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Infinite sum of harmonic number

4

3

$begingroup$

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

share|cite|improve this question

asked Apr 2 at 14:47

S. YooS. Yoo

675

$endgroup$

add a comment | 

4

3

$begingroup$

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

share|cite|improve this question

asked Apr 2 at 14:47

S. YooS. Yoo

675

$endgroup$

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

share|cite|improve this question

asked Apr 2 at 14:47

S. YooS. Yoo

675

$endgroup$

share|cite|improve this question

asked Apr 2 at 14:47

S. YooS. Yoo

675

share|cite|improve this question

asked Apr 2 at 14:47

S. YooS. Yoo

675

asked Apr 2 at 14:47

S. YooS. Yoo

675

asked Apr 2 at 14:47

S. YooS. Yoo

675

1 Answer
1

active

oldest

votes

8

$begingroup$

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
$$H_k=sum_n=1^kfrac1n$$

share|cite|improve this answer

edited Apr 2 at 18:22

Théophile

20.5k13047

answered Apr 2 at 15:08

Peter ForemanPeter Foreman

8,5061321

$endgroup$

add a comment | 

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8

$begingroup$

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
$$H_k=sum_n=1^kfrac1n$$

share|cite|improve this answer

edited Apr 2 at 18:22

Théophile

20.5k13047

answered Apr 2 at 15:08

Peter ForemanPeter Foreman

8,5061321

$endgroup$

add a comment | 

8

$begingroup$

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
$$H_k=sum_n=1^kfrac1n$$

share|cite|improve this answer

edited Apr 2 at 18:22

Théophile

20.5k13047

answered Apr 2 at 15:08

Peter ForemanPeter Foreman

8,5061321

$endgroup$

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
$$H_k=sum_n=1^kfrac1n$$

share|cite|improve this answer

edited Apr 2 at 18:22

Théophile

20.5k13047

answered Apr 2 at 15:08

Peter ForemanPeter Foreman

8,5061321

$endgroup$

share|cite|improve this answer

edited Apr 2 at 18:22

Théophile

20.5k13047

answered Apr 2 at 15:08

Peter ForemanPeter Foreman

8,5061321

edited Apr 2 at 18:22

Théophile

20.5k13047

edited Apr 2 at 18:22

Théophile

20.5k13047

answered Apr 2 at 15:08

Peter ForemanPeter Foreman

8,5061321

answered Apr 2 at 15:08

Peter ForemanPeter Foreman

8,5061321