it $$lim_{n to infty}cos(1/n)=lim_{n to infty} frac{e^{i/n}+e^{-i/n}}{2}=(1+1)/2=1$$ this means that the series diverges is it right thought.
sequences-and-series convergence
asked Dec 14 '15 at 6:50
Nebo Alex
1,1241028
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show 2 more comments
up vote
2
down vote
favorite
for checking it $$lim_{n to infty}cos(1/n)=lim_{n to infty} frac{e^{i/n}+e^{-i/n}}{2}=(1+1)/2=1$$ this means that the series diverges is it right thought.
sequences-and-series convergence
asked Dec 14 '15 at 6:50
Nebo Alex
1,1241028
You are right ;It diverges
– Learnmore
Dec 14 '15 at 6:52
but what happens if it would be $sin(1/n) $ the above procedure is right
– Nebo Alex
Dec 14 '15 at 6:54
yes sure it goes to zero
– Nebo Alex
Dec 14 '15 at 6:57
$cos(1/n^2)$ converges .
– Nebo Alex
Dec 14 '15 at 6:59
If it is $sin(1/n)$ the series $sum sin (1/n)$ is divergent, as it is comparable to $sum 1/n$.
– user99914
Dec 14 '15 at 7:03
|
show 2 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
for checking it $$lim_{n to infty}cos(1/n)=lim_{n to infty} frac{e^{i/n}+e^{-i/n}}{2}=(1+1)/2=1$$ this means that the series diverges is it right thought.
sequences-and-series convergence
asked Dec 14 '15 at 6:50
Nebo Alex
1,1241028
for checking it $$lim_{n to infty}cos(1/n)=lim_{n to infty} frac{e^{i/n}+e^{-i/n}}{2}=(1+1)/2=1$$ this means that the series diverges is it right thought.
sequences-and-series convergence
sequences-and-series convergence
asked Dec 14 '15 at 6:50
Nebo Alex
1,1241028
asked Dec 14 '15 at 6:50
Nebo Alex
1,1241028
edited Dec 14 '15 at 6:52
user228113
asked Dec 14 '15 at 6:50
Nebo Alex
1,1241028
asked Dec 14 '15 at 6:50
Nebo Alex
1,1241028
asked Dec 14 '15 at 6:50
Nebo Alex
1,1241028
1,1241028
You are right ;It diverges
– Learnmore
Dec 14 '15 at 6:52
but what happens if it would be $sin(1/n) $ the above procedure is right
– Nebo Alex
Dec 14 '15 at 6:54
yes sure it goes to zero
– Nebo Alex
Dec 14 '15 at 6:57
$cos(1/n^2)$ converges .
– Nebo Alex
Dec 14 '15 at 6:59
If it is $sin(1/n)$ the series $sum sin (1/n)$ is divergent, as it is comparable to $sum 1/n$.
– user99914
Dec 14 '15 at 7:03
|
show 2 more comments
You are right ;It diverges
– Learnmore
Dec 14 '15 at 6:52
but what happens if it would be $sin(1/n) $ the above procedure is right
– Nebo Alex
Dec 14 '15 at 6:54
yes sure it goes to zero
– Nebo Alex
Dec 14 '15 at 6:57
$cos(1/n^2)$ converges .
– Nebo Alex
Dec 14 '15 at 6:59
If it is $sin(1/n)$ the series $sum sin (1/n)$ is divergent, as it is comparable to $sum 1/n$.
– user99914
Dec 14 '15 at 7:03
You are right ;It diverges
– Learnmore
Dec 14 '15 at 6:52
You are right ;It diverges
– Learnmore
Dec 14 '15 at 6:52
but what happens if it would be $sin(1/n) $ the above procedure is right
– Nebo Alex
Dec 14 '15 at 6:54
but what happens if it would be $sin(1/n) $ the above procedure is right
– Nebo Alex
Dec 14 '15 at 6:54
yes sure it goes to zero
– Nebo Alex
Dec 14 '15 at 6:57
yes sure it goes to zero
– Nebo Alex
Dec 14 '15 at 6:57
$cos(1/n^2)$ converges .
– Nebo Alex
Dec 14 '15 at 6:59
$cos(1/n^2)$ converges .
– Nebo Alex
Dec 14 '15 at 6:59
If it is $sin(1/n)$ the series $sum sin (1/n)$ is divergent, as it is comparable to $sum 1/n$.
– user99914
Dec 14 '15 at 7:03
If it is $sin(1/n)$ the series $sum sin (1/n)$ is divergent, as it is comparable to $sum 1/n$.
– user99914
Dec 14 '15 at 7:03
|
show 2 more comments
1 Answer
1
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up vote
0
down vote
The easiest way to do this sort of limit is to use continuity. Note that the limit of the sequence is the same as the limit of the corresponding function (assuming the latter limit exists):
$$
lim_{ntoinfty} cos(tfrac1n) = lim_{xtoinfty} cos(tfrac1x).
$$
Now we know that $lim_{xtoinfty} frac1x = 0$, and that $cos(u)$ is continuous at $u=0$. Therefore
$$
lim_{xtoinfty} cos(tfrac1x) = cosbig( lim_{xtoinfty} tfrac1x big) = cos 0 = 1.
$$
answered Dec 14 '15 at 7:55
Greg Martin
34.4k23161
I'd be interested to hear the critiques implied by the downvotes.
– Greg Martin
Dec 15 '15 at 18:05
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The easiest way to do this sort of limit is to use continuity. Note that the limit of the sequence is the same as the limit of the corresponding function (assuming the latter limit exists):
$$
lim_{ntoinfty} cos(tfrac1n) = lim_{xtoinfty} cos(tfrac1x).
$$
Now we know that $lim_{xtoinfty} frac1x = 0$, and that $cos(u)$ is continuous at $u=0$. Therefore
$$
lim_{xtoinfty} cos(tfrac1x) = cosbig( lim_{xtoinfty} tfrac1x big) = cos 0 = 1.
$$
answered Dec 14 '15 at 7:55
Greg Martin
34.4k23161
I'd be interested to hear the critiques implied by the downvotes.
– Greg Martin
Dec 15 '15 at 18:05
add a comment |
up vote
0
down vote
The easiest way to do this sort of limit is to use continuity. Note that the limit of the sequence is the same as the limit of the corresponding function (assuming the latter limit exists):
$$
lim_{ntoinfty} cos(tfrac1n) = lim_{xtoinfty} cos(tfrac1x).
$$
Now we know that $lim_{xtoinfty} frac1x = 0$, and that $cos(u)$ is continuous at $u=0$. Therefore
$$
lim_{xtoinfty} cos(tfrac1x) = cosbig( lim_{xtoinfty} tfrac1x big) = cos 0 = 1.
$$
answered Dec 14 '15 at 7:55
Greg Martin
34.4k23161
I'd be interested to hear the critiques implied by the downvotes.
– Greg Martin
Dec 15 '15 at 18:05
add a comment |
up vote
0
down vote
up vote
0
down vote
The easiest way to do this sort of limit is to use continuity. Note that the limit of the sequence is the same as the limit of the corresponding function (assuming the latter limit exists):
$$
lim_{ntoinfty} cos(tfrac1n) = lim_{xtoinfty} cos(tfrac1x).
$$
Now we know that $lim_{xtoinfty} frac1x = 0$, and that $cos(u)$ is continuous at $u=0$. Therefore
$$
lim_{xtoinfty} cos(tfrac1x) = cosbig( lim_{xtoinfty} tfrac1x big) = cos 0 = 1.
$$
answered Dec 14 '15 at 7:55
Greg Martin
34.4k23161
The easiest way to do this sort of limit is to use continuity. Note that the limit of the sequence is the same as the limit of the corresponding function (assuming the latter limit exists):
$$
lim_{ntoinfty} cos(tfrac1n) = lim_{xtoinfty} cos(tfrac1x).
$$
Now we know that $lim_{xtoinfty} frac1x = 0$, and that $cos(u)$ is continuous at $u=0$. Therefore
$$
lim_{xtoinfty} cos(tfrac1x) = cosbig( lim_{xtoinfty} tfrac1x big) = cos 0 = 1.
$$
answered Dec 14 '15 at 7:55
Greg Martin
34.4k23161
answered Dec 14 '15 at 7:55
Greg Martin
34.4k23161
answered Dec 14 '15 at 7:55
Greg Martin
34.4k23161
answered Dec 14 '15 at 7:55
Greg Martin
34.4k23161
34.4k23161
I'd be interested to hear the critiques implied by the downvotes.
– Greg Martin
Dec 15 '15 at 18:05
add a comment |
I'd be interested to hear the critiques implied by the downvotes.
– Greg Martin
Dec 15 '15 at 18:05
I'd be interested to hear the critiques implied by the downvotes.
– Greg Martin
Dec 15 '15 at 18:05
I'd be interested to hear the critiques implied by the downvotes.
– Greg Martin
Dec 15 '15 at 18:05
add a comment |
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You are right ;It diverges
– Learnmore
Dec 14 '15 at 6:52
but what happens if it would be $sin(1/n) $ the above procedure is right
– Nebo Alex
Dec 14 '15 at 6:54
yes sure it goes to zero
– Nebo Alex
Dec 14 '15 at 6:57
$cos(1/n^2)$ converges .
– Nebo Alex
Dec 14 '15 at 6:59
If it is $sin(1/n)$ the series $sum sin (1/n)$ is divergent, as it is comparable to $sum 1/n$.
– user99914
Dec 14 '15 at 7:03