Converging series and converging alternative series implies absolute convergence?
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It is known that that harmonic series diverges, but the alternating form of the harmonic series converges. However, I am not sure if there are examples of series $a_n$ that
$$ sum_{n=1}^{infty} a_n space text{converges} space text{and} space sum_{n=1}^{infty} (-1)^{n+1}a_n space text{also converges}$$
but the series does not converge absolutely. A candidate $a_n$ I can think of is
$$ sum_{n=1}^{infty} frac{(-1)^n}{n}$$, but this may be problematic because the signs do not agree. Is there more inspiring examples?
real-analysis sequences-and-series divergent-series
edited Dec 6 '18 at 15:46
José Carlos Santos
158k22126228
asked Dec 6 '18 at 14:16
hephaeshephaes
1709
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$begingroup$
It is known that that harmonic series diverges, but the alternating form of the harmonic series converges. However, I am not sure if there are examples of series $a_n$ that
$$ sum_{n=1}^{infty} a_n space text{converges} space text{and} space sum_{n=1}^{infty} (-1)^{n+1}a_n space text{also converges}$$
but the series does not converge absolutely. A candidate $a_n$ I can think of is
$$ sum_{n=1}^{infty} frac{(-1)^n}{n}$$, but this may be problematic because the signs do not agree. Is there more inspiring examples?
real-analysis sequences-and-series divergent-series
edited Dec 6 '18 at 15:46
José Carlos Santos
158k22126228
asked Dec 6 '18 at 14:16
hephaeshephaes
1709
$endgroup$
add a comment |
$begingroup$
It is known that that harmonic series diverges, but the alternating form of the harmonic series converges. However, I am not sure if there are examples of series $a_n$ that
$$ sum_{n=1}^{infty} a_n space text{converges} space text{and} space sum_{n=1}^{infty} (-1)^{n+1}a_n space text{also converges}$$
but the series does not converge absolutely. A candidate $a_n$ I can think of is
$$ sum_{n=1}^{infty} frac{(-1)^n}{n}$$, but this may be problematic because the signs do not agree. Is there more inspiring examples?
real-analysis sequences-and-series divergent-series
edited Dec 6 '18 at 15:46
José Carlos Santos
158k22126228
asked Dec 6 '18 at 14:16
hephaeshephaes
1709
$endgroup$
It is known that that harmonic series diverges, but the alternating form of the harmonic series converges. However, I am not sure if there are examples of series $a_n$ that
$$ sum_{n=1}^{infty} a_n space text{converges} space text{and} space sum_{n=1}^{infty} (-1)^{n+1}a_n space text{also converges}$$
but the series does not converge absolutely. A candidate $a_n$ I can think of is
$$ sum_{n=1}^{infty} frac{(-1)^n}{n}$$, but this may be problematic because the signs do not agree. Is there more inspiring examples?
real-analysis sequences-and-series divergent-series
real-analysis sequences-and-series divergent-series
edited Dec 6 '18 at 15:46
José Carlos Santos
158k22126228
asked Dec 6 '18 at 14:16
hephaeshephaes
1709
edited Dec 6 '18 at 15:46
José Carlos Santos
158k22126228
asked Dec 6 '18 at 14:16
hephaeshephaes
1709
edited Dec 6 '18 at 15:46
José Carlos Santos
158k22126228
edited Dec 6 '18 at 15:46
José Carlos Santos
158k22126228
edited Dec 6 '18 at 15:46
José Carlos Santos
158k22126228
158k22126228
asked Dec 6 '18 at 14:16
hephaeshephaes
1709
asked Dec 6 '18 at 14:16
hephaeshephaes
1709
asked Dec 6 '18 at 14:16
hephaeshephaes
1709
1709
add a comment |
add a comment |
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That candidate doesn't work, because$$sum_{n=1}^infty(-1)^{n+1}frac{(-1)^n}n=sum_{n-1}^infty-frac1n,$$which diverges.
However, that property holds for the seriesbegin{multline}1+1+(-1)+(-1)+frac12+frac12+left(-frac12right)+left(-frac12right)+\+frac13+frac13+left(-frac13right)+left(-frac13right)+cdotsend{multline}
answered Dec 6 '18 at 14:28
José Carlos SantosJosé Carlos Santos
158k22126228
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You can modify your example by taking complex values: Let $x= exp(2 pi it)$ for some $t in (0,pi)$ and consider the series representation of the main branch of the logarithm, which by Abel-summation converges also on the boundary of the unit ball - except in zero. We have
$$-log(1-x) = sum_{k=1}^infty frac{x^k}{k}$$
and
$$log(1+x) = sum_{k=1}^infty frac{(-1)^{k+1} x^k}{k}.$$
Both series are convergent, but we don't have absolutely convergence.
answered Dec 6 '18 at 14:35
p4schp4sch
5,210217
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2 Answers
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$begingroup$
That candidate doesn't work, because$$sum_{n=1}^infty(-1)^{n+1}frac{(-1)^n}n=sum_{n-1}^infty-frac1n,$$which diverges.
However, that property holds for the seriesbegin{multline}1+1+(-1)+(-1)+frac12+frac12+left(-frac12right)+left(-frac12right)+\+frac13+frac13+left(-frac13right)+left(-frac13right)+cdotsend{multline}
answered Dec 6 '18 at 14:28
José Carlos SantosJosé Carlos Santos
158k22126228
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add a comment |
$begingroup$
That candidate doesn't work, because$$sum_{n=1}^infty(-1)^{n+1}frac{(-1)^n}n=sum_{n-1}^infty-frac1n,$$which diverges.
However, that property holds for the seriesbegin{multline}1+1+(-1)+(-1)+frac12+frac12+left(-frac12right)+left(-frac12right)+\+frac13+frac13+left(-frac13right)+left(-frac13right)+cdotsend{multline}
answered Dec 6 '18 at 14:28
José Carlos SantosJosé Carlos Santos
158k22126228
$endgroup$
add a comment |
$begingroup$
That candidate doesn't work, because$$sum_{n=1}^infty(-1)^{n+1}frac{(-1)^n}n=sum_{n-1}^infty-frac1n,$$which diverges.
However, that property holds for the seriesbegin{multline}1+1+(-1)+(-1)+frac12+frac12+left(-frac12right)+left(-frac12right)+\+frac13+frac13+left(-frac13right)+left(-frac13right)+cdotsend{multline}
answered Dec 6 '18 at 14:28
José Carlos SantosJosé Carlos Santos
158k22126228
$endgroup$
That candidate doesn't work, because$$sum_{n=1}^infty(-1)^{n+1}frac{(-1)^n}n=sum_{n-1}^infty-frac1n,$$which diverges.
However, that property holds for the seriesbegin{multline}1+1+(-1)+(-1)+frac12+frac12+left(-frac12right)+left(-frac12right)+\+frac13+frac13+left(-frac13right)+left(-frac13right)+cdotsend{multline}
answered Dec 6 '18 at 14:28
José Carlos SantosJosé Carlos Santos
158k22126228
answered Dec 6 '18 at 14:28
José Carlos SantosJosé Carlos Santos
158k22126228
answered Dec 6 '18 at 14:28
José Carlos SantosJosé Carlos Santos
158k22126228
answered Dec 6 '18 at 14:28
José Carlos SantosJosé Carlos Santos
158k22126228
158k22126228
add a comment |
add a comment |
$begingroup$
You can modify your example by taking complex values: Let $x= exp(2 pi it)$ for some $t in (0,pi)$ and consider the series representation of the main branch of the logarithm, which by Abel-summation converges also on the boundary of the unit ball - except in zero. We have
$$-log(1-x) = sum_{k=1}^infty frac{x^k}{k}$$
and
$$log(1+x) = sum_{k=1}^infty frac{(-1)^{k+1} x^k}{k}.$$
Both series are convergent, but we don't have absolutely convergence.
answered Dec 6 '18 at 14:35
p4schp4sch
5,210217
$endgroup$
add a comment |
$begingroup$
You can modify your example by taking complex values: Let $x= exp(2 pi it)$ for some $t in (0,pi)$ and consider the series representation of the main branch of the logarithm, which by Abel-summation converges also on the boundary of the unit ball - except in zero. We have
$$-log(1-x) = sum_{k=1}^infty frac{x^k}{k}$$
and
$$log(1+x) = sum_{k=1}^infty frac{(-1)^{k+1} x^k}{k}.$$
Both series are convergent, but we don't have absolutely convergence.
answered Dec 6 '18 at 14:35
p4schp4sch
5,210217
$endgroup$
add a comment |
$begingroup$
You can modify your example by taking complex values: Let $x= exp(2 pi it)$ for some $t in (0,pi)$ and consider the series representation of the main branch of the logarithm, which by Abel-summation converges also on the boundary of the unit ball - except in zero. We have
$$-log(1-x) = sum_{k=1}^infty frac{x^k}{k}$$
and
$$log(1+x) = sum_{k=1}^infty frac{(-1)^{k+1} x^k}{k}.$$
Both series are convergent, but we don't have absolutely convergence.
answered Dec 6 '18 at 14:35
p4schp4sch
5,210217
$endgroup$
You can modify your example by taking complex values: Let $x= exp(2 pi it)$ for some $t in (0,pi)$ and consider the series representation of the main branch of the logarithm, which by Abel-summation converges also on the boundary of the unit ball - except in zero. We have
$$-log(1-x) = sum_{k=1}^infty frac{x^k}{k}$$
and
$$log(1+x) = sum_{k=1}^infty frac{(-1)^{k+1} x^k}{k}.$$
Both series are convergent, but we don't have absolutely convergence.
answered Dec 6 '18 at 14:35
p4schp4sch
5,210217
edited Dec 6 '18 at 16:11
answered Dec 6 '18 at 14:35
p4schp4sch
5,210217
answered Dec 6 '18 at 14:35
p4schp4sch
5,210217
answered Dec 6 '18 at 14:35
p4schp4sch
5,210217
5,210217
add a comment |
add a comment |
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