Does an infinite series from n=$-infty$ to $infty$ converge or diverge?
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
I came across the following series in my math homework (Fourier Series):
Does the following series converge or diverge? If converges, does it converge absolutely?
$sum_n=-infty^inftyfrac(-1)^nn^2+3$
Typically, I would be well equipped to answer the question, however the "n=$-infty$" is giving me trouble. Normally, if "n=$0$", the alternating series test could show convergence, and a direct comparison test with a p-series could show absolute convergence. How does the "$-infty$" change the problem, if at all?
sequences-and-series
edited Sep 4 at 12:10
Ahmad Bazzi
5,8991623
asked Sep 4 at 12:00
Jake McGrath
11
add a comment |Â
up vote
0
down vote
favorite
I came across the following series in my math homework (Fourier Series):
Does the following series converge or diverge? If converges, does it converge absolutely?
$sum_n=-infty^inftyfrac(-1)^nn^2+3$
Typically, I would be well equipped to answer the question, however the "n=$-infty$" is giving me trouble. Normally, if "n=$0$", the alternating series test could show convergence, and a direct comparison test with a p-series could show absolute convergence. How does the "$-infty$" change the problem, if at all?
sequences-and-series
edited Sep 4 at 12:10
Ahmad Bazzi
5,8991623
asked Sep 4 at 12:00
Jake McGrath
11
1
Break the sum into two pieces, $sum_n=-infty^-1 frac(-1)^nn^2+3$ and $sum_n=0^infty frac(-1)^nn^2+3$. If both pieces are absolutely convergent, then so is the entire sum.
– bangs
Sep 4 at 12:05
Here's a note on what you have with you : ncatlab.org/nlab/show/doubly+infinite+series
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 4 at 12:08
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I came across the following series in my math homework (Fourier Series):
Does the following series converge or diverge? If converges, does it converge absolutely?
$sum_n=-infty^inftyfrac(-1)^nn^2+3$
Typically, I would be well equipped to answer the question, however the "n=$-infty$" is giving me trouble. Normally, if "n=$0$", the alternating series test could show convergence, and a direct comparison test with a p-series could show absolute convergence. How does the "$-infty$" change the problem, if at all?
sequences-and-series
edited Sep 4 at 12:10
Ahmad Bazzi
5,8991623
asked Sep 4 at 12:00
Jake McGrath
11
I came across the following series in my math homework (Fourier Series):
Does the following series converge or diverge? If converges, does it converge absolutely?
$sum_n=-infty^inftyfrac(-1)^nn^2+3$
Typically, I would be well equipped to answer the question, however the "n=$-infty$" is giving me trouble. Normally, if "n=$0$", the alternating series test could show convergence, and a direct comparison test with a p-series could show absolute convergence. How does the "$-infty$" change the problem, if at all?
sequences-and-series
sequences-and-series
edited Sep 4 at 12:10
Ahmad Bazzi
5,8991623
asked Sep 4 at 12:00
Jake McGrath
11
edited Sep 4 at 12:10
Ahmad Bazzi
5,8991623
asked Sep 4 at 12:00
Jake McGrath
11
edited Sep 4 at 12:10
Ahmad Bazzi
5,8991623
edited Sep 4 at 12:10
Ahmad Bazzi
5,8991623
edited Sep 4 at 12:10
Ahmad Bazzi
5,8991623
5,8991623
asked Sep 4 at 12:00
Jake McGrath
11
asked Sep 4 at 12:00
Jake McGrath
11
asked Sep 4 at 12:00
Jake McGrath
11
11
1
Break the sum into two pieces, $sum_n=-infty^-1 frac(-1)^nn^2+3$ and $sum_n=0^infty frac(-1)^nn^2+3$. If both pieces are absolutely convergent, then so is the entire sum.
– bangs
Sep 4 at 12:05
Here's a note on what you have with you : ncatlab.org/nlab/show/doubly+infinite+series
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 4 at 12:08
add a comment |Â
1
Break the sum into two pieces, $sum_n=-infty^-1 frac(-1)^nn^2+3$ and $sum_n=0^infty frac(-1)^nn^2+3$. If both pieces are absolutely convergent, then so is the entire sum.
– bangs
Sep 4 at 12:05
Here's a note on what you have with you : ncatlab.org/nlab/show/doubly+infinite+series
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 4 at 12:08
1
1
Break the sum into two pieces, $sum_n=-infty^-1 frac(-1)^nn^2+3$ and $sum_n=0^infty frac(-1)^nn^2+3$. If both pieces are absolutely convergent, then so is the entire sum.
– bangs
Sep 4 at 12:05
Break the sum into two pieces, $sum_n=-infty^-1 frac(-1)^nn^2+3$ and $sum_n=0^infty frac(-1)^nn^2+3$. If both pieces are absolutely convergent, then so is the entire sum.
– bangs
Sep 4 at 12:05
Here's a note on what you have with you : ncatlab.org/nlab/show/doubly+infinite+series
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 4 at 12:08
Here's a note on what you have with you : ncatlab.org/nlab/show/doubly+infinite+series
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 4 at 12:08
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
Since the summand $f(n)$ satisfies
$$
f(-n)=-f(n)
$$ one may just study the convergence over $n in [0,infty)$.
Then the series is absolutely convergent by the $p$-test and the given series is convergent.
answered Sep 4 at 12:06
Olivier Oloa
106k17174293
add a comment |Â
up vote
1
down vote
beginequation
sumlimits_n= -infty^0
frac(-1)^nn^2 + 3
=
sumlimits_n= 0^+infty
frac(-1)^nn^2 + 3
endequation
So
beginequation
sumlimits_n= -infty^+infty
frac(-1)^nn^2 + 3
=
underbrace
sumlimits_n= -infty^0
frac(-1)^nn^2 + 3
_n = -infty ldots 0
+
underbrace
sumlimits_n= 0^+infty
frac(-1)^nn^2 + 3
_n = 0 ldots +infty
-
underbrace
frac(-1)^00^2 + 3
_n=0
=
2sumlimits_n= 0^+infty
frac(-1)^nn^2 + 3
-
frac13
endequation
The series
beginequation
sumlimits_n= 0^+infty
frac(-1)^nn^2 + 3
endequation
is absolutely convergent due to the $p-$test. So your series converges.
answered Sep 4 at 12:06
Ahmad Bazzi
5,8991623
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Since the summand $f(n)$ satisfies
$$
f(-n)=-f(n)
$$ one may just study the convergence over $n in [0,infty)$.
Then the series is absolutely convergent by the $p$-test and the given series is convergent.
answered Sep 4 at 12:06
Olivier Oloa
106k17174293
add a comment |Â
up vote
2
down vote
Since the summand $f(n)$ satisfies
$$
f(-n)=-f(n)
$$ one may just study the convergence over $n in [0,infty)$.
Then the series is absolutely convergent by the $p$-test and the given series is convergent.
answered Sep 4 at 12:06
Olivier Oloa
106k17174293
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Since the summand $f(n)$ satisfies
$$
f(-n)=-f(n)
$$ one may just study the convergence over $n in [0,infty)$.
Then the series is absolutely convergent by the $p$-test and the given series is convergent.
answered Sep 4 at 12:06
Olivier Oloa
106k17174293
Since the summand $f(n)$ satisfies
$$
f(-n)=-f(n)
$$ one may just study the convergence over $n in [0,infty)$.
Then the series is absolutely convergent by the $p$-test and the given series is convergent.
answered Sep 4 at 12:06
Olivier Oloa
106k17174293
answered Sep 4 at 12:06
Olivier Oloa
106k17174293
answered Sep 4 at 12:06
Olivier Oloa
106k17174293
answered Sep 4 at 12:06
Olivier Oloa
106k17174293
106k17174293
add a comment |Â
add a comment |Â
up vote
1
down vote
beginequation
sumlimits_n= -infty^0
frac(-1)^nn^2 + 3
=
sumlimits_n= 0^+infty
frac(-1)^nn^2 + 3
endequation
So
beginequation
sumlimits_n= -infty^+infty
frac(-1)^nn^2 + 3
=
underbrace
sumlimits_n= -infty^0
frac(-1)^nn^2 + 3
_n = -infty ldots 0
+
underbrace
sumlimits_n= 0^+infty
frac(-1)^nn^2 + 3
_n = 0 ldots +infty
-
underbrace
frac(-1)^00^2 + 3
_n=0
=
2sumlimits_n= 0^+infty
frac(-1)^nn^2 + 3
-
frac13
endequation
The series
beginequation
sumlimits_n= 0^+infty
frac(-1)^nn^2 + 3
endequation
is absolutely convergent due to the $p-$test. So your series converges.
answered Sep 4 at 12:06
Ahmad Bazzi
5,8991623
add a comment |Â
up vote
1
down vote
beginequation
sumlimits_n= -infty^0
frac(-1)^nn^2 + 3
=
sumlimits_n= 0^+infty
frac(-1)^nn^2 + 3
endequation
So
beginequation
sumlimits_n= -infty^+infty
frac(-1)^nn^2 + 3
=
underbrace
sumlimits_n= -infty^0
frac(-1)^nn^2 + 3
_n = -infty ldots 0
+
underbrace
sumlimits_n= 0^+infty
frac(-1)^nn^2 + 3
_n = 0 ldots +infty
-
underbrace
frac(-1)^00^2 + 3
_n=0
=
2sumlimits_n= 0^+infty
frac(-1)^nn^2 + 3
-
frac13
endequation
The series
beginequation
sumlimits_n= 0^+infty
frac(-1)^nn^2 + 3
endequation
is absolutely convergent due to the $p-$test. So your series converges.
answered Sep 4 at 12:06
Ahmad Bazzi
5,8991623
add a comment |Â
up vote
1
down vote
up vote
1
down vote
beginequation
sumlimits_n= -infty^0
frac(-1)^nn^2 + 3
=
sumlimits_n= 0^+infty
frac(-1)^nn^2 + 3
endequation
So
beginequation
sumlimits_n= -infty^+infty
frac(-1)^nn^2 + 3
=
underbrace
sumlimits_n= -infty^0
frac(-1)^nn^2 + 3
_n = -infty ldots 0
+
underbrace
sumlimits_n= 0^+infty
frac(-1)^nn^2 + 3
_n = 0 ldots +infty
-
underbrace
frac(-1)^00^2 + 3
_n=0
=
2sumlimits_n= 0^+infty
frac(-1)^nn^2 + 3
-
frac13
endequation
The series
beginequation
sumlimits_n= 0^+infty
frac(-1)^nn^2 + 3
endequation
is absolutely convergent due to the $p-$test. So your series converges.
answered Sep 4 at 12:06
Ahmad Bazzi
5,8991623
beginequation
sumlimits_n= -infty^0
frac(-1)^nn^2 + 3
=
sumlimits_n= 0^+infty
frac(-1)^nn^2 + 3
endequation
So
beginequation
sumlimits_n= -infty^+infty
frac(-1)^nn^2 + 3
=
underbrace
sumlimits_n= -infty^0
frac(-1)^nn^2 + 3
_n = -infty ldots 0
+
underbrace
sumlimits_n= 0^+infty
frac(-1)^nn^2 + 3
_n = 0 ldots +infty
-
underbrace
frac(-1)^00^2 + 3
_n=0
=
2sumlimits_n= 0^+infty
frac(-1)^nn^2 + 3
-
frac13
endequation
The series
beginequation
sumlimits_n= 0^+infty
frac(-1)^nn^2 + 3
endequation
is absolutely convergent due to the $p-$test. So your series converges.
answered Sep 4 at 12:06
Ahmad Bazzi
5,8991623
answered Sep 4 at 12:06
Ahmad Bazzi
5,8991623
answered Sep 4 at 12:06
Ahmad Bazzi
5,8991623
answered Sep 4 at 12:06
Ahmad Bazzi
5,8991623
5,8991623
add a comment |Â
add a comment |Â
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2904963%2fdoes-an-infinite-series-from-n-infty-to-infty-converge-or-diverge%23new-answer', 'question_page');
);
Post as a guest
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
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
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
1
Break the sum into two pieces, $sum_n=-infty^-1 frac(-1)^nn^2+3$ and $sum_n=0^infty frac(-1)^nn^2+3$. If both pieces are absolutely convergent, then so is the entire sum.
– bangs
Sep 4 at 12:05
Here's a note on what you have with you : ncatlab.org/nlab/show/doubly+infinite+series
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 4 at 12:08