Evaluation of a series involving the harmonic number

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Let $textH_n$ denote the n-th harmonic number, does the following sequence have a closed-form ? As an approximation I have got $0.0922514...$
$$ frac32+lim_ntoinfty bigg(sum_k=3^nfrac2k+sqrtk^2-4-textH_nbigg)$$

sequences-and-series zeta-functions eulers-constant euler-sums

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asked Jul 19 at 21:43

Kays Tomy

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Let $textH_n$ denote the n-th harmonic number, does the following sequence have a closed-form ? As an approximation I have got $0.0922514...$
$$ frac32+lim_ntoinfty bigg(sum_k=3^nfrac2k+sqrtk^2-4-textH_nbigg)$$

sequences-and-series zeta-functions eulers-constant euler-sums

share|cite|improve this question

asked Jul 19 at 21:43

Kays Tomy

934

add a comment | 

up vote
0
down vote

favorite

1

up vote
0
down vote

favorite

1

1

Let $textH_n$ denote the n-th harmonic number, does the following sequence have a closed-form ? As an approximation I have got $0.0922514...$
$$ frac32+lim_ntoinfty bigg(sum_k=3^nfrac2k+sqrtk^2-4-textH_nbigg)$$

sequences-and-series zeta-functions eulers-constant euler-sums

share|cite|improve this question

asked Jul 19 at 21:43

Kays Tomy

934

Let $textH_n$ denote the n-th harmonic number, does the following sequence have a closed-form ? As an approximation I have got $0.0922514...$
$$ frac32+lim_ntoinfty bigg(sum_k=3^nfrac2k+sqrtk^2-4-textH_nbigg)$$

sequences-and-series zeta-functions eulers-constant euler-sums

share|cite|improve this question

asked Jul 19 at 21:43

Kays Tomy

934

share|cite|improve this question

share|cite|improve this question

asked Jul 19 at 21:43

Kays Tomy

934

asked Jul 19 at 21:43

Kays Tomy

934

asked Jul 19 at 21:43

Kays Tomy

934

934

add a comment | 

add a comment | 

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$$S(n)=sum^n_k=3frac2k+sqrtk^2-4=sum^n_k=3frac2k+sqrtk^2-4cdotfrack-sqrtk^2-4k-sqrtk^2-4=frac12sum^n_k=3k-sqrtk^2-4$$ which can be simplified to $$frac12left(fracn(n+1)2-3-sum^n_k=3sqrtk^2-4right)$$

By generalized binomial theorem,

$$beginalign
sum^n_k=3sqrtk^2-4 & = sum^n_k=3ksqrt1-frac4 k^2 \
&=sum^n_k=3ksum^infty_m=0binom1/2m(-4)^mfrac1k^2m \
&=underbracesum^n_k=3k_textwhen m=0+underbrace(-2)sum^n_k=3frac1k_textwhen m=1+sum^n_k=3ksum^infty_m=2binom1/2m(-4)^mfrac1k^2m\
& =fracn(n+1)2-2(H_n-frac52)+sum^n_k=3sum^infty_m=2binom1/2m(-4)^mfrac1k^2m-1\
endalign
$$

$$S(n)=frac12left(-3+2H_n-5-sum^n_k=3sum^infty_m=2binom1/2m(-4)^mfrac1k^2m-1right)$$

Therefore,
$$lim_ntoinftyS(n)-H_n=-4-frac12sum^infty_k=3sum^infty_m=2binom1/2m(-4)^mfrac1k^2m-1$$

You may try to express it in terms of zeta function.

EDIT:
$$binom1/2m(-4)^m=-fracC^2m_m2m-1$$
So, $$lim_ntoinftyS(n)-H_n=-4+frac12sum^infty_k=3sum^infty_m=2fracC^2m_m(2m-1)k^2m-1$$

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answered Jul 20 at 4:04

Szeto

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1 Answer
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1 Answer
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$$S(n)=sum^n_k=3frac2k+sqrtk^2-4=sum^n_k=3frac2k+sqrtk^2-4cdotfrack-sqrtk^2-4k-sqrtk^2-4=frac12sum^n_k=3k-sqrtk^2-4$$ which can be simplified to $$frac12left(fracn(n+1)2-3-sum^n_k=3sqrtk^2-4right)$$

By generalized binomial theorem,

$$beginalign
sum^n_k=3sqrtk^2-4 & = sum^n_k=3ksqrt1-frac4 k^2 \
&=sum^n_k=3ksum^infty_m=0binom1/2m(-4)^mfrac1k^2m \
&=underbracesum^n_k=3k_textwhen m=0+underbrace(-2)sum^n_k=3frac1k_textwhen m=1+sum^n_k=3ksum^infty_m=2binom1/2m(-4)^mfrac1k^2m\
& =fracn(n+1)2-2(H_n-frac52)+sum^n_k=3sum^infty_m=2binom1/2m(-4)^mfrac1k^2m-1\
endalign
$$

$$S(n)=frac12left(-3+2H_n-5-sum^n_k=3sum^infty_m=2binom1/2m(-4)^mfrac1k^2m-1right)$$

Therefore,
$$lim_ntoinftyS(n)-H_n=-4-frac12sum^infty_k=3sum^infty_m=2binom1/2m(-4)^mfrac1k^2m-1$$

You may try to express it in terms of zeta function.

EDIT:
$$binom1/2m(-4)^m=-fracC^2m_m2m-1$$
So, $$lim_ntoinftyS(n)-H_n=-4+frac12sum^infty_k=3sum^infty_m=2fracC^2m_m(2m-1)k^2m-1$$

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answered Jul 20 at 4:04

Szeto

4,1731521

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up vote
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$$S(n)=sum^n_k=3frac2k+sqrtk^2-4=sum^n_k=3frac2k+sqrtk^2-4cdotfrack-sqrtk^2-4k-sqrtk^2-4=frac12sum^n_k=3k-sqrtk^2-4$$ which can be simplified to $$frac12left(fracn(n+1)2-3-sum^n_k=3sqrtk^2-4right)$$

By generalized binomial theorem,

$$beginalign
sum^n_k=3sqrtk^2-4 & = sum^n_k=3ksqrt1-frac4 k^2 \
&=sum^n_k=3ksum^infty_m=0binom1/2m(-4)^mfrac1k^2m \
&=underbracesum^n_k=3k_textwhen m=0+underbrace(-2)sum^n_k=3frac1k_textwhen m=1+sum^n_k=3ksum^infty_m=2binom1/2m(-4)^mfrac1k^2m\
& =fracn(n+1)2-2(H_n-frac52)+sum^n_k=3sum^infty_m=2binom1/2m(-4)^mfrac1k^2m-1\
endalign
$$

$$S(n)=frac12left(-3+2H_n-5-sum^n_k=3sum^infty_m=2binom1/2m(-4)^mfrac1k^2m-1right)$$

Therefore,
$$lim_ntoinftyS(n)-H_n=-4-frac12sum^infty_k=3sum^infty_m=2binom1/2m(-4)^mfrac1k^2m-1$$

You may try to express it in terms of zeta function.

EDIT:
$$binom1/2m(-4)^m=-fracC^2m_m2m-1$$
So, $$lim_ntoinftyS(n)-H_n=-4+frac12sum^infty_k=3sum^infty_m=2fracC^2m_m(2m-1)k^2m-1$$

share|cite|improve this answer

answered Jul 20 at 4:04

Szeto

4,1731521

add a comment | 

up vote
0
down vote

up vote
0
down vote

$$S(n)=sum^n_k=3frac2k+sqrtk^2-4=sum^n_k=3frac2k+sqrtk^2-4cdotfrack-sqrtk^2-4k-sqrtk^2-4=frac12sum^n_k=3k-sqrtk^2-4$$ which can be simplified to $$frac12left(fracn(n+1)2-3-sum^n_k=3sqrtk^2-4right)$$

By generalized binomial theorem,

$$beginalign
sum^n_k=3sqrtk^2-4 & = sum^n_k=3ksqrt1-frac4 k^2 \
&=sum^n_k=3ksum^infty_m=0binom1/2m(-4)^mfrac1k^2m \
&=underbracesum^n_k=3k_textwhen m=0+underbrace(-2)sum^n_k=3frac1k_textwhen m=1+sum^n_k=3ksum^infty_m=2binom1/2m(-4)^mfrac1k^2m\
& =fracn(n+1)2-2(H_n-frac52)+sum^n_k=3sum^infty_m=2binom1/2m(-4)^mfrac1k^2m-1\
endalign
$$

$$S(n)=frac12left(-3+2H_n-5-sum^n_k=3sum^infty_m=2binom1/2m(-4)^mfrac1k^2m-1right)$$

Therefore,
$$lim_ntoinftyS(n)-H_n=-4-frac12sum^infty_k=3sum^infty_m=2binom1/2m(-4)^mfrac1k^2m-1$$

You may try to express it in terms of zeta function.

EDIT:
$$binom1/2m(-4)^m=-fracC^2m_m2m-1$$
So, $$lim_ntoinftyS(n)-H_n=-4+frac12sum^infty_k=3sum^infty_m=2fracC^2m_m(2m-1)k^2m-1$$

share|cite|improve this answer

answered Jul 20 at 4:04

Szeto

4,1731521

$$S(n)=sum^n_k=3frac2k+sqrtk^2-4=sum^n_k=3frac2k+sqrtk^2-4cdotfrack-sqrtk^2-4k-sqrtk^2-4=frac12sum^n_k=3k-sqrtk^2-4$$ which can be simplified to $$frac12left(fracn(n+1)2-3-sum^n_k=3sqrtk^2-4right)$$

By generalized binomial theorem,

$$beginalign
sum^n_k=3sqrtk^2-4 & = sum^n_k=3ksqrt1-frac4 k^2 \
&=sum^n_k=3ksum^infty_m=0binom1/2m(-4)^mfrac1k^2m \
&=underbracesum^n_k=3k_textwhen m=0+underbrace(-2)sum^n_k=3frac1k_textwhen m=1+sum^n_k=3ksum^infty_m=2binom1/2m(-4)^mfrac1k^2m\
& =fracn(n+1)2-2(H_n-frac52)+sum^n_k=3sum^infty_m=2binom1/2m(-4)^mfrac1k^2m-1\
endalign
$$

$$S(n)=frac12left(-3+2H_n-5-sum^n_k=3sum^infty_m=2binom1/2m(-4)^mfrac1k^2m-1right)$$

Therefore,
$$lim_ntoinftyS(n)-H_n=-4-frac12sum^infty_k=3sum^infty_m=2binom1/2m(-4)^mfrac1k^2m-1$$

You may try to express it in terms of zeta function.

EDIT:
$$binom1/2m(-4)^m=-fracC^2m_m2m-1$$
So, $$lim_ntoinftyS(n)-H_n=-4+frac12sum^infty_k=3sum^infty_m=2fracC^2m_m(2m-1)k^2m-1$$

share|cite|improve this answer

answered Jul 20 at 4:04

Szeto

4,1731521

share|cite|improve this answer

share|cite|improve this answer

answered Jul 20 at 4:04

Szeto

4,1731521

answered Jul 20 at 4:04

Szeto

4,1731521

answered Jul 20 at 4:04

Szeto

4,1731521

4,1731521

add a comment | 

add a comment | 

 

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