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Infinite sum of harmonic number

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4

3

$begingroup$

I learned that I can find the value of some infinite sum.

Then what is the value of this sum?
$$frac12 + left(1+frac12right)frac1{2^2}+left(1+frac12 +frac13right)frac1{2^3}+left(1+frac12 +frac13 +frac14right)frac1{2^4} + cdots $$
And I want to know How to find the value of the infinite sum of this-like form.

sequences-and-series

share|cite|improve this question

asked Apr 2 at 14:47

S. YooS. Yoo

424

$endgroup$

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?
$$frac12 + left(1+frac12right)frac1{2^2}+left(1+frac12 +frac13right)frac1{2^3}+left(1+frac12 +frac13 +frac14right)frac1{2^4} + cdots $$
And I want to know How to find the value of the infinite sum of this-like form.

sequences-and-series

share|cite|improve this question

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

share|cite|improve this question

asked Apr 2 at 14:47

S. YooS. Yoo

424

$endgroup$

share|cite|improve this question

asked Apr 2 at 14:47

S. YooS. Yoo

424

share|cite|improve this question

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

1 Answer
1

active

oldest

votes

8

$begingroup$

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
$$H_k=sum_{n=1}^kfrac1n$$

share|cite|improve this answer

edited Apr 2 at 18:22

Théophile

20.4k13047

answered Apr 2 at 15:08

Peter ForemanPeter Foreman

6,2611317

$endgroup$

add a comment | 

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8

$begingroup$

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
$$H_k=sum_{n=1}^kfrac1n$$

share|cite|improve this answer

edited Apr 2 at 18:22

Théophile

20.4k13047

answered Apr 2 at 15:08

Peter ForemanPeter Foreman

6,2611317

$endgroup$

add a comment | 

8

$begingroup$

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
$$H_k=sum_{n=1}^kfrac1n$$

share|cite|improve this answer

edited Apr 2 at 18:22

Théophile

20.4k13047

answered Apr 2 at 15:08

Peter ForemanPeter Foreman

6,2611317

$endgroup$

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
$$H_k=sum_{n=1}^kfrac1n$$

share|cite|improve this answer

edited Apr 2 at 18:22

Théophile

20.4k13047

answered Apr 2 at 15:08

Peter ForemanPeter Foreman

6,2611317

$endgroup$

share|cite|improve this answer

edited Apr 2 at 18:22

Théophile

20.4k13047

answered Apr 2 at 15:08

Peter ForemanPeter Foreman

6,2611317

edited Apr 2 at 18:22

Théophile

20.4k13047

edited Apr 2 at 18:22

Théophile

20.4k13047

answered Apr 2 at 15:08

Peter ForemanPeter Foreman

6,2611317

answered Apr 2 at 15:08

Peter ForemanPeter Foreman

6,2611317