find all real values of $x$ for which the series converges absolutely?
The problem:
Find all real values of $x$ for which the series $$sum_{n=2}^{infty} frac{x^n}{n(log n)^2}$$
converges absolutely.
My attempt:
My answer is $|x| <1$ because when $n to infty$, $x^ n$ tends to $0$ if $-1<x<1$
Is it correct?
Any hints/solution will be appreciated. Thank you.
real-analysis
edited Nov 28 at 23:26
Mason
1,8581529
asked Nov 28 at 23:05
jasmine
1,524416
add a comment |
The problem:
Find all real values of $x$ for which the series $$sum_{n=2}^{infty} frac{x^n}{n(log n)^2}$$
converges absolutely.
My attempt:
My answer is $|x| <1$ because when $n to infty$, $x^ n$ tends to $0$ if $-1<x<1$
Is it correct?
Any hints/solution will be appreciated. Thank you.
real-analysis
edited Nov 28 at 23:26
Mason
1,8581529
asked Nov 28 at 23:05
jasmine
1,524416
3
I guess the series is also absolutely convergent when $x=1$ or $x=-1$ because $sum_{n=2}^infty frac{1}{n(log n)^2}$ is convergent.
– John_Wick
Nov 28 at 23:09
add a comment |
The problem:
Find all real values of $x$ for which the series $$sum_{n=2}^{infty} frac{x^n}{n(log n)^2}$$
converges absolutely.
My attempt:
My answer is $|x| <1$ because when $n to infty$, $x^ n$ tends to $0$ if $-1<x<1$
Is it correct?
Any hints/solution will be appreciated. Thank you.
real-analysis
edited Nov 28 at 23:26
Mason
1,8581529
asked Nov 28 at 23:05
jasmine
1,524416
The problem:
Find all real values of $x$ for which the series $$sum_{n=2}^{infty} frac{x^n}{n(log n)^2}$$
converges absolutely.
My attempt:
My answer is $|x| <1$ because when $n to infty$, $x^ n$ tends to $0$ if $-1<x<1$
Is it correct?
Any hints/solution will be appreciated. Thank you.
real-analysis
real-analysis
edited Nov 28 at 23:26
Mason
1,8581529
asked Nov 28 at 23:05
jasmine
1,524416
edited Nov 28 at 23:26
Mason
1,8581529
asked Nov 28 at 23:05
jasmine
1,524416
edited Nov 28 at 23:26
Mason
1,8581529
edited Nov 28 at 23:26
Mason
1,8581529
edited Nov 28 at 23:26
Mason
1,8581529
1,8581529
asked Nov 28 at 23:05
jasmine
1,524416
asked Nov 28 at 23:05
jasmine
1,524416
asked Nov 28 at 23:05
jasmine
1,524416
1,524416
3
I guess the series is also absolutely convergent when $x=1$ or $x=-1$ because $sum_{n=2}^infty frac{1}{n(log n)^2}$ is convergent.
– John_Wick
Nov 28 at 23:09
add a comment |
3
I guess the series is also absolutely convergent when $x=1$ or $x=-1$ because $sum_{n=2}^infty frac{1}{n(log n)^2}$ is convergent.
– John_Wick
Nov 28 at 23:09
3
3
I guess the series is also absolutely convergent when $x=1$ or $x=-1$ because $sum_{n=2}^infty frac{1}{n(log n)^2}$ is convergent.
– John_Wick
Nov 28 at 23:09
I guess the series is also absolutely convergent when $x=1$ or $x=-1$ because $sum_{n=2}^infty frac{1}{n(log n)^2}$ is convergent.
– John_Wick
Nov 28 at 23:09
add a comment |
1 Answer
1
active
oldest
votes
Yes that's correct indeed by root test we obtain
$$|x|left(frac{1}{n(log n)^2}right)^frac1nto |x|<1$$
and it converges also for $x=pm 1$, can you see why?
answered Nov 28 at 23:08
gimusi
1
if $x = -1$, then it will converges by leibnitz test???? Am i right
– jasmine
Nov 28 at 23:12
1
@jasmine Once we know that it converges for $x=1$ it converges also absolutely for $x=-1$ therefore we don't need to invoke Leibnitz.
– gimusi
Nov 28 at 23:13
1
@jasmine Mmmhhhh...that's does not seem a good method.You are claiming that the LHS is greater than something that converges, how can we conclude from here?
– gimusi
Nov 28 at 23:19
1
@jasmine We need integral test to prove that or Cauchy condensation test. Refer to the related OP
– gimusi
Nov 28 at 23:23
1
@jasmine You are welcome! Be
– gimusi
Nov 28 at 23:24
|
show 9 more comments
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1 Answer
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active
oldest
votes
1 Answer
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active
oldest
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oldest
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active
oldest
votes
Yes that's correct indeed by root test we obtain
$$|x|left(frac{1}{n(log n)^2}right)^frac1nto |x|<1$$
and it converges also for $x=pm 1$, can you see why?
answered Nov 28 at 23:08
gimusi
1
if $x = -1$, then it will converges by leibnitz test???? Am i right
– jasmine
Nov 28 at 23:12
1
@jasmine Once we know that it converges for $x=1$ it converges also absolutely for $x=-1$ therefore we don't need to invoke Leibnitz.
– gimusi
Nov 28 at 23:13
1
@jasmine Mmmhhhh...that's does not seem a good method.You are claiming that the LHS is greater than something that converges, how can we conclude from here?
– gimusi
Nov 28 at 23:19
1
@jasmine We need integral test to prove that or Cauchy condensation test. Refer to the related OP
– gimusi
Nov 28 at 23:23
1
@jasmine You are welcome! Be
– gimusi
Nov 28 at 23:24
|
show 9 more comments
Yes that's correct indeed by root test we obtain
$$|x|left(frac{1}{n(log n)^2}right)^frac1nto |x|<1$$
and it converges also for $x=pm 1$, can you see why?
answered Nov 28 at 23:08
gimusi
1
if $x = -1$, then it will converges by leibnitz test???? Am i right
– jasmine
Nov 28 at 23:12
1
@jasmine Once we know that it converges for $x=1$ it converges also absolutely for $x=-1$ therefore we don't need to invoke Leibnitz.
– gimusi
Nov 28 at 23:13
1
@jasmine Mmmhhhh...that's does not seem a good method.You are claiming that the LHS is greater than something that converges, how can we conclude from here?
– gimusi
Nov 28 at 23:19
1
@jasmine We need integral test to prove that or Cauchy condensation test. Refer to the related OP
– gimusi
Nov 28 at 23:23
1
@jasmine You are welcome! Be
– gimusi
Nov 28 at 23:24
|
show 9 more comments
Yes that's correct indeed by root test we obtain
$$|x|left(frac{1}{n(log n)^2}right)^frac1nto |x|<1$$
and it converges also for $x=pm 1$, can you see why?
answered Nov 28 at 23:08
gimusi
1
Yes that's correct indeed by root test we obtain
$$|x|left(frac{1}{n(log n)^2}right)^frac1nto |x|<1$$
and it converges also for $x=pm 1$, can you see why?
answered Nov 28 at 23:08
gimusi
1
answered Nov 28 at 23:08
gimusi
1
answered Nov 28 at 23:08
gimusi
1
answered Nov 28 at 23:08
gimusi
1
1
if $x = -1$, then it will converges by leibnitz test???? Am i right
– jasmine
Nov 28 at 23:12
1
@jasmine Once we know that it converges for $x=1$ it converges also absolutely for $x=-1$ therefore we don't need to invoke Leibnitz.
– gimusi
Nov 28 at 23:13
1
@jasmine Mmmhhhh...that's does not seem a good method.You are claiming that the LHS is greater than something that converges, how can we conclude from here?
– gimusi
Nov 28 at 23:19
1
@jasmine We need integral test to prove that or Cauchy condensation test. Refer to the related OP
– gimusi
Nov 28 at 23:23
1
@jasmine You are welcome! Be
– gimusi
Nov 28 at 23:24
|
show 9 more comments
if $x = -1$, then it will converges by leibnitz test???? Am i right
– jasmine
Nov 28 at 23:12
1
@jasmine Once we know that it converges for $x=1$ it converges also absolutely for $x=-1$ therefore we don't need to invoke Leibnitz.
– gimusi
Nov 28 at 23:13
1
@jasmine Mmmhhhh...that's does not seem a good method.You are claiming that the LHS is greater than something that converges, how can we conclude from here?
– gimusi
Nov 28 at 23:19
1
@jasmine We need integral test to prove that or Cauchy condensation test. Refer to the related OP
– gimusi
Nov 28 at 23:23
1
@jasmine You are welcome! Be
– gimusi
Nov 28 at 23:24
if $x = -1$, then it will converges by leibnitz test???? Am i right
– jasmine
Nov 28 at 23:12
if $x = -1$, then it will converges by leibnitz test???? Am i right
– jasmine
Nov 28 at 23:12
1
1
@jasmine Once we know that it converges for $x=1$ it converges also absolutely for $x=-1$ therefore we don't need to invoke Leibnitz.
– gimusi
Nov 28 at 23:13
@jasmine Once we know that it converges for $x=1$ it converges also absolutely for $x=-1$ therefore we don't need to invoke Leibnitz.
– gimusi
Nov 28 at 23:13
1
1
@jasmine Mmmhhhh...that's does not seem a good method.You are claiming that the LHS is greater than something that converges, how can we conclude from here?
– gimusi
Nov 28 at 23:19
@jasmine Mmmhhhh...that's does not seem a good method.You are claiming that the LHS is greater than something that converges, how can we conclude from here?
– gimusi
Nov 28 at 23:19
1
1
@jasmine We need integral test to prove that or Cauchy condensation test. Refer to the related OP
– gimusi
Nov 28 at 23:23
@jasmine We need integral test to prove that or Cauchy condensation test. Refer to the related OP
– gimusi
Nov 28 at 23:23
1
1
@jasmine You are welcome! Be
– gimusi
Nov 28 at 23:24
@jasmine You are welcome! Be
– gimusi
Nov 28 at 23:24
|
show 9 more comments
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3
I guess the series is also absolutely convergent when $x=1$ or $x=-1$ because $sum_{n=2}^infty frac{1}{n(log n)^2}$ is convergent.
– John_Wick
Nov 28 at 23:09