Convergence/Divergence of infinite series
Clash Royale CLAN TAG#URR8PPP
4
$begingroup$
I'm trying to figure out if the following series converges or diverges. I have spent hours on it and can't figure it out. Tried to use the comparison test, Dirichlet, and Abel, but none of it worked. Any hints?
$$
sum_n=1^infty fraccos^2(n)sqrtn
$$
real-analysis sequences-and-series convergence divergent-series
asked Feb 6 at 19:21
peroxisome7peroxisome7
234
$endgroup$
add a comment |
4
$begingroup$
I'm trying to figure out if the following series converges or diverges. I have spent hours on it and can't figure it out. Tried to use the comparison test, Dirichlet, and Abel, but none of it worked. Any hints?
$$
sum_n=1^infty fraccos^2(n)sqrtn
$$
real-analysis sequences-and-series convergence divergent-series
asked Feb 6 at 19:21
peroxisome7peroxisome7
234
$endgroup$
1
$begingroup$
Hey can you show what you did for the Abel comparasin test?
$endgroup$
– Max0815
Feb 6 at 19:38
add a comment |
4
4
4
2
$begingroup$
I'm trying to figure out if the following series converges or diverges. I have spent hours on it and can't figure it out. Tried to use the comparison test, Dirichlet, and Abel, but none of it worked. Any hints?
$$
sum_n=1^infty fraccos^2(n)sqrtn
$$
real-analysis sequences-and-series convergence divergent-series
asked Feb 6 at 19:21
peroxisome7peroxisome7
234
$endgroup$
I'm trying to figure out if the following series converges or diverges. I have spent hours on it and can't figure it out. Tried to use the comparison test, Dirichlet, and Abel, but none of it worked. Any hints?
$$
sum_n=1^infty fraccos^2(n)sqrtn
$$
real-analysis sequences-and-series convergence divergent-series
real-analysis sequences-and-series convergence divergent-series
asked Feb 6 at 19:21
peroxisome7peroxisome7
234
asked Feb 6 at 19:21
peroxisome7peroxisome7
234
asked Feb 6 at 19:21
peroxisome7peroxisome7
234
asked Feb 6 at 19:21
peroxisome7peroxisome7
234
asked Feb 6 at 19:21
peroxisome7peroxisome7
234
234
1
$begingroup$
Hey can you show what you did for the Abel comparasin test?
$endgroup$
– Max0815
Feb 6 at 19:38
add a comment |
1
$begingroup$
Hey can you show what you did for the Abel comparasin test?
$endgroup$
– Max0815
Feb 6 at 19:38
1
1
$begingroup$
Hey can you show what you did for the Abel comparasin test?
$endgroup$
– Max0815
Feb 6 at 19:38
$begingroup$
Hey can you show what you did for the Abel comparasin test?
$endgroup$
– Max0815
Feb 6 at 19:38
add a comment |
2 Answers
2
active
oldest
votes
8
$begingroup$
HINT:
Recall that $cos^2(x)=frac1+cos(2x)2$.
answered Feb 6 at 19:24
Mark ViolaMark Viola
133k1277174
$endgroup$
$begingroup$
Is using this and splitting the series into two helpful? Is the addition of a divergent and a convergent series also divergent?
$endgroup$
– peroxisome7
Feb 6 at 19:57
$begingroup$
@peroxisome7, Yes. That is a version of contrapositive statement of the fact that difference of two convergent series is again convergent.
$endgroup$
– Sangchul Lee
Feb 6 at 20:07
1
$begingroup$
@peroxisome Note that $$sum_n=1^N fraccos^2(n)sqrt n =frac12 sum_N=1^N frac1sqrt n+frac12 sum_n=1^N fraccos(2n)sqrt n$$Now, Dirichelt's test shows that the second partial sum on the right-hand side converges. If the partial sum on the left-hand side converged, then the first partial sum on the right-hand side would be the sum (difference) of two convergent partial sums and hence would also converge. Inasmuch as the first partial sum on the right-hand side diverges, what can one conclude?
$endgroup$
– Mark Viola
Feb 6 at 21:20
add a comment |
8
$begingroup$
Another approach is to observe that among $e^in, n=1,2,dots, 7,$ at least one of these points lies in the arc $e^it: tin (-pi/4,pi/4).$ Thus
$$sum_n=1^7 fraccos^2 nsqrt n ge frac1 2frac1sqrt 7.$$
The same thing happens for $n=8,dots,14.$ etc. So the series in question is at least
$$sum_m=1^infty frac1 2frac1sqrt 7m = infty.$$
Note that this idea will work if $cos^2 n$ is replaced by $|cos n|^p$ for any $p>0$.
answered Feb 6 at 19:57
zhw.zhw.
73.6k43175
$endgroup$
$begingroup$
(+1) I remember having used this trick to solve a previous question on the site. Under its apparence of simplicity, it is indeed rather powerful.
$endgroup$
– Did
Feb 6 at 20:05
2
$begingroup$
@zhw. Inasmuch as at least one of the points $1,2,dots, 7$ lies in that arc, $cos^2(n)ge 1/2$ for such a point. Then, $sum_n=1^7 fraccos^2(n)sqrt nge frac12sqrt 7$, would it not?
$endgroup$
– Mark Viola
Feb 6 at 20:42
$begingroup$
@MarkViola Right you are. Now edited.
$endgroup$
– zhw.
Feb 6 at 23:20
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
8
$begingroup$
HINT:
Recall that $cos^2(x)=frac1+cos(2x)2$.
answered Feb 6 at 19:24
Mark ViolaMark Viola
133k1277174
$endgroup$
$begingroup$
Is using this and splitting the series into two helpful? Is the addition of a divergent and a convergent series also divergent?
$endgroup$
– peroxisome7
Feb 6 at 19:57
$begingroup$
@peroxisome7, Yes. That is a version of contrapositive statement of the fact that difference of two convergent series is again convergent.
$endgroup$
– Sangchul Lee
Feb 6 at 20:07
1
$begingroup$
@peroxisome Note that $$sum_n=1^N fraccos^2(n)sqrt n =frac12 sum_N=1^N frac1sqrt n+frac12 sum_n=1^N fraccos(2n)sqrt n$$Now, Dirichelt's test shows that the second partial sum on the right-hand side converges. If the partial sum on the left-hand side converged, then the first partial sum on the right-hand side would be the sum (difference) of two convergent partial sums and hence would also converge. Inasmuch as the first partial sum on the right-hand side diverges, what can one conclude?
$endgroup$
– Mark Viola
Feb 6 at 21:20
add a comment |
8
$begingroup$
HINT:
Recall that $cos^2(x)=frac1+cos(2x)2$.
answered Feb 6 at 19:24
Mark ViolaMark Viola
133k1277174
$endgroup$
$begingroup$
Is using this and splitting the series into two helpful? Is the addition of a divergent and a convergent series also divergent?
$endgroup$
– peroxisome7
Feb 6 at 19:57
$begingroup$
@peroxisome7, Yes. That is a version of contrapositive statement of the fact that difference of two convergent series is again convergent.
$endgroup$
– Sangchul Lee
Feb 6 at 20:07
1
$begingroup$
@peroxisome Note that $$sum_n=1^N fraccos^2(n)sqrt n =frac12 sum_N=1^N frac1sqrt n+frac12 sum_n=1^N fraccos(2n)sqrt n$$Now, Dirichelt's test shows that the second partial sum on the right-hand side converges. If the partial sum on the left-hand side converged, then the first partial sum on the right-hand side would be the sum (difference) of two convergent partial sums and hence would also converge. Inasmuch as the first partial sum on the right-hand side diverges, what can one conclude?
$endgroup$
– Mark Viola
Feb 6 at 21:20
add a comment |
8
8
8
$begingroup$
HINT:
Recall that $cos^2(x)=frac1+cos(2x)2$.
answered Feb 6 at 19:24
Mark ViolaMark Viola
133k1277174
$endgroup$
HINT:
Recall that $cos^2(x)=frac1+cos(2x)2$.
answered Feb 6 at 19:24
Mark ViolaMark Viola
133k1277174
answered Feb 6 at 19:24
Mark ViolaMark Viola
133k1277174
answered Feb 6 at 19:24
Mark ViolaMark Viola
133k1277174
answered Feb 6 at 19:24
Mark ViolaMark Viola
133k1277174
133k1277174
$begingroup$
Is using this and splitting the series into two helpful? Is the addition of a divergent and a convergent series also divergent?
$endgroup$
– peroxisome7
Feb 6 at 19:57
$begingroup$
@peroxisome7, Yes. That is a version of contrapositive statement of the fact that difference of two convergent series is again convergent.
$endgroup$
– Sangchul Lee
Feb 6 at 20:07
1
$begingroup$
@peroxisome Note that $$sum_n=1^N fraccos^2(n)sqrt n =frac12 sum_N=1^N frac1sqrt n+frac12 sum_n=1^N fraccos(2n)sqrt n$$Now, Dirichelt's test shows that the second partial sum on the right-hand side converges. If the partial sum on the left-hand side converged, then the first partial sum on the right-hand side would be the sum (difference) of two convergent partial sums and hence would also converge. Inasmuch as the first partial sum on the right-hand side diverges, what can one conclude?
$endgroup$
– Mark Viola
Feb 6 at 21:20
add a comment |
$begingroup$
Is using this and splitting the series into two helpful? Is the addition of a divergent and a convergent series also divergent?
$endgroup$
– peroxisome7
Feb 6 at 19:57
$begingroup$
@peroxisome7, Yes. That is a version of contrapositive statement of the fact that difference of two convergent series is again convergent.
$endgroup$
– Sangchul Lee
Feb 6 at 20:07
1
$begingroup$
@peroxisome Note that $$sum_n=1^N fraccos^2(n)sqrt n =frac12 sum_N=1^N frac1sqrt n+frac12 sum_n=1^N fraccos(2n)sqrt n$$Now, Dirichelt's test shows that the second partial sum on the right-hand side converges. If the partial sum on the left-hand side converged, then the first partial sum on the right-hand side would be the sum (difference) of two convergent partial sums and hence would also converge. Inasmuch as the first partial sum on the right-hand side diverges, what can one conclude?
$endgroup$
– Mark Viola
Feb 6 at 21:20
$begingroup$
Is using this and splitting the series into two helpful? Is the addition of a divergent and a convergent series also divergent?
$endgroup$
– peroxisome7
Feb 6 at 19:57
$begingroup$
Is using this and splitting the series into two helpful? Is the addition of a divergent and a convergent series also divergent?
$endgroup$
– peroxisome7
Feb 6 at 19:57
$begingroup$
@peroxisome7, Yes. That is a version of contrapositive statement of the fact that difference of two convergent series is again convergent.
$endgroup$
– Sangchul Lee
Feb 6 at 20:07
$begingroup$
@peroxisome7, Yes. That is a version of contrapositive statement of the fact that difference of two convergent series is again convergent.
$endgroup$
– Sangchul Lee
Feb 6 at 20:07
1
1
$begingroup$
@peroxisome Note that $$sum_n=1^N fraccos^2(n)sqrt n =frac12 sum_N=1^N frac1sqrt n+frac12 sum_n=1^N fraccos(2n)sqrt n$$Now, Dirichelt's test shows that the second partial sum on the right-hand side converges. If the partial sum on the left-hand side converged, then the first partial sum on the right-hand side would be the sum (difference) of two convergent partial sums and hence would also converge. Inasmuch as the first partial sum on the right-hand side diverges, what can one conclude?
$endgroup$
– Mark Viola
Feb 6 at 21:20
$begingroup$
@peroxisome Note that $$sum_n=1^N fraccos^2(n)sqrt n =frac12 sum_N=1^N frac1sqrt n+frac12 sum_n=1^N fraccos(2n)sqrt n$$Now, Dirichelt's test shows that the second partial sum on the right-hand side converges. If the partial sum on the left-hand side converged, then the first partial sum on the right-hand side would be the sum (difference) of two convergent partial sums and hence would also converge. Inasmuch as the first partial sum on the right-hand side diverges, what can one conclude?
$endgroup$
– Mark Viola
Feb 6 at 21:20
add a comment |
8
$begingroup$
Another approach is to observe that among $e^in, n=1,2,dots, 7,$ at least one of these points lies in the arc $e^it: tin (-pi/4,pi/4).$ Thus
$$sum_n=1^7 fraccos^2 nsqrt n ge frac1 2frac1sqrt 7.$$
The same thing happens for $n=8,dots,14.$ etc. So the series in question is at least
$$sum_m=1^infty frac1 2frac1sqrt 7m = infty.$$
Note that this idea will work if $cos^2 n$ is replaced by $|cos n|^p$ for any $p>0$.
answered Feb 6 at 19:57
zhw.zhw.
73.6k43175
$endgroup$
$begingroup$
(+1) I remember having used this trick to solve a previous question on the site. Under its apparence of simplicity, it is indeed rather powerful.
$endgroup$
– Did
Feb 6 at 20:05
2
$begingroup$
@zhw. Inasmuch as at least one of the points $1,2,dots, 7$ lies in that arc, $cos^2(n)ge 1/2$ for such a point. Then, $sum_n=1^7 fraccos^2(n)sqrt nge frac12sqrt 7$, would it not?
$endgroup$
– Mark Viola
Feb 6 at 20:42
$begingroup$
@MarkViola Right you are. Now edited.
$endgroup$
– zhw.
Feb 6 at 23:20
add a comment |
8
$begingroup$
Another approach is to observe that among $e^in, n=1,2,dots, 7,$ at least one of these points lies in the arc $e^it: tin (-pi/4,pi/4).$ Thus
$$sum_n=1^7 fraccos^2 nsqrt n ge frac1 2frac1sqrt 7.$$
The same thing happens for $n=8,dots,14.$ etc. So the series in question is at least
$$sum_m=1^infty frac1 2frac1sqrt 7m = infty.$$
Note that this idea will work if $cos^2 n$ is replaced by $|cos n|^p$ for any $p>0$.
answered Feb 6 at 19:57
zhw.zhw.
73.6k43175
$endgroup$
$begingroup$
(+1) I remember having used this trick to solve a previous question on the site. Under its apparence of simplicity, it is indeed rather powerful.
$endgroup$
– Did
Feb 6 at 20:05
2
$begingroup$
@zhw. Inasmuch as at least one of the points $1,2,dots, 7$ lies in that arc, $cos^2(n)ge 1/2$ for such a point. Then, $sum_n=1^7 fraccos^2(n)sqrt nge frac12sqrt 7$, would it not?
$endgroup$
– Mark Viola
Feb 6 at 20:42
$begingroup$
@MarkViola Right you are. Now edited.
$endgroup$
– zhw.
Feb 6 at 23:20
add a comment |
8
8
8
$begingroup$
Another approach is to observe that among $e^in, n=1,2,dots, 7,$ at least one of these points lies in the arc $e^it: tin (-pi/4,pi/4).$ Thus
$$sum_n=1^7 fraccos^2 nsqrt n ge frac1 2frac1sqrt 7.$$
The same thing happens for $n=8,dots,14.$ etc. So the series in question is at least
$$sum_m=1^infty frac1 2frac1sqrt 7m = infty.$$
Note that this idea will work if $cos^2 n$ is replaced by $|cos n|^p$ for any $p>0$.
answered Feb 6 at 19:57
zhw.zhw.
73.6k43175
$endgroup$
Another approach is to observe that among $e^in, n=1,2,dots, 7,$ at least one of these points lies in the arc $e^it: tin (-pi/4,pi/4).$ Thus
$$sum_n=1^7 fraccos^2 nsqrt n ge frac1 2frac1sqrt 7.$$
The same thing happens for $n=8,dots,14.$ etc. So the series in question is at least
$$sum_m=1^infty frac1 2frac1sqrt 7m = infty.$$
Note that this idea will work if $cos^2 n$ is replaced by $|cos n|^p$ for any $p>0$.
answered Feb 6 at 19:57
zhw.zhw.
73.6k43175
edited Feb 6 at 23:20
answered Feb 6 at 19:57
zhw.zhw.
73.6k43175
answered Feb 6 at 19:57
zhw.zhw.
73.6k43175
answered Feb 6 at 19:57
zhw.zhw.
73.6k43175
73.6k43175
$begingroup$
(+1) I remember having used this trick to solve a previous question on the site. Under its apparence of simplicity, it is indeed rather powerful.
$endgroup$
– Did
Feb 6 at 20:05
2
$begingroup$
@zhw. Inasmuch as at least one of the points $1,2,dots, 7$ lies in that arc, $cos^2(n)ge 1/2$ for such a point. Then, $sum_n=1^7 fraccos^2(n)sqrt nge frac12sqrt 7$, would it not?
$endgroup$
– Mark Viola
Feb 6 at 20:42
$begingroup$
@MarkViola Right you are. Now edited.
$endgroup$
– zhw.
Feb 6 at 23:20
add a comment |
$begingroup$
(+1) I remember having used this trick to solve a previous question on the site. Under its apparence of simplicity, it is indeed rather powerful.
$endgroup$
– Did
Feb 6 at 20:05
2
$begingroup$
@zhw. Inasmuch as at least one of the points $1,2,dots, 7$ lies in that arc, $cos^2(n)ge 1/2$ for such a point. Then, $sum_n=1^7 fraccos^2(n)sqrt nge frac12sqrt 7$, would it not?
$endgroup$
– Mark Viola
Feb 6 at 20:42
$begingroup$
@MarkViola Right you are. Now edited.
$endgroup$
– zhw.
Feb 6 at 23:20
$begingroup$
(+1) I remember having used this trick to solve a previous question on the site. Under its apparence of simplicity, it is indeed rather powerful.
$endgroup$
– Did
Feb 6 at 20:05
$begingroup$
(+1) I remember having used this trick to solve a previous question on the site. Under its apparence of simplicity, it is indeed rather powerful.
$endgroup$
– Did
Feb 6 at 20:05
2
2
$begingroup$
@zhw. Inasmuch as at least one of the points $1,2,dots, 7$ lies in that arc, $cos^2(n)ge 1/2$ for such a point. Then, $sum_n=1^7 fraccos^2(n)sqrt nge frac12sqrt 7$, would it not?
$endgroup$
– Mark Viola
Feb 6 at 20:42
$begingroup$
@zhw. Inasmuch as at least one of the points $1,2,dots, 7$ lies in that arc, $cos^2(n)ge 1/2$ for such a point. Then, $sum_n=1^7 fraccos^2(n)sqrt nge frac12sqrt 7$, would it not?
$endgroup$
– Mark Viola
Feb 6 at 20:42
$begingroup$
@MarkViola Right you are. Now edited.
$endgroup$
– zhw.
Feb 6 at 23:20
$begingroup$
@MarkViola Right you are. Now edited.
$endgroup$
– zhw.
Feb 6 at 23:20
add a comment |
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$begingroup$
Hey can you show what you did for the Abel comparasin test?
$endgroup$
– Max0815
Feb 6 at 19:38