Infinite sum of harmonic number
$begingroup$
I learned that I can find the value of some infinite sum.
Then what is the value of this sum?
$$frac12 + left(1+frac12right)frac1{2^2}+left(1+frac12 +frac13right)frac1{2^3}+left(1+frac12 +frac13 +frac14right)frac1{2^4} + cdots $$
And I want to know How to find the value of the infinite sum of this-like form.
sequences-and-series
asked Apr 2 at 14:47
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?
$$frac12 + left(1+frac12right)frac1{2^2}+left(1+frac12 +frac13right)frac1{2^3}+left(1+frac12 +frac13 +frac14right)frac1{2^4} + cdots $$
And I want to know How to find the value of the infinite sum of this-like form.
sequences-and-series
asked Apr 2 at 14:47
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?
$$frac12 + left(1+frac12right)frac1{2^2}+left(1+frac12 +frac13right)frac1{2^3}+left(1+frac12 +frac13 +frac14right)frac1{2^4} + cdots $$
And I want to know How to find the value of the infinite sum of this-like form.
sequences-and-series
asked Apr 2 at 14:47
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?
$$frac12 + left(1+frac12right)frac1{2^2}+left(1+frac12 +frac13right)frac1{2^3}+left(1+frac12 +frac13 +frac14right)frac1{2^4} + cdots $$
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 Apr 2 at 14:47
S. YooS. Yoo
424
asked Apr 2 at 14:47
S. YooS. Yoo
424
asked Apr 2 at 14:47
S. YooS. Yoo
424
asked Apr 2 at 14:47
S. YooS. Yoo
424
asked Apr 2 at 14:47
S. YooS. Yoo
424
424
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
One can split this summation into
$$left(frac12+frac1{2^2}+frac1{2^3}+cdotsright)+frac12left(frac1{2^2}+frac1{2^3}+cdotsright)+frac13left(frac1{2^3}+frac1{2^4}+cdotsright)+cdots$$
$$begin{align}
&=sum_{k=1}^infty frac1k sum_{j=k}^infty frac1{2^j}\
&=sum_{k=1}^infty frac1k left(frac{2^{-k}}{1-2^{-1}}right)\
&=sum_{k=1}^infty frac{2^{1-k}}k\
&=2sum_{k=1}^infty frac{left(frac12right)^k}k\
&=2left(-ln{left(1-frac12right)}right)\
&boxed{=2ln{(2)}}
end{align}$$
By using the fact that
$$ln{(1-x)}=-sum_{k=1}^infty frac{x^k}k$$
for all $|x|lt 1$.
In fact one can use a similar method to prove that
$$sum_{k=1}^infty x^kH_k=frac1{1-x}ln{left(frac1{1-x}right)}$$
for $|x|lt1$. Where $H_k$ is the $k$th harmonic number given by
$$H_k=sum_{n=1}^kfrac1n$$
edited Apr 2 at 18:22
Théophile
20.4k13047
answered Apr 2 at 15:08
Peter ForemanPeter Foreman
5,8141317
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add a comment |
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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
$$left(frac12+frac1{2^2}+frac1{2^3}+cdotsright)+frac12left(frac1{2^2}+frac1{2^3}+cdotsright)+frac13left(frac1{2^3}+frac1{2^4}+cdotsright)+cdots$$
$$begin{align}
&=sum_{k=1}^infty frac1k sum_{j=k}^infty frac1{2^j}\
&=sum_{k=1}^infty frac1k left(frac{2^{-k}}{1-2^{-1}}right)\
&=sum_{k=1}^infty frac{2^{1-k}}k\
&=2sum_{k=1}^infty frac{left(frac12right)^k}k\
&=2left(-ln{left(1-frac12right)}right)\
&boxed{=2ln{(2)}}
end{align}$$
By using the fact that
$$ln{(1-x)}=-sum_{k=1}^infty frac{x^k}k$$
for all $|x|lt 1$.
In fact one can use a similar method to prove that
$$sum_{k=1}^infty x^kH_k=frac1{1-x}ln{left(frac1{1-x}right)}$$
for $|x|lt1$. Where $H_k$ is the $k$th harmonic number given by
$$H_k=sum_{n=1}^kfrac1n$$
edited Apr 2 at 18:22
Théophile
20.4k13047
answered Apr 2 at 15:08
Peter ForemanPeter Foreman
5,8141317
$endgroup$
add a comment |
$begingroup$
One can split this summation into
$$left(frac12+frac1{2^2}+frac1{2^3}+cdotsright)+frac12left(frac1{2^2}+frac1{2^3}+cdotsright)+frac13left(frac1{2^3}+frac1{2^4}+cdotsright)+cdots$$
$$begin{align}
&=sum_{k=1}^infty frac1k sum_{j=k}^infty frac1{2^j}\
&=sum_{k=1}^infty frac1k left(frac{2^{-k}}{1-2^{-1}}right)\
&=sum_{k=1}^infty frac{2^{1-k}}k\
&=2sum_{k=1}^infty frac{left(frac12right)^k}k\
&=2left(-ln{left(1-frac12right)}right)\
&boxed{=2ln{(2)}}
end{align}$$
By using the fact that
$$ln{(1-x)}=-sum_{k=1}^infty frac{x^k}k$$
for all $|x|lt 1$.
In fact one can use a similar method to prove that
$$sum_{k=1}^infty x^kH_k=frac1{1-x}ln{left(frac1{1-x}right)}$$
for $|x|lt1$. Where $H_k$ is the $k$th harmonic number given by
$$H_k=sum_{n=1}^kfrac1n$$
edited Apr 2 at 18:22
Théophile
20.4k13047
answered Apr 2 at 15:08
Peter ForemanPeter Foreman
5,8141317
$endgroup$
add a comment |
$begingroup$
One can split this summation into
$$left(frac12+frac1{2^2}+frac1{2^3}+cdotsright)+frac12left(frac1{2^2}+frac1{2^3}+cdotsright)+frac13left(frac1{2^3}+frac1{2^4}+cdotsright)+cdots$$
$$begin{align}
&=sum_{k=1}^infty frac1k sum_{j=k}^infty frac1{2^j}\
&=sum_{k=1}^infty frac1k left(frac{2^{-k}}{1-2^{-1}}right)\
&=sum_{k=1}^infty frac{2^{1-k}}k\
&=2sum_{k=1}^infty frac{left(frac12right)^k}k\
&=2left(-ln{left(1-frac12right)}right)\
&boxed{=2ln{(2)}}
end{align}$$
By using the fact that
$$ln{(1-x)}=-sum_{k=1}^infty frac{x^k}k$$
for all $|x|lt 1$.
In fact one can use a similar method to prove that
$$sum_{k=1}^infty x^kH_k=frac1{1-x}ln{left(frac1{1-x}right)}$$
for $|x|lt1$. Where $H_k$ is the $k$th harmonic number given by
$$H_k=sum_{n=1}^kfrac1n$$
edited Apr 2 at 18:22
Théophile
20.4k13047
answered Apr 2 at 15:08
Peter ForemanPeter Foreman
5,8141317
$endgroup$
One can split this summation into
$$left(frac12+frac1{2^2}+frac1{2^3}+cdotsright)+frac12left(frac1{2^2}+frac1{2^3}+cdotsright)+frac13left(frac1{2^3}+frac1{2^4}+cdotsright)+cdots$$
$$begin{align}
&=sum_{k=1}^infty frac1k sum_{j=k}^infty frac1{2^j}\
&=sum_{k=1}^infty frac1k left(frac{2^{-k}}{1-2^{-1}}right)\
&=sum_{k=1}^infty frac{2^{1-k}}k\
&=2sum_{k=1}^infty frac{left(frac12right)^k}k\
&=2left(-ln{left(1-frac12right)}right)\
&boxed{=2ln{(2)}}
end{align}$$
By using the fact that
$$ln{(1-x)}=-sum_{k=1}^infty frac{x^k}k$$
for all $|x|lt 1$.
In fact one can use a similar method to prove that
$$sum_{k=1}^infty x^kH_k=frac1{1-x}ln{left(frac1{1-x}right)}$$
for $|x|lt1$. Where $H_k$ is the $k$th harmonic number given by
$$H_k=sum_{n=1}^kfrac1n$$
edited Apr 2 at 18:22
Théophile
20.4k13047
answered Apr 2 at 15:08
Peter ForemanPeter Foreman
5,8141317
edited Apr 2 at 18:22
Théophile
20.4k13047
edited Apr 2 at 18:22
Théophile
20.4k13047
edited Apr 2 at 18:22
Théophile
20.4k13047
20.4k13047
answered Apr 2 at 15:08
Peter ForemanPeter Foreman
5,8141317
answered Apr 2 at 15:08
Peter ForemanPeter Foreman
5,8141317
answered Apr 2 at 15:08
Peter ForemanPeter Foreman
5,8141317
5,8141317
add a comment |
add a comment |
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