determine series convergence.
up vote
1
down vote
favorite
Determine whether this series converges:
$sum_{n=1}^infty cos(n^2pi) (sqrt{n+11} -sqrt{n+2}) $
I know that $lim a_{n} = 0$ and that this series alternates because of $cos(n^2pi)$, but don't know where to go from here.
sequences-and-series convergence conditional-convergence
edited Nov 26 at 20:55
gimusi
88.5k74394
asked Nov 26 at 20:49
J. Lastin
294
add a comment |
up vote
1
down vote
favorite
Determine whether this series converges:
$sum_{n=1}^infty cos(n^2pi) (sqrt{n+11} -sqrt{n+2}) $
I know that $lim a_{n} = 0$ and that this series alternates because of $cos(n^2pi)$, but don't know where to go from here.
sequences-and-series convergence conditional-convergence
edited Nov 26 at 20:55
gimusi
88.5k74394
asked Nov 26 at 20:49
J. Lastin
294
1
Squares of even and odd numbers are even and odd numbers, respectively
– Mark Viola
Nov 26 at 20:57
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Determine whether this series converges:
$sum_{n=1}^infty cos(n^2pi) (sqrt{n+11} -sqrt{n+2}) $
I know that $lim a_{n} = 0$ and that this series alternates because of $cos(n^2pi)$, but don't know where to go from here.
sequences-and-series convergence conditional-convergence
edited Nov 26 at 20:55
gimusi
88.5k74394
asked Nov 26 at 20:49
J. Lastin
294
Determine whether this series converges:
$sum_{n=1}^infty cos(n^2pi) (sqrt{n+11} -sqrt{n+2}) $
I know that $lim a_{n} = 0$ and that this series alternates because of $cos(n^2pi)$, but don't know where to go from here.
sequences-and-series convergence conditional-convergence
sequences-and-series convergence conditional-convergence
edited Nov 26 at 20:55
gimusi
88.5k74394
asked Nov 26 at 20:49
J. Lastin
294
edited Nov 26 at 20:55
gimusi
88.5k74394
asked Nov 26 at 20:49
J. Lastin
294
edited Nov 26 at 20:55
gimusi
88.5k74394
edited Nov 26 at 20:55
gimusi
88.5k74394
edited Nov 26 at 20:55
gimusi
88.5k74394
88.5k74394
asked Nov 26 at 20:49
J. Lastin
294
asked Nov 26 at 20:49
J. Lastin
294
asked Nov 26 at 20:49
J. Lastin
294
294
1
Squares of even and odd numbers are even and odd numbers, respectively
– Mark Viola
Nov 26 at 20:57
add a comment |
1
Squares of even and odd numbers are even and odd numbers, respectively
– Mark Viola
Nov 26 at 20:57
1
1
Squares of even and odd numbers are even and odd numbers, respectively
– Mark Viola
Nov 26 at 20:57
Squares of even and odd numbers are even and odd numbers, respectively
– Mark Viola
Nov 26 at 20:57
add a comment |
3 Answers
3
active
oldest
votes
up vote
4
down vote
accepted
Observe that $$|a_n|=sqrt{n+11}-sqrt{n+2}=frac{n+11-n-2}{sqrt{n+11}+sqrt{n+2}}=frac{9}{sqrt{n+11}+sqrt{n+2}}$$ and that this sequence is decreasing and converges to zero. So, Leibniz criterion for alternate series says that it is convergent.
edited Nov 27 at 1:43
NoseKnowsAll
1,8821020
answered Nov 26 at 20:54
Tito Eliatron
1,077621
I'm also supposed to determine whether the series converges conditionally or absolutely. Am I supposed to use the D'Alembert criterion for this?
– J. Lastin
Nov 26 at 21:39
2
The same calculation about $|a_n|$, says that, if you compare it with $1/sqrt{n}$, your series is NOT absolutely convergent.
– Tito Eliatron
Nov 26 at 21:41
add a comment |
up vote
1
down vote
Try to use the alternating series test. Note that $a_{n} = sqrt{n+11} - sqrt{n+2}$ is a decreasing sequence (why?) and converges to 0. Also, we have $cos(n^{2}pi) = (-1)^{n}$.
answered Nov 26 at 20:52
Seewoo Lee
6,036826
add a comment |
up vote
1
down vote
Since the square of an even number is an even number and the square of an odd number is an odd number, then $cos(n^2pi)=(-1)^n$. Now appeal to Leibniz's Test.
Can you finish?
answered Nov 26 at 21:00
Mark Viola
129k1273170
Good hint, even if I think that the asker was already aware about that! :)
– gimusi
Nov 26 at 21:38
@gimusi Thanks. It seemed that the OP had all of the conditions except monotonicity to apply Leibniz's Test
– Mark Viola
Nov 26 at 21:42
Yes indeed I think that the main doubt was on that! Cheers
– gimusi
Nov 26 at 21:43
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Observe that $$|a_n|=sqrt{n+11}-sqrt{n+2}=frac{n+11-n-2}{sqrt{n+11}+sqrt{n+2}}=frac{9}{sqrt{n+11}+sqrt{n+2}}$$ and that this sequence is decreasing and converges to zero. So, Leibniz criterion for alternate series says that it is convergent.
edited Nov 27 at 1:43
NoseKnowsAll
1,8821020
answered Nov 26 at 20:54
Tito Eliatron
1,077621
I'm also supposed to determine whether the series converges conditionally or absolutely. Am I supposed to use the D'Alembert criterion for this?
– J. Lastin
Nov 26 at 21:39
2
The same calculation about $|a_n|$, says that, if you compare it with $1/sqrt{n}$, your series is NOT absolutely convergent.
– Tito Eliatron
Nov 26 at 21:41
add a comment |
up vote
4
down vote
accepted
Observe that $$|a_n|=sqrt{n+11}-sqrt{n+2}=frac{n+11-n-2}{sqrt{n+11}+sqrt{n+2}}=frac{9}{sqrt{n+11}+sqrt{n+2}}$$ and that this sequence is decreasing and converges to zero. So, Leibniz criterion for alternate series says that it is convergent.
edited Nov 27 at 1:43
NoseKnowsAll
1,8821020
answered Nov 26 at 20:54
Tito Eliatron
1,077621
I'm also supposed to determine whether the series converges conditionally or absolutely. Am I supposed to use the D'Alembert criterion for this?
– J. Lastin
Nov 26 at 21:39
2
The same calculation about $|a_n|$, says that, if you compare it with $1/sqrt{n}$, your series is NOT absolutely convergent.
– Tito Eliatron
Nov 26 at 21:41
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Observe that $$|a_n|=sqrt{n+11}-sqrt{n+2}=frac{n+11-n-2}{sqrt{n+11}+sqrt{n+2}}=frac{9}{sqrt{n+11}+sqrt{n+2}}$$ and that this sequence is decreasing and converges to zero. So, Leibniz criterion for alternate series says that it is convergent.
edited Nov 27 at 1:43
NoseKnowsAll
1,8821020
answered Nov 26 at 20:54
Tito Eliatron
1,077621
Observe that $$|a_n|=sqrt{n+11}-sqrt{n+2}=frac{n+11-n-2}{sqrt{n+11}+sqrt{n+2}}=frac{9}{sqrt{n+11}+sqrt{n+2}}$$ and that this sequence is decreasing and converges to zero. So, Leibniz criterion for alternate series says that it is convergent.
edited Nov 27 at 1:43
NoseKnowsAll
1,8821020
answered Nov 26 at 20:54
Tito Eliatron
1,077621
edited Nov 27 at 1:43
NoseKnowsAll
1,8821020
edited Nov 27 at 1:43
NoseKnowsAll
1,8821020
edited Nov 27 at 1:43
NoseKnowsAll
1,8821020
1,8821020
answered Nov 26 at 20:54
Tito Eliatron
1,077621
answered Nov 26 at 20:54
Tito Eliatron
1,077621
answered Nov 26 at 20:54
Tito Eliatron
1,077621
1,077621
I'm also supposed to determine whether the series converges conditionally or absolutely. Am I supposed to use the D'Alembert criterion for this?
– J. Lastin
Nov 26 at 21:39
2
The same calculation about $|a_n|$, says that, if you compare it with $1/sqrt{n}$, your series is NOT absolutely convergent.
– Tito Eliatron
Nov 26 at 21:41
add a comment |
I'm also supposed to determine whether the series converges conditionally or absolutely. Am I supposed to use the D'Alembert criterion for this?
– J. Lastin
Nov 26 at 21:39
2
The same calculation about $|a_n|$, says that, if you compare it with $1/sqrt{n}$, your series is NOT absolutely convergent.
– Tito Eliatron
Nov 26 at 21:41
I'm also supposed to determine whether the series converges conditionally or absolutely. Am I supposed to use the D'Alembert criterion for this?
– J. Lastin
Nov 26 at 21:39
I'm also supposed to determine whether the series converges conditionally or absolutely. Am I supposed to use the D'Alembert criterion for this?
– J. Lastin
Nov 26 at 21:39
2
2
The same calculation about $|a_n|$, says that, if you compare it with $1/sqrt{n}$, your series is NOT absolutely convergent.
– Tito Eliatron
Nov 26 at 21:41
The same calculation about $|a_n|$, says that, if you compare it with $1/sqrt{n}$, your series is NOT absolutely convergent.
– Tito Eliatron
Nov 26 at 21:41
add a comment |
up vote
1
down vote
Try to use the alternating series test. Note that $a_{n} = sqrt{n+11} - sqrt{n+2}$ is a decreasing sequence (why?) and converges to 0. Also, we have $cos(n^{2}pi) = (-1)^{n}$.
answered Nov 26 at 20:52
Seewoo Lee
6,036826
add a comment |
up vote
1
down vote
Try to use the alternating series test. Note that $a_{n} = sqrt{n+11} - sqrt{n+2}$ is a decreasing sequence (why?) and converges to 0. Also, we have $cos(n^{2}pi) = (-1)^{n}$.
answered Nov 26 at 20:52
Seewoo Lee
6,036826
add a comment |
up vote
1
down vote
up vote
1
down vote
Try to use the alternating series test. Note that $a_{n} = sqrt{n+11} - sqrt{n+2}$ is a decreasing sequence (why?) and converges to 0. Also, we have $cos(n^{2}pi) = (-1)^{n}$.
answered Nov 26 at 20:52
Seewoo Lee
6,036826
Try to use the alternating series test. Note that $a_{n} = sqrt{n+11} - sqrt{n+2}$ is a decreasing sequence (why?) and converges to 0. Also, we have $cos(n^{2}pi) = (-1)^{n}$.
answered Nov 26 at 20:52
Seewoo Lee
6,036826
answered Nov 26 at 20:52
Seewoo Lee
6,036826
answered Nov 26 at 20:52
Seewoo Lee
6,036826
answered Nov 26 at 20:52
Seewoo Lee
6,036826
6,036826
add a comment |
add a comment |
up vote
1
down vote
Since the square of an even number is an even number and the square of an odd number is an odd number, then $cos(n^2pi)=(-1)^n$. Now appeal to Leibniz's Test.
Can you finish?
answered Nov 26 at 21:00
Mark Viola
129k1273170
Good hint, even if I think that the asker was already aware about that! :)
– gimusi
Nov 26 at 21:38
@gimusi Thanks. It seemed that the OP had all of the conditions except monotonicity to apply Leibniz's Test
– Mark Viola
Nov 26 at 21:42
Yes indeed I think that the main doubt was on that! Cheers
– gimusi
Nov 26 at 21:43
add a comment |
up vote
1
down vote
Since the square of an even number is an even number and the square of an odd number is an odd number, then $cos(n^2pi)=(-1)^n$. Now appeal to Leibniz's Test.
Can you finish?
answered Nov 26 at 21:00
Mark Viola
129k1273170
Good hint, even if I think that the asker was already aware about that! :)
– gimusi
Nov 26 at 21:38
@gimusi Thanks. It seemed that the OP had all of the conditions except monotonicity to apply Leibniz's Test
– Mark Viola
Nov 26 at 21:42
Yes indeed I think that the main doubt was on that! Cheers
– gimusi
Nov 26 at 21:43
add a comment |
up vote
1
down vote
up vote
1
down vote
Since the square of an even number is an even number and the square of an odd number is an odd number, then $cos(n^2pi)=(-1)^n$. Now appeal to Leibniz's Test.
Can you finish?
answered Nov 26 at 21:00
Mark Viola
129k1273170
Since the square of an even number is an even number and the square of an odd number is an odd number, then $cos(n^2pi)=(-1)^n$. Now appeal to Leibniz's Test.
Can you finish?
answered Nov 26 at 21:00
Mark Viola
129k1273170
answered Nov 26 at 21:00
Mark Viola
129k1273170
answered Nov 26 at 21:00
Mark Viola
129k1273170
answered Nov 26 at 21:00
Mark Viola
129k1273170
129k1273170
Good hint, even if I think that the asker was already aware about that! :)
– gimusi
Nov 26 at 21:38
@gimusi Thanks. It seemed that the OP had all of the conditions except monotonicity to apply Leibniz's Test
– Mark Viola
Nov 26 at 21:42
Yes indeed I think that the main doubt was on that! Cheers
– gimusi
Nov 26 at 21:43
add a comment |
Good hint, even if I think that the asker was already aware about that! :)
– gimusi
Nov 26 at 21:38
@gimusi Thanks. It seemed that the OP had all of the conditions except monotonicity to apply Leibniz's Test
– Mark Viola
Nov 26 at 21:42
Yes indeed I think that the main doubt was on that! Cheers
– gimusi
Nov 26 at 21:43
Good hint, even if I think that the asker was already aware about that! :)
– gimusi
Nov 26 at 21:38
Good hint, even if I think that the asker was already aware about that! :)
– gimusi
Nov 26 at 21:38
@gimusi Thanks. It seemed that the OP had all of the conditions except monotonicity to apply Leibniz's Test
– Mark Viola
Nov 26 at 21:42
@gimusi Thanks. It seemed that the OP had all of the conditions except monotonicity to apply Leibniz's Test
– Mark Viola
Nov 26 at 21:42
Yes indeed I think that the main doubt was on that! Cheers
– gimusi
Nov 26 at 21:43
Yes indeed I think that the main doubt was on that! Cheers
– gimusi
Nov 26 at 21:43
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3014897%2fdetermine-series-convergence%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
Squares of even and odd numbers are even and odd numbers, respectively
– Mark Viola
Nov 26 at 20:57