Infinite sum of harmonic numberSum of reciprocals of numbers with certain terms omittedWhat is the closed form sum of this series?Is there a name for infinite series of this type?Infinite amount of additions, finite sum?Why can infinite series be summed different ways to get different results?Why does the order of summation of the terms of an infinite series influence its value?Infinite Series $left(frac12+frac14-frac23right)+left(frac15+frac17-frac26right)+left(frac18+frac110-frac29right)+cdots$Sum of the recripocals of the Harmonic NumbersFind the sum of the infinite seriesAlternating harmonic series convergence
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Infinite sum of harmonic number
Sum of reciprocals of numbers with certain terms omittedWhat is the closed form sum of this series?Is there a name for infinite series of this type?Infinite amount of additions, finite sum?Why can infinite series be summed different ways to get different results?Why does the order of summation of the terms of an infinite series influence its value?Infinite Series $left(frac12+frac14-frac23right)+left(frac15+frac17-frac26right)+left(frac18+frac110-frac29right)+cdots$Sum of the recripocals of the Harmonic NumbersFind the sum of the infinite seriesAlternating harmonic series convergence
$begingroup$
I learned that I can find the value of some infinite sum.
Then what is the value of this sum?
$$frac12 + left(1+frac12right)frac12^2+left(1+frac12 +frac13right)frac12^3+left(1+frac12 +frac13 +frac14right)frac12^4 + cdots $$
And I want to know How to find the value of the infinite sum of this-like form.
sequences-and-series
asked 7 hours ago
S. YooS. Yoo
424
$endgroup$
add a comment |
$begingroup$
I learned that I can find the value of some infinite sum.
Then what is the value of this sum?
$$frac12 + left(1+frac12right)frac12^2+left(1+frac12 +frac13right)frac12^3+left(1+frac12 +frac13 +frac14right)frac12^4 + cdots $$
And I want to know How to find the value of the infinite sum of this-like form.
sequences-and-series
asked 7 hours ago
S. YooS. Yoo
424
$endgroup$
add a comment |
$begingroup$
I learned that I can find the value of some infinite sum.
Then what is the value of this sum?
$$frac12 + left(1+frac12right)frac12^2+left(1+frac12 +frac13right)frac12^3+left(1+frac12 +frac13 +frac14right)frac12^4 + cdots $$
And I want to know How to find the value of the infinite sum of this-like form.
sequences-and-series
asked 7 hours ago
S. YooS. Yoo
424
$endgroup$
I learned that I can find the value of some infinite sum.
Then what is the value of this sum?
$$frac12 + left(1+frac12right)frac12^2+left(1+frac12 +frac13right)frac12^3+left(1+frac12 +frac13 +frac14right)frac12^4 + cdots $$
And I want to know How to find the value of the infinite sum of this-like form.
sequences-and-series
sequences-and-series
asked 7 hours ago
S. YooS. Yoo
424
asked 7 hours ago
S. YooS. Yoo
424
asked 7 hours ago
S. YooS. Yoo
424
asked 7 hours ago
S. YooS. Yoo
424
asked 7 hours ago
S. YooS. Yoo
424
424
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
One can split this summation into
$$left(frac12+frac12^2+frac12^3+cdotsright)+frac12left(frac12^2+frac12^3+cdotsright)+frac13left(frac12^3+frac12^4+cdotsright)+cdots$$
$$beginalign
&=sum_k=1^infty frac1k sum_j=k^infty frac12^j\
&=sum_k=1^infty frac1k left(frac2^-k1-2^-1right)\
&=sum_k=1^infty frac2^1-kk\
&=2sum_k=1^infty fracleft(frac12right)^kk\
&=2left(-lnleft(1-frac12right)right)\
&boxed=2ln(2)
endalign$$
By using the fact that
$$ln(1-x)=-sum_k=1^infty fracx^kk$$
for all $|x|lt 1$.
In fact one can use a similar method to prove that
$$sum_k=1^infty x^kH_k=frac11-xlnleft(frac11-xright)$$
for $|x|lt1$. Where $H_k$ is the $k$th harmonic number given by
$$H_k=sum_n=1^kfrac1n$$
edited 3 hours ago
Théophile
20.3k13047
answered 7 hours ago
Peter ForemanPeter Foreman
5,5291216
$endgroup$
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
One can split this summation into
$$left(frac12+frac12^2+frac12^3+cdotsright)+frac12left(frac12^2+frac12^3+cdotsright)+frac13left(frac12^3+frac12^4+cdotsright)+cdots$$
$$beginalign
&=sum_k=1^infty frac1k sum_j=k^infty frac12^j\
&=sum_k=1^infty frac1k left(frac2^-k1-2^-1right)\
&=sum_k=1^infty frac2^1-kk\
&=2sum_k=1^infty fracleft(frac12right)^kk\
&=2left(-lnleft(1-frac12right)right)\
&boxed=2ln(2)
endalign$$
By using the fact that
$$ln(1-x)=-sum_k=1^infty fracx^kk$$
for all $|x|lt 1$.
In fact one can use a similar method to prove that
$$sum_k=1^infty x^kH_k=frac11-xlnleft(frac11-xright)$$
for $|x|lt1$. Where $H_k$ is the $k$th harmonic number given by
$$H_k=sum_n=1^kfrac1n$$
edited 3 hours ago
Théophile
20.3k13047
answered 7 hours ago
Peter ForemanPeter Foreman
5,5291216
$endgroup$
add a comment |
$begingroup$
One can split this summation into
$$left(frac12+frac12^2+frac12^3+cdotsright)+frac12left(frac12^2+frac12^3+cdotsright)+frac13left(frac12^3+frac12^4+cdotsright)+cdots$$
$$beginalign
&=sum_k=1^infty frac1k sum_j=k^infty frac12^j\
&=sum_k=1^infty frac1k left(frac2^-k1-2^-1right)\
&=sum_k=1^infty frac2^1-kk\
&=2sum_k=1^infty fracleft(frac12right)^kk\
&=2left(-lnleft(1-frac12right)right)\
&boxed=2ln(2)
endalign$$
By using the fact that
$$ln(1-x)=-sum_k=1^infty fracx^kk$$
for all $|x|lt 1$.
In fact one can use a similar method to prove that
$$sum_k=1^infty x^kH_k=frac11-xlnleft(frac11-xright)$$
for $|x|lt1$. Where $H_k$ is the $k$th harmonic number given by
$$H_k=sum_n=1^kfrac1n$$
edited 3 hours ago
Théophile
20.3k13047
answered 7 hours ago
Peter ForemanPeter Foreman
5,5291216
$endgroup$
add a comment |
$begingroup$
One can split this summation into
$$left(frac12+frac12^2+frac12^3+cdotsright)+frac12left(frac12^2+frac12^3+cdotsright)+frac13left(frac12^3+frac12^4+cdotsright)+cdots$$
$$beginalign
&=sum_k=1^infty frac1k sum_j=k^infty frac12^j\
&=sum_k=1^infty frac1k left(frac2^-k1-2^-1right)\
&=sum_k=1^infty frac2^1-kk\
&=2sum_k=1^infty fracleft(frac12right)^kk\
&=2left(-lnleft(1-frac12right)right)\
&boxed=2ln(2)
endalign$$
By using the fact that
$$ln(1-x)=-sum_k=1^infty fracx^kk$$
for all $|x|lt 1$.
In fact one can use a similar method to prove that
$$sum_k=1^infty x^kH_k=frac11-xlnleft(frac11-xright)$$
for $|x|lt1$. Where $H_k$ is the $k$th harmonic number given by
$$H_k=sum_n=1^kfrac1n$$
edited 3 hours ago
Théophile
20.3k13047
answered 7 hours ago
Peter ForemanPeter Foreman
5,5291216
$endgroup$
One can split this summation into
$$left(frac12+frac12^2+frac12^3+cdotsright)+frac12left(frac12^2+frac12^3+cdotsright)+frac13left(frac12^3+frac12^4+cdotsright)+cdots$$
$$beginalign
&=sum_k=1^infty frac1k sum_j=k^infty frac12^j\
&=sum_k=1^infty frac1k left(frac2^-k1-2^-1right)\
&=sum_k=1^infty frac2^1-kk\
&=2sum_k=1^infty fracleft(frac12right)^kk\
&=2left(-lnleft(1-frac12right)right)\
&boxed=2ln(2)
endalign$$
By using the fact that
$$ln(1-x)=-sum_k=1^infty fracx^kk$$
for all $|x|lt 1$.
In fact one can use a similar method to prove that
$$sum_k=1^infty x^kH_k=frac11-xlnleft(frac11-xright)$$
for $|x|lt1$. Where $H_k$ is the $k$th harmonic number given by
$$H_k=sum_n=1^kfrac1n$$
edited 3 hours ago
Théophile
20.3k13047
answered 7 hours ago
Peter ForemanPeter Foreman
5,5291216
edited 3 hours ago
Théophile
20.3k13047
edited 3 hours ago
Théophile
20.3k13047
edited 3 hours ago
Théophile
20.3k13047
20.3k13047
answered 7 hours ago
Peter ForemanPeter Foreman
5,5291216
answered 7 hours ago
Peter ForemanPeter Foreman
5,5291216
answered 7 hours ago
Peter ForemanPeter Foreman
5,5291216
5,5291216
add a comment |
add a comment |
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