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
90.6k74495
asked Nov 26 at 20:49
J. Lastin
534
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
90.6k74495
asked Nov 26 at 20:49
J. Lastin
534
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
90.6k74495
asked Nov 26 at 20:49
J. Lastin
534
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
90.6k74495
asked Nov 26 at 20:49
J. Lastin
534
edited Nov 26 at 20:55
gimusi
90.6k74495
asked Nov 26 at 20:49
J. Lastin
534
edited Nov 26 at 20:55
gimusi
90.6k74495
edited Nov 26 at 20:55
gimusi
90.6k74495
edited Nov 26 at 20:55
gimusi
90.6k74495
90.6k74495
asked Nov 26 at 20:49
J. Lastin
534
asked Nov 26 at 20:49
J. Lastin
534
asked Nov 26 at 20:49
J. Lastin
534
534
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
1,317622
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
6,066826
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
1,317622
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
1,317622
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
1,317622
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
1,317622
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
1,317622
answered Nov 26 at 20:54
Tito Eliatron
1,317622
answered Nov 26 at 20:54
Tito Eliatron
1,317622
1,317622
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
6,066826
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
6,066826
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
6,066826
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
6,066826
answered Nov 26 at 20:52
Seewoo Lee
6,066826
answered Nov 26 at 20:52
Seewoo Lee
6,066826
answered Nov 26 at 20:52
Seewoo Lee
6,066826
6,066826
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 |
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1
Squares of even and odd numbers are even and odd numbers, respectively
– Mark Viola
Nov 26 at 20:57