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Studying the convergence of the series $sum_n=1^inftysinfrac1nlogleft(1+sinfrac1nright)$
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up vote
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Study the convergence of the series
$$sum_n=1^inftysinfrac1nlogleft(1+sinfrac1nright)$$
This is what I came up with
$$lim_xto inftyfracsinfrac1nlogleft(1+sinfrac1nright)
sin^2frac1n= 1 $$
This implies that
$$sinfrac1nlogleft(1+sinfrac1nright) sim sin^2frac1n$$
using the inequality $sinxlt x$ $left(0le x lt piright)$
$$sin^2frac1n lt frac1n^2$$
Since $sum_n=1^inftyfrac1n^2$ converges so does $sum_n=1^inftysin^2frac1n$ this implies the convergence of $$sum_n=1^inftysinfrac1nlogleft(1+sinfrac1nright)$$
Is this right?
real-analysis sequences-and-series convergence
edited Jul 25 at 17:41
Lorenzo B.
1,5402318
asked Jul 25 at 16:48
J.Dane
159112
add a comment |Â
up vote
7
down vote
favorite
Study the convergence of the series
$$sum_n=1^inftysinfrac1nlogleft(1+sinfrac1nright)$$
This is what I came up with
$$lim_xto inftyfracsinfrac1nlogleft(1+sinfrac1nright)
sin^2frac1n= 1 $$
This implies that
$$sinfrac1nlogleft(1+sinfrac1nright) sim sin^2frac1n$$
using the inequality $sinxlt x$ $left(0le x lt piright)$
$$sin^2frac1n lt frac1n^2$$
Since $sum_n=1^inftyfrac1n^2$ converges so does $sum_n=1^inftysin^2frac1n$ this implies the convergence of $$sum_n=1^inftysinfrac1nlogleft(1+sinfrac1nright)$$
Is this right?
real-analysis sequences-and-series convergence
edited Jul 25 at 17:41
Lorenzo B.
1,5402318
asked Jul 25 at 16:48
J.Dane
159112
1
You are correct!
– Parcly Taxel
Jul 25 at 16:51
Thank you. I was not very sure.
– J.Dane
Jul 25 at 16:55
2
Note that $log(1+x)le x$. Hence, for $nge 1$$$0le sinleft(frac1nright)logleft(1+sinleft(frac1nright)right)le frac1n^2$$
– Mark Viola
Jul 25 at 17:23
@J.Dane Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
2 days ago
add a comment |Â
up vote
7
down vote
favorite
up vote
7
down vote
favorite
Study the convergence of the series
$$sum_n=1^inftysinfrac1nlogleft(1+sinfrac1nright)$$
This is what I came up with
$$lim_xto inftyfracsinfrac1nlogleft(1+sinfrac1nright)
sin^2frac1n= 1 $$
This implies that
$$sinfrac1nlogleft(1+sinfrac1nright) sim sin^2frac1n$$
using the inequality $sinxlt x$ $left(0le x lt piright)$
$$sin^2frac1n lt frac1n^2$$
Since $sum_n=1^inftyfrac1n^2$ converges so does $sum_n=1^inftysin^2frac1n$ this implies the convergence of $$sum_n=1^inftysinfrac1nlogleft(1+sinfrac1nright)$$
Is this right?
real-analysis sequences-and-series convergence
edited Jul 25 at 17:41
Lorenzo B.
1,5402318
asked Jul 25 at 16:48
J.Dane
159112
Study the convergence of the series
$$sum_n=1^inftysinfrac1nlogleft(1+sinfrac1nright)$$
This is what I came up with
$$lim_xto inftyfracsinfrac1nlogleft(1+sinfrac1nright)
sin^2frac1n= 1 $$
This implies that
$$sinfrac1nlogleft(1+sinfrac1nright) sim sin^2frac1n$$
using the inequality $sinxlt x$ $left(0le x lt piright)$
$$sin^2frac1n lt frac1n^2$$
Since $sum_n=1^inftyfrac1n^2$ converges so does $sum_n=1^inftysin^2frac1n$ this implies the convergence of $$sum_n=1^inftysinfrac1nlogleft(1+sinfrac1nright)$$
Is this right?
real-analysis sequences-and-series convergence
edited Jul 25 at 17:41
Lorenzo B.
1,5402318
asked Jul 25 at 16:48
J.Dane
159112
edited Jul 25 at 17:41
Lorenzo B.
1,5402318
edited Jul 25 at 17:41
Lorenzo B.
1,5402318
edited Jul 25 at 17:41
Lorenzo B.
1,5402318
1,5402318
asked Jul 25 at 16:48
J.Dane
159112
asked Jul 25 at 16:48
J.Dane
159112
asked Jul 25 at 16:48
J.Dane
159112
159112
1
You are correct!
– Parcly Taxel
Jul 25 at 16:51
Thank you. I was not very sure.
– J.Dane
Jul 25 at 16:55
2
Note that $log(1+x)le x$. Hence, for $nge 1$$$0le sinleft(frac1nright)logleft(1+sinleft(frac1nright)right)le frac1n^2$$
– Mark Viola
Jul 25 at 17:23
@J.Dane Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
2 days ago
add a comment |Â
1
You are correct!
– Parcly Taxel
Jul 25 at 16:51
Thank you. I was not very sure.
– J.Dane
Jul 25 at 16:55
2
Note that $log(1+x)le x$. Hence, for $nge 1$$$0le sinleft(frac1nright)logleft(1+sinleft(frac1nright)right)le frac1n^2$$
– Mark Viola
Jul 25 at 17:23
@J.Dane Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
2 days ago
1
1
You are correct!
– Parcly Taxel
Jul 25 at 16:51
You are correct!
– Parcly Taxel
Jul 25 at 16:51
Thank you. I was not very sure.
– J.Dane
Jul 25 at 16:55
Thank you. I was not very sure.
– J.Dane
Jul 25 at 16:55
2
2
Note that $log(1+x)le x$. Hence, for $nge 1$$$0le sinleft(frac1nright)logleft(1+sinleft(frac1nright)right)le frac1n^2$$
– Mark Viola
Jul 25 at 17:23
Note that $log(1+x)le x$. Hence, for $nge 1$$$0le sinleft(frac1nright)logleft(1+sinleft(frac1nright)right)le frac1n^2$$
– Mark Viola
Jul 25 at 17:23
@J.Dane Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
2 days ago
@J.Dane Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
2 days ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
0
down vote
Thanks to Mark Viola comment, we know $dfrac1n<1$ is in first quadrant so $sindfrac1n>0$, then
$$sum_n=1^inftyleft(sinfrac1nright)lnleft(1+sinfrac1nright)<sum_n=1^inftyleft(sinfrac1nright)^2<sum_n=1^inftyfrac1n^2=zeta(2)$$
answered Jul 25 at 18:19
Nosrati
19.3k41544
add a comment |Â
up vote
0
down vote
Your conclusion is right but we need to clarify some issue.
You have shown that
$$sinfrac1nlogleft(1+sinfrac1nright) sim sin^2frac1n$$
and then
$$sin^2frac1n lt frac1n^2$$
thus by comparison test we have that since $sum_n=1^inftyfrac1n^2$ also $sum_n=1^inftysin^2frac1n$ converges.
Now to conclude we have two choice:
- Refer to limit comparison test to show the convergence of the given series given the convergence of $sum_n=1^inftysin^2frac1n$
- Refer again to the comparison test showing that (see the comment by Mark Viola)
$$sinfrac1nlogleft(1+sinfrac1nright)le sin frac1n^2le frac1n^2$$
As a simpler alternative, in my opinion, from here
$$sinfrac1nlogleft(1+sinfrac1nright)sim frac1n^2$$
you can directly conclude by limit comparison test with the convergent $sum_n=1^inftyfrac1n^2$.
answered Jul 25 at 17:04
gimusi
65k73583
How you can write $sinfrac1nlogleft(1+sinfrac1nright)le sin frac1n^2$ ?
– Empty
Jul 25 at 17:27
Refer to the hint given here above from Mark Viola
– gimusi
Jul 25 at 17:37
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Thanks to Mark Viola comment, we know $dfrac1n<1$ is in first quadrant so $sindfrac1n>0$, then
$$sum_n=1^inftyleft(sinfrac1nright)lnleft(1+sinfrac1nright)<sum_n=1^inftyleft(sinfrac1nright)^2<sum_n=1^inftyfrac1n^2=zeta(2)$$
answered Jul 25 at 18:19
Nosrati
19.3k41544
add a comment |Â
up vote
0
down vote
Thanks to Mark Viola comment, we know $dfrac1n<1$ is in first quadrant so $sindfrac1n>0$, then
$$sum_n=1^inftyleft(sinfrac1nright)lnleft(1+sinfrac1nright)<sum_n=1^inftyleft(sinfrac1nright)^2<sum_n=1^inftyfrac1n^2=zeta(2)$$
answered Jul 25 at 18:19
Nosrati
19.3k41544
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Thanks to Mark Viola comment, we know $dfrac1n<1$ is in first quadrant so $sindfrac1n>0$, then
$$sum_n=1^inftyleft(sinfrac1nright)lnleft(1+sinfrac1nright)<sum_n=1^inftyleft(sinfrac1nright)^2<sum_n=1^inftyfrac1n^2=zeta(2)$$
answered Jul 25 at 18:19
Nosrati
19.3k41544
Thanks to Mark Viola comment, we know $dfrac1n<1$ is in first quadrant so $sindfrac1n>0$, then
$$sum_n=1^inftyleft(sinfrac1nright)lnleft(1+sinfrac1nright)<sum_n=1^inftyleft(sinfrac1nright)^2<sum_n=1^inftyfrac1n^2=zeta(2)$$
answered Jul 25 at 18:19
Nosrati
19.3k41544
answered Jul 25 at 18:19
Nosrati
19.3k41544
answered Jul 25 at 18:19
Nosrati
19.3k41544
answered Jul 25 at 18:19
Nosrati
19.3k41544
19.3k41544
add a comment |Â
add a comment |Â
up vote
0
down vote
Your conclusion is right but we need to clarify some issue.
You have shown that
$$sinfrac1nlogleft(1+sinfrac1nright) sim sin^2frac1n$$
and then
$$sin^2frac1n lt frac1n^2$$
thus by comparison test we have that since $sum_n=1^inftyfrac1n^2$ also $sum_n=1^inftysin^2frac1n$ converges.
Now to conclude we have two choice:
- Refer to limit comparison test to show the convergence of the given series given the convergence of $sum_n=1^inftysin^2frac1n$
- Refer again to the comparison test showing that (see the comment by Mark Viola)
$$sinfrac1nlogleft(1+sinfrac1nright)le sin frac1n^2le frac1n^2$$
As a simpler alternative, in my opinion, from here
$$sinfrac1nlogleft(1+sinfrac1nright)sim frac1n^2$$
you can directly conclude by limit comparison test with the convergent $sum_n=1^inftyfrac1n^2$.
answered Jul 25 at 17:04
gimusi
65k73583
How you can write $sinfrac1nlogleft(1+sinfrac1nright)le sin frac1n^2$ ?
– Empty
Jul 25 at 17:27
Refer to the hint given here above from Mark Viola
– gimusi
Jul 25 at 17:37
add a comment |Â
up vote
0
down vote
Your conclusion is right but we need to clarify some issue.
You have shown that
$$sinfrac1nlogleft(1+sinfrac1nright) sim sin^2frac1n$$
and then
$$sin^2frac1n lt frac1n^2$$
thus by comparison test we have that since $sum_n=1^inftyfrac1n^2$ also $sum_n=1^inftysin^2frac1n$ converges.
Now to conclude we have two choice:
- Refer to limit comparison test to show the convergence of the given series given the convergence of $sum_n=1^inftysin^2frac1n$
- Refer again to the comparison test showing that (see the comment by Mark Viola)
$$sinfrac1nlogleft(1+sinfrac1nright)le sin frac1n^2le frac1n^2$$
As a simpler alternative, in my opinion, from here
$$sinfrac1nlogleft(1+sinfrac1nright)sim frac1n^2$$
you can directly conclude by limit comparison test with the convergent $sum_n=1^inftyfrac1n^2$.
answered Jul 25 at 17:04
gimusi
65k73583
How you can write $sinfrac1nlogleft(1+sinfrac1nright)le sin frac1n^2$ ?
– Empty
Jul 25 at 17:27
Refer to the hint given here above from Mark Viola
– gimusi
Jul 25 at 17:37
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Your conclusion is right but we need to clarify some issue.
You have shown that
$$sinfrac1nlogleft(1+sinfrac1nright) sim sin^2frac1n$$
and then
$$sin^2frac1n lt frac1n^2$$
thus by comparison test we have that since $sum_n=1^inftyfrac1n^2$ also $sum_n=1^inftysin^2frac1n$ converges.
Now to conclude we have two choice:
- Refer to limit comparison test to show the convergence of the given series given the convergence of $sum_n=1^inftysin^2frac1n$
- Refer again to the comparison test showing that (see the comment by Mark Viola)
$$sinfrac1nlogleft(1+sinfrac1nright)le sin frac1n^2le frac1n^2$$
As a simpler alternative, in my opinion, from here
$$sinfrac1nlogleft(1+sinfrac1nright)sim frac1n^2$$
you can directly conclude by limit comparison test with the convergent $sum_n=1^inftyfrac1n^2$.
answered Jul 25 at 17:04
gimusi
65k73583
Your conclusion is right but we need to clarify some issue.
You have shown that
$$sinfrac1nlogleft(1+sinfrac1nright) sim sin^2frac1n$$
and then
$$sin^2frac1n lt frac1n^2$$
thus by comparison test we have that since $sum_n=1^inftyfrac1n^2$ also $sum_n=1^inftysin^2frac1n$ converges.
Now to conclude we have two choice:
- Refer to limit comparison test to show the convergence of the given series given the convergence of $sum_n=1^inftysin^2frac1n$
- Refer again to the comparison test showing that (see the comment by Mark Viola)
$$sinfrac1nlogleft(1+sinfrac1nright)le sin frac1n^2le frac1n^2$$
As a simpler alternative, in my opinion, from here
$$sinfrac1nlogleft(1+sinfrac1nright)sim frac1n^2$$
you can directly conclude by limit comparison test with the convergent $sum_n=1^inftyfrac1n^2$.
answered Jul 25 at 17:04
gimusi
65k73583
edited Jul 25 at 19:32
answered Jul 25 at 17:04
gimusi
65k73583
answered Jul 25 at 17:04
gimusi
65k73583
answered Jul 25 at 17:04
gimusi
65k73583
65k73583
How you can write $sinfrac1nlogleft(1+sinfrac1nright)le sin frac1n^2$ ?
– Empty
Jul 25 at 17:27
Refer to the hint given here above from Mark Viola
– gimusi
Jul 25 at 17:37
add a comment |Â
How you can write $sinfrac1nlogleft(1+sinfrac1nright)le sin frac1n^2$ ?
– Empty
Jul 25 at 17:27
Refer to the hint given here above from Mark Viola
– gimusi
Jul 25 at 17:37
How you can write $sinfrac1nlogleft(1+sinfrac1nright)le sin frac1n^2$ ?
– Empty
Jul 25 at 17:27
How you can write $sinfrac1nlogleft(1+sinfrac1nright)le sin frac1n^2$ ?
– Empty
Jul 25 at 17:27
Refer to the hint given here above from Mark Viola
– gimusi
Jul 25 at 17:37
Refer to the hint given here above from Mark Viola
– gimusi
Jul 25 at 17:37
add a comment |Â
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1
You are correct!
– Parcly Taxel
Jul 25 at 16:51
Thank you. I was not very sure.
– J.Dane
Jul 25 at 16:55
2
Note that $log(1+x)le x$. Hence, for $nge 1$$$0le sinleft(frac1nright)logleft(1+sinleft(frac1nright)right)le frac1n^2$$
– Mark Viola
Jul 25 at 17:23
@J.Dane Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
2 days ago