a********e
发帖数: 508 | 1 20. Is there a sequence a_n such that, a_n is convergent but not absolutely
convergent and (a_n)^3 is not convergent?
这个是不存在吗？各位高手有何高见？ | s******s
发帖数: 63 | 2 a_n^3...
真不好证 感觉像是不存在
要是是平方就容易了。。。 | M****i
发帖数: 58 | 3 If your post is correct, the answer is "NO" because every convergent sequence is also absolutely convergent.
If the "sequence" in your question is changed into "series", then it will be
more interesting and a partial answer is "YES" provided that the complex
numbers are allowed.
Example: a_n=exp(i2n\pi/3)/n^(1/3). Then the convergence of \Sigma_n a_n is
ensured by Dirichlet's criterion. Moreover, it is easily seen that |a_n|=n^(-1/3) and (a_
n)^3=1/n, hence the series \Sigma_n |a_n| and \Sigma_n (a_n)^3 are divergent.
Can anyone give some ideas for the real number case? | w*********i
发帖数: 77 | 4 "every convergent sequence is also absolutely convergent" wrong.
1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6.... converges but
1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6.... diverges.
sequence is also absolutely convergent.
will be
complex
a_n is
|a_n|=n^(-1/3) and (a_
divergent.
【在 M****i 的大作中提到】
: If your post is correct, the answer is "NO" because every convergent sequence is also absolutely convergent.
: If the "sequence" in your question is changed into "series", then it will be
: more interesting and a partial answer is "YES" provided that the complex
: numbers are allowed.
: Example: a_n=exp(i2n\pi/3)/n^(1/3). Then the convergence of \Sigma_n a_n is
: ensured by Dirichlet's criterion. Moreover, it is easily seen that |a_n|=n^(-1/3) and (a_
: n)^3=1/n, hence the series \Sigma_n |a_n| and \Sigma_n (a_n)^3 are divergent.
: Can anyone give some ideas for the real number case?
| M****i
发帖数: 58 | 5 Hello woshipangzi, it seems that you confuse the definition of a "sequence"
and that of a "series": http://en.wikipedia.org/wiki/Sequence
【在 w*********i 的大作中提到】
: "every convergent sequence is also absolutely convergent" wrong.
: 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6.... converges but
: 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6.... diverges.
:
: sequence is also absolutely convergent.
: will be
: complex
: a_n is
: |a_n|=n^(-1/3) and (a_
: divergent.
| a********e
发帖数: 508 | 6 nice insight! I am not sure what the original interviewer means, but
I agree that the case for real series is quite difficult
sequence is also absolutely convergent.
be
is
^(-1/3) and (a_
divergent.
【在 M****i 的大作中提到】
: If your post is correct, the answer is "NO" because every convergent sequence is also absolutely convergent.
: If the "sequence" in your question is changed into "series", then it will be
: more interesting and a partial answer is "YES" provided that the complex
: numbers are allowed.
: Example: a_n=exp(i2n\pi/3)/n^(1/3). Then the convergence of \Sigma_n a_n is
: ensured by Dirichlet's criterion. Moreover, it is easily seen that |a_n|=n^(-1/3) and (a_
: n)^3=1/n, hence the series \Sigma_n |a_n| and \Sigma_n (a_n)^3 are divergent.
: Can anyone give some ideas for the real number case?
| l******i
发帖数: 104 | 7 Yes:
1, -1/2, 1, -3/4, 1, -7/8 ...
a_n ~ (1/2)^n converge
abs(a_n) ~ 1 diverge
(a_n)^3 ~ 3/4n diverge | M****i
发帖数: 58 | 8 Recall that a necessary condition for the series \Sigma_n a_n being
convergent is that a_n converges to 0.
【在 l******i 的大作中提到】
: Yes:
: 1, -1/2, 1, -3/4, 1, -7/8 ...
: a_n ~ (1/2)^n converge
: abs(a_n) ~ 1 diverge
: (a_n)^3 ~ 3/4n diverge
| a********e
发帖数: 508 | 9 thank you. It's a smart way to think about the example,
but it might need some improvement
note that such sequence must converge to 0 for the sum series to converge
【在 l******i 的大作中提到】
: Yes:
: 1, -1/2, 1, -3/4, 1, -7/8 ...
: a_n ~ (1/2)^n converge
: abs(a_n) ~ 1 diverge
: (a_n)^3 ~ 3/4n diverge
| y****e
发帖数: 28 | 10 Here is an counter example.
Define a_{3k-2}=a_{3k-1}=a_{3k}=(3k)^{-1/3} for all k=1,2,...
Define b={-2,1,1,-2,1,1,-2,1,1,...}
let c_n = a_n * b_n
By using Dirichlet's test http://en.wikipedia.org/wiki/Dirichlet_test,
we have c_n converges, since the series \sum_{n=1}^N b_n is bounded.
It's easy to see that |c_n| and c_n^3 do not converge | | | a********e
发帖数: 508 | 11 That works! brilliant!
【在 y****e 的大作中提到】
: Here is an counter example.
: Define a_{3k-2}=a_{3k-1}=a_{3k}=(3k)^{-1/3} for all k=1,2,...
: Define b={-2,1,1,-2,1,1,-2,1,1,...}
: let c_n = a_n * b_n
: By using Dirichlet's test http://en.wikipedia.org/wiki/Dirichlet_test,
: we have c_n converges, since the series \sum_{n=1}^N b_n is bounded.
: It's easy to see that |c_n| and c_n^3 do not converge
| M****i
发帖数: 58 | 12 It can be checked that the real part of my previous example also answers
your question in the field of real numbers: \Sigma_n cos(2n\pi/3)/n^{1/3}.
【在 a********e 的大作中提到】
: nice insight! I am not sure what the original interviewer means, but
: I agree that the case for real series is quite difficult
:
: sequence is also absolutely convergent.
: be
: is
: ^(-1/3) and (a_
: divergent.
| a********e
发帖数: 508 | 13 Right!!! It is indeed an intereating approach of using complex number
to derive the convergence property of real number series!
【在 M****i 的大作中提到】
: It can be checked that the real part of my previous example also answers
: your question in the field of real numbers: \Sigma_n cos(2n\pi/3)/n^{1/3}.
| y****e
发帖数: 28 | 14 In fact, MathFi's real number series and my proposed solution are very
similar to each other except that his real number series do not get
immediate attention due to a little bit more computation. After seeing this
, I probably got some some inspirations from his solution for complex series
without realizing it.
In the spirit of giving a simple and clean answer, I rewrite my solution as
the following,
Define a={1,1,1, 1/2,1/2,1/2, 1/3,1/3,1/3, ...}
b={-2,1,1, -2,1,1, -2,1,1, ...}
Let c_n=a_n*b_n. It is very trivial to verity that series c_n converges and
that series |c_n| and series c_n^3 do not converge.
For the convergence of the c_n series, we do no even need the Dirichlet's
test, since every three terms have been canceled out.
【在 a********e 的大作中提到】
: Right!!! It is indeed an intereating approach of using complex number
: to derive the convergence property of real number series!
| y****e
发帖数: 28 | 15 A typo in define the series a
It should be
a^3={1,1,1, 1/2,1/2,1/2, 1/3,1/3,1/3, ...}
or the uglier form
a={1,1,1, 1/2^{1/3}, 1/2^{1/3}, 1/2^{1/3}, 1/3^{1/3},1/3^{1/3},1/3^{1/3}, ..
.}
this
series
as
and
【在 y****e 的大作中提到】
: In fact, MathFi's real number series and my proposed solution are very
: similar to each other except that his real number series do not get
: immediate attention due to a little bit more computation. After seeing this
: , I probably got some some inspirations from his solution for complex series
: without realizing it.
: In the spirit of giving a simple and clean answer, I rewrite my solution as
: the following,
: Define a={1,1,1, 1/2,1/2,1/2, 1/3,1/3,1/3, ...}
: b={-2,1,1, -2,1,1, -2,1,1, ...}
: Let c_n=a_n*b_n. It is very trivial to verity that series c_n converges and
| a********e
发帖数: 508 | 16 nice disscussion. I think the key of this prob is to constrcut the sequence
using Dirichlet's criterion instead of construct it directly
this
series
as
and
【在 y****e 的大作中提到】
: In fact, MathFi's real number series and my proposed solution are very
: similar to each other except that his real number series do not get
: immediate attention due to a little bit more computation. After seeing this
: , I probably got some some inspirations from his solution for complex series
: without realizing it.
: In the spirit of giving a simple and clean answer, I rewrite my solution as
: the following,
: Define a={1,1,1, 1/2,1/2,1/2, 1/3,1/3,1/3, ...}
: b={-2,1,1, -2,1,1, -2,1,1, ...}
: Let c_n=a_n*b_n. It is very trivial to verity that series c_n converges and
| D**********d
发帖数: 849 | 17 \sum a_n is convergent ==> |a_n| --> 0 ==> |a_n|^2 --> 0
a_n^3 = |a_n|^2 * a_n and \sum a_n is convergent
\sum a_n^3 must be convergent.
absolutely
【在 a********e 的大作中提到】
: 20. Is there a sequence a_n such that, a_n is convergent but not absolutely
: convergent and (a_n)^3 is not convergent?
: 这个是不存在吗？各位高手有何高见？
|
|
