Does the series for $cos(x)/x$ converges? [duplicate]
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For what values of $theta$ does $sum_1^infty frace^i nthetan$ Converge?
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The sequence of
$$
a_x =cos (x)over x
$$
does converge to zero.
As a result, intuitively
$$
sum_x=1^infty cos (x)over x
$$
should also converge right? But I've been told that the series diverges. This shouldn't be true... right?
sequences-and-series
edited Aug 19 at 12:24
Bernard
111k635103
asked Aug 19 at 11:38
A man with a hat
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marked as duplicate by Isaac Browne, Nosrati, Lord Shark the Unknown, Arnaud D., Jack D'Aurizio♦ sequences-and-series
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Aug 19 at 13:18
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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up vote
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This question already has an answer here:
For what values of $theta$ does $sum_1^infty frace^i nthetan$ Converge?
2 answers
The sequence of
$$
a_x =cos (x)over x
$$
does converge to zero.
As a result, intuitively
$$
sum_x=1^infty cos (x)over x
$$
should also converge right? But I've been told that the series diverges. This shouldn't be true... right?
sequences-and-series
edited Aug 19 at 12:24
Bernard
111k635103
asked Aug 19 at 11:38
A man with a hat
204
marked as duplicate by Isaac Browne, Nosrati, Lord Shark the Unknown, Arnaud D., Jack D'Aurizio♦ sequences-and-series
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Aug 19 at 13:18
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
What is $n$ in your definition of $a_n$ ? Do you mean $cos(n)/n$?
– horace
Aug 19 at 11:40
1
Are you sure you formatted the question right? It isn't making much sense to me at the moment.
– prog_SAHIL
Aug 19 at 11:41
sorry, i'll edit it
– A man with a hat
Aug 19 at 11:41
When $n=0,$ the term $cos(0)/0$ is not defined. Do you mean to sum from $n=1$ to $infty$?
– coffeemath
Aug 19 at 11:46
1
If a series converges, then the limit of the general term must be $0$. The converse is not true, as the usual example of the harmonic series shows.
– Bernard Massé
Aug 19 at 11:54
add a comment |Â
up vote
1
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up vote
1
down vote
favorite
This question already has an answer here:
For what values of $theta$ does $sum_1^infty frace^i nthetan$ Converge?
2 answers
The sequence of
$$
a_x =cos (x)over x
$$
does converge to zero.
As a result, intuitively
$$
sum_x=1^infty cos (x)over x
$$
should also converge right? But I've been told that the series diverges. This shouldn't be true... right?
sequences-and-series
edited Aug 19 at 12:24
Bernard
111k635103
asked Aug 19 at 11:38
A man with a hat
204
This question already has an answer here:
For what values of $theta$ does $sum_1^infty frace^i nthetan$ Converge?
2 answers
The sequence of
$$
a_x =cos (x)over x
$$
does converge to zero.
As a result, intuitively
$$
sum_x=1^infty cos (x)over x
$$
should also converge right? But I've been told that the series diverges. This shouldn't be true... right?
This question already has an answer here:
For what values of $theta$ does $sum_1^infty frace^i nthetan$ Converge?
2 answers
sequences-and-series
edited Aug 19 at 12:24
Bernard
111k635103
asked Aug 19 at 11:38
A man with a hat
204
edited Aug 19 at 12:24
Bernard
111k635103
edited Aug 19 at 12:24
Bernard
111k635103
edited Aug 19 at 12:24
Bernard
111k635103
111k635103
asked Aug 19 at 11:38
A man with a hat
204
asked Aug 19 at 11:38
A man with a hat
204
asked Aug 19 at 11:38
A man with a hat
204
204
marked as duplicate by Isaac Browne, Nosrati, Lord Shark the Unknown, Arnaud D., Jack D'Aurizio♦ sequences-and-series
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Aug 19 at 13:18
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marked as duplicate by Isaac Browne, Nosrati, Lord Shark the Unknown, Arnaud D., Jack D'Aurizio♦ sequences-and-series
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This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
What is $n$ in your definition of $a_n$ ? Do you mean $cos(n)/n$?
– horace
Aug 19 at 11:40
1
Are you sure you formatted the question right? It isn't making much sense to me at the moment.
– prog_SAHIL
Aug 19 at 11:41
sorry, i'll edit it
– A man with a hat
Aug 19 at 11:41
When $n=0,$ the term $cos(0)/0$ is not defined. Do you mean to sum from $n=1$ to $infty$?
– coffeemath
Aug 19 at 11:46
1
If a series converges, then the limit of the general term must be $0$. The converse is not true, as the usual example of the harmonic series shows.
– Bernard Massé
Aug 19 at 11:54
add a comment |Â
2
What is $n$ in your definition of $a_n$ ? Do you mean $cos(n)/n$?
– horace
Aug 19 at 11:40
1
Are you sure you formatted the question right? It isn't making much sense to me at the moment.
– prog_SAHIL
Aug 19 at 11:41
sorry, i'll edit it
– A man with a hat
Aug 19 at 11:41
When $n=0,$ the term $cos(0)/0$ is not defined. Do you mean to sum from $n=1$ to $infty$?
– coffeemath
Aug 19 at 11:46
1
If a series converges, then the limit of the general term must be $0$. The converse is not true, as the usual example of the harmonic series shows.
– Bernard Massé
Aug 19 at 11:54
2
2
What is $n$ in your definition of $a_n$ ? Do you mean $cos(n)/n$?
– horace
Aug 19 at 11:40
What is $n$ in your definition of $a_n$ ? Do you mean $cos(n)/n$?
– horace
Aug 19 at 11:40
1
1
Are you sure you formatted the question right? It isn't making much sense to me at the moment.
– prog_SAHIL
Aug 19 at 11:41
Are you sure you formatted the question right? It isn't making much sense to me at the moment.
– prog_SAHIL
Aug 19 at 11:41
sorry, i'll edit it
– A man with a hat
Aug 19 at 11:41
sorry, i'll edit it
– A man with a hat
Aug 19 at 11:41
When $n=0,$ the term $cos(0)/0$ is not defined. Do you mean to sum from $n=1$ to $infty$?
– coffeemath
Aug 19 at 11:46
When $n=0,$ the term $cos(0)/0$ is not defined. Do you mean to sum from $n=1$ to $infty$?
– coffeemath
Aug 19 at 11:46
1
1
If a series converges, then the limit of the general term must be $0$. The converse is not true, as the usual example of the harmonic series shows.
– Bernard Massé
Aug 19 at 11:54
If a series converges, then the limit of the general term must be $0$. The converse is not true, as the usual example of the harmonic series shows.
– Bernard Massé
Aug 19 at 11:54
add a comment |Â
1 Answer
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It is not at all intuitive to me that the series ought to converge simply because the terms go to zero. For example $sum_n=1^infty frac1n$ is well known to diverge even though $frac1nto 0$. Or, as an even easier example, consider
$$ 1 + underbracefrac12+ frac12_2text halves +
underbracefrac13 + frac13+ frac13_3text thirds +
underbracefrac14 + frac14 + frac14+ frac14_4text fourths+
underbracefrac15 + frac15+ frac15+ frac15+frac15_5text fifths+
cdots $$
It does look like your particular series converges (conditionally), by Dirichlet's test, though.
answered Aug 19 at 11:55
Henning Makholm
229k16294525
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1 Answer
1
active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
It is not at all intuitive to me that the series ought to converge simply because the terms go to zero. For example $sum_n=1^infty frac1n$ is well known to diverge even though $frac1nto 0$. Or, as an even easier example, consider
$$ 1 + underbracefrac12+ frac12_2text halves +
underbracefrac13 + frac13+ frac13_3text thirds +
underbracefrac14 + frac14 + frac14+ frac14_4text fourths+
underbracefrac15 + frac15+ frac15+ frac15+frac15_5text fifths+
cdots $$
It does look like your particular series converges (conditionally), by Dirichlet's test, though.
answered Aug 19 at 11:55
Henning Makholm
229k16294525
add a comment |Â
up vote
2
down vote
It is not at all intuitive to me that the series ought to converge simply because the terms go to zero. For example $sum_n=1^infty frac1n$ is well known to diverge even though $frac1nto 0$. Or, as an even easier example, consider
$$ 1 + underbracefrac12+ frac12_2text halves +
underbracefrac13 + frac13+ frac13_3text thirds +
underbracefrac14 + frac14 + frac14+ frac14_4text fourths+
underbracefrac15 + frac15+ frac15+ frac15+frac15_5text fifths+
cdots $$
It does look like your particular series converges (conditionally), by Dirichlet's test, though.
answered Aug 19 at 11:55
Henning Makholm
229k16294525
add a comment |Â
up vote
2
down vote
up vote
2
down vote
It is not at all intuitive to me that the series ought to converge simply because the terms go to zero. For example $sum_n=1^infty frac1n$ is well known to diverge even though $frac1nto 0$. Or, as an even easier example, consider
$$ 1 + underbracefrac12+ frac12_2text halves +
underbracefrac13 + frac13+ frac13_3text thirds +
underbracefrac14 + frac14 + frac14+ frac14_4text fourths+
underbracefrac15 + frac15+ frac15+ frac15+frac15_5text fifths+
cdots $$
It does look like your particular series converges (conditionally), by Dirichlet's test, though.
answered Aug 19 at 11:55
Henning Makholm
229k16294525
It is not at all intuitive to me that the series ought to converge simply because the terms go to zero. For example $sum_n=1^infty frac1n$ is well known to diverge even though $frac1nto 0$. Or, as an even easier example, consider
$$ 1 + underbracefrac12+ frac12_2text halves +
underbracefrac13 + frac13+ frac13_3text thirds +
underbracefrac14 + frac14 + frac14+ frac14_4text fourths+
underbracefrac15 + frac15+ frac15+ frac15+frac15_5text fifths+
cdots $$
It does look like your particular series converges (conditionally), by Dirichlet's test, though.
answered Aug 19 at 11:55
Henning Makholm
229k16294525
answered Aug 19 at 11:55
Henning Makholm
229k16294525
answered Aug 19 at 11:55
Henning Makholm
229k16294525
answered Aug 19 at 11:55
Henning Makholm
229k16294525
229k16294525
add a comment |Â
add a comment |Â
2
What is $n$ in your definition of $a_n$ ? Do you mean $cos(n)/n$?
– horace
Aug 19 at 11:40
1
Are you sure you formatted the question right? It isn't making much sense to me at the moment.
– prog_SAHIL
Aug 19 at 11:41
sorry, i'll edit it
– A man with a hat
Aug 19 at 11:41
When $n=0,$ the term $cos(0)/0$ is not defined. Do you mean to sum from $n=1$ to $infty$?
– coffeemath
Aug 19 at 11:46
1
If a series converges, then the limit of the general term must be $0$. The converse is not true, as the usual example of the harmonic series shows.
– Bernard Massé
Aug 19 at 11:54