sum of $sum_{n=1}^{infty}(-1)^{n+1}(2^{1/n}-1)$
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Is there any way to analytically find the sum of an infinite series if the infinite series is not geometric, nor can it be set up as residues for a convenient meromorphic function?
For instance, if I have the function
$f(x)=1+sum_{n=1}^{infty}(-1)^{n+1}(x^{1/n}-1)$, I have a function of x that will converge conditionally for all $xgeq1$, because it is an alternating series where the limit of terms converges to zero. However, I don't see any way of solving the sum using residues.
Are there any other methods to attempt to find the value of the series (and therefore of $f(x)$)?
sequences-and-series
edited Nov 21 at 11:22
Tianlalu
2,674632
asked Nov 21 at 11:14
MoKo19
813
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MoKo19 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
up vote
-1
down vote
favorite
Is there any way to analytically find the sum of an infinite series if the infinite series is not geometric, nor can it be set up as residues for a convenient meromorphic function?
For instance, if I have the function
$f(x)=1+sum_{n=1}^{infty}(-1)^{n+1}(x^{1/n}-1)$, I have a function of x that will converge conditionally for all $xgeq1$, because it is an alternating series where the limit of terms converges to zero. However, I don't see any way of solving the sum using residues.
Are there any other methods to attempt to find the value of the series (and therefore of $f(x)$)?
sequences-and-series
edited Nov 21 at 11:22
Tianlalu
2,674632
asked Nov 21 at 11:14
MoKo19
813
New contributor
MoKo19 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
You can definitely approximate it using the fact that $$lim_{x to 0} left( frac{a^x - 1}{x} right) = ln{a}, qquad forall~a > 0$$
– rtybase
Nov 21 at 13:36
@rtybase: How does that help?
– MoKo19
Nov 21 at 13:46
E.g. $$sumlimits_{n=1}^{infty}(-1)^{n+1}(2^{1/n}-1) approx sum_{n=1}^{N(varepsilon)}(-1)^{n+1}(2^{1/n}-1)+ln{2}left(sum_{n=N(varepsilon)+1}^{infty}(-1)^{n+1}frac{1}{n}right)$$
– rtybase
Nov 21 at 14:02
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Is there any way to analytically find the sum of an infinite series if the infinite series is not geometric, nor can it be set up as residues for a convenient meromorphic function?
For instance, if I have the function
$f(x)=1+sum_{n=1}^{infty}(-1)^{n+1}(x^{1/n}-1)$, I have a function of x that will converge conditionally for all $xgeq1$, because it is an alternating series where the limit of terms converges to zero. However, I don't see any way of solving the sum using residues.
Are there any other methods to attempt to find the value of the series (and therefore of $f(x)$)?
sequences-and-series
edited Nov 21 at 11:22
Tianlalu
2,674632
asked Nov 21 at 11:14
MoKo19
813
New contributor
MoKo19 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Is there any way to analytically find the sum of an infinite series if the infinite series is not geometric, nor can it be set up as residues for a convenient meromorphic function?
For instance, if I have the function
$f(x)=1+sum_{n=1}^{infty}(-1)^{n+1}(x^{1/n}-1)$, I have a function of x that will converge conditionally for all $xgeq1$, because it is an alternating series where the limit of terms converges to zero. However, I don't see any way of solving the sum using residues.
Are there any other methods to attempt to find the value of the series (and therefore of $f(x)$)?
sequences-and-series
sequences-and-series
edited Nov 21 at 11:22
Tianlalu
2,674632
asked Nov 21 at 11:14
MoKo19
813
New contributor
MoKo19 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 21 at 11:22
Tianlalu
2,674632
asked Nov 21 at 11:14
MoKo19
813
New contributor
MoKo19 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 21 at 11:22
Tianlalu
2,674632
edited Nov 21 at 11:22
Tianlalu
2,674632
edited Nov 21 at 11:22
Tianlalu
2,674632
2,674632
asked Nov 21 at 11:14
MoKo19
813
New contributor
MoKo19 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 21 at 11:14
MoKo19
813
asked Nov 21 at 11:14
MoKo19
813
813
New contributor
MoKo19 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
MoKo19 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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You can definitely approximate it using the fact that $$lim_{x to 0} left( frac{a^x - 1}{x} right) = ln{a}, qquad forall~a > 0$$
– rtybase
Nov 21 at 13:36
@rtybase: How does that help?
– MoKo19
Nov 21 at 13:46
E.g. $$sumlimits_{n=1}^{infty}(-1)^{n+1}(2^{1/n}-1) approx sum_{n=1}^{N(varepsilon)}(-1)^{n+1}(2^{1/n}-1)+ln{2}left(sum_{n=N(varepsilon)+1}^{infty}(-1)^{n+1}frac{1}{n}right)$$
– rtybase
Nov 21 at 14:02
add a comment |
You can definitely approximate it using the fact that $$lim_{x to 0} left( frac{a^x - 1}{x} right) = ln{a}, qquad forall~a > 0$$
– rtybase
Nov 21 at 13:36
@rtybase: How does that help?
– MoKo19
Nov 21 at 13:46
E.g. $$sumlimits_{n=1}^{infty}(-1)^{n+1}(2^{1/n}-1) approx sum_{n=1}^{N(varepsilon)}(-1)^{n+1}(2^{1/n}-1)+ln{2}left(sum_{n=N(varepsilon)+1}^{infty}(-1)^{n+1}frac{1}{n}right)$$
– rtybase
Nov 21 at 14:02
You can definitely approximate it using the fact that $$lim_{x to 0} left( frac{a^x - 1}{x} right) = ln{a}, qquad forall~a > 0$$
– rtybase
Nov 21 at 13:36
You can definitely approximate it using the fact that $$lim_{x to 0} left( frac{a^x - 1}{x} right) = ln{a}, qquad forall~a > 0$$
– rtybase
Nov 21 at 13:36
@rtybase: How does that help?
– MoKo19
Nov 21 at 13:46
@rtybase: How does that help?
– MoKo19
Nov 21 at 13:46
E.g. $$sumlimits_{n=1}^{infty}(-1)^{n+1}(2^{1/n}-1) approx sum_{n=1}^{N(varepsilon)}(-1)^{n+1}(2^{1/n}-1)+ln{2}left(sum_{n=N(varepsilon)+1}^{infty}(-1)^{n+1}frac{1}{n}right)$$
– rtybase
Nov 21 at 14:02
E.g. $$sumlimits_{n=1}^{infty}(-1)^{n+1}(2^{1/n}-1) approx sum_{n=1}^{N(varepsilon)}(-1)^{n+1}(2^{1/n}-1)+ln{2}left(sum_{n=N(varepsilon)+1}^{infty}(-1)^{n+1}frac{1}{n}right)$$
– rtybase
Nov 21 at 14:02
add a comment |
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MoKo19 is a new contributor. Be nice, and check out our Code of Conduct.
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You can definitely approximate it using the fact that $$lim_{x to 0} left( frac{a^x - 1}{x} right) = ln{a}, qquad forall~a > 0$$
– rtybase
Nov 21 at 13:36
@rtybase: How does that help?
– MoKo19
Nov 21 at 13:46
E.g. $$sumlimits_{n=1}^{infty}(-1)^{n+1}(2^{1/n}-1) approx sum_{n=1}^{N(varepsilon)}(-1)^{n+1}(2^{1/n}-1)+ln{2}left(sum_{n=N(varepsilon)+1}^{infty}(-1)^{n+1}frac{1}{n}right)$$
– rtybase
Nov 21 at 14:02