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  • Euler Sums系列(一)

    [Largesum_{n=1}^{infty} frac{H_{n}}{2^nn^4} ]


    (Largemathbf{Solution:})
    Let

    [mathcal{S}=sum^infty_{n=1}frac{H_n}{n^42^n} ]

    We first consider a slightly different yet related sum. The main idea is to solve this sum with two different methods, one of which involves the sum in question. This then allows us to determine the value of the desired sum.

    [egin{align*} &sum^infty_{n=1}frac{(-1)^nH_n}{n^4} =frac{1}{6}sum^infty_{n=1}(-1)^{n-1}H_nint^1_0x^{n-1}ln^3{x} { m d}x =frac{1}{6}int^1_0frac{ln^3{x}ln(1+x)}{x(1+x)}{ m d}x\ =&frac{1}{6}int^1_0frac{ln^3{x}ln(1+x)}{x}{ m d}x-frac{1}{6}int^1_0frac{ln^3{x}ln(1+x)}{1+x}{ m d}x =frac{1}{6}sum^infty_{n=1}frac{(-1)^{n-1}}{n}int^1_0x^{n-1}ln^3{x} { m d}x\ -&frac{1}{6}int^2_1frac{ln{x}ln^3(x-1)}{x}{ m d}x =sum^infty_{n=1}frac{(-1)^{n}}{n^5}+int^1_{frac{1}{2}}frac{ln{x}ln^3(1-x)}{6x}-int^1_{frac{1}{2}}frac{ln^2{x}ln^2(1-x)}{2x}{ m d}x\+&int^1_{frac{1}{2}}frac{ln^3{x}ln(1-x)}{2x}{ m d}x-int^1_{frac{1}{2}}frac{ln^4{x}}{6x}{ m d}x =-frac{15}{16}zeta(5)+mathcal{I}_1-mathcal{I}_2+mathcal{I}_3-mathcal{I}_4 end{align*}]

    Starting with the easiest integral,

    [egin{align*} mathcal{I}_4=frac{1}{30}ln^5{2} end{align*}]

    For (mathcal{I}_3),

    [egin{align*} mathcal{I}_3 =&-frac{1}{2}sum^infty_{n=1}frac{1}{n}int^1_{frac{1}{2}}x^{n-1}ln^3{x} { m d}x =-frac{1}{2}sum^infty_{n=1}frac{1}{n}frac{partial^3}{partial n^3}left(frac{1}{n}-frac{1}{n2^n} ight)\ =&sum^infty_{n=1}left(frac{3}{n^5}-frac{3}{n^52^n}-frac{3ln{2}}{n^42^n}-frac{3ln^2{2}}{n^32^{n+1}}-frac{ln^3{2}}{n^22^{n+1}} ight)\ =&3zeta(5)-3{ m Li}_5left(dfrac{1}{2} ight)-3{ m Li}_4left(dfrac{1}{2} ight)ln{2}-frac{3}{2}ln^2{2}left(frac{7}{8}zeta(3)-frac{pi^2}{12}ln{2}+frac{1}{6}ln^3{2} ight)\&-frac{1}{2}ln^3{2}left(frac{pi^2}{12}-frac{1}{2}ln^2{2} ight)\ =&3zeta(5)-3{ m Li}_5left(dfrac{1}{2} ight)-3{ m Li}_4left(dfrac{1}{2} ight)ln{2}-frac{21}{16}zeta(3)ln^2{2}+frac{pi^2}{12}ln^3{2} end{align*}]

    For (mathcal{I}_2),

    [egin{align*} mathcal{I}_2 =&frac{1}{6}ln^5{2}+frac{1}{3}int^1_{frac{1}{2}}frac{ln^3{x}ln(1-x)}{1-x}{ m d}x =frac{1}{6}ln^5{2}-frac{1}{3}sum^infty_{n=1}H_nfrac{partial^3}{partial n^3}left(frac{1}{n+1}-frac{1}{(n+1)2^{n+1}} ight)\ =&frac{1}{6}ln^5{2}+sum^infty_{n=1}frac{2H_n}{(n+1)^4}-sum^infty_{n=1}frac{2H_n}{(n+1)^42^{n+1}}-sum^infty_{n=1}frac{2ln{2}H_n}{(n+1)^32^{n+1}}\ &-sum^infty_{n=1}frac{ln^2{2}H_n}{(n+1)^22^{n+1}}-sum^infty_{n=1}frac{ln^3{2}H_n}{3(n+1)2^{n+1}}\ =&frac{1}{6}ln^5{2}+4zeta(5)-frac{pi^2}{3}zeta(3)-2mathcal{S}+2{ m Li}_5left(dfrac{1}{2} ight)-frac{pi^4}{360}ln{2}+frac{1}{4}zeta(3)ln^2{2}-frac{1}{12}ln^5{2}\ &-frac{1}{8}zeta(3)ln^2{2}+frac{1}{6}ln^5{2}-frac{1}{6}ln^5{2}\ =&-2mathcal{S}+2{ m Li}_5left(dfrac{1}{2} ight)+4zeta(5)-frac{pi^4}{360}ln{2}+frac{1}{8}zeta(3)ln^2{2}-frac{pi^2}{3}zeta(3)+frac{1}{12}ln^5{2} end{align*}]

    For (mathcal{I}_1),

    [egin{align*} mathcal{I}_1 =&frac{1}{6}int^{frac{1}{2}}_0frac{ln^3{x}ln(1-x)}{1-x}{ m d}x =-frac{1}{6}sum^infty_{n=1}H_nfrac{partial^3}{partial n^3}left(frac{1}{(n+1)2^{n+1}} ight)\ =&sum^infty_{n=1}frac{H_n}{(n+1)^42^{n+1}}+sum^infty_{n=1}frac{ln{2}H_n}{(n+1)^32^{n+1}}+sum^infty_{n=1}frac{ln^2{2}H_n}{2(n+1)^22^{n+1}}+sum^infty_{n=1}frac{ln^3{2}H_n}{6(n+1)2^{n+1}}\ =&mathcal{S}-{ m Li}_5left(dfrac{1}{2} ight)+frac{pi^4}{720}ln{2}-frac{1}{16}zeta(3)ln^2{2}+frac{1}{24}ln^5{2} end{align*}]

    Combining these four integrals as (mathcal{I}_1-mathcal{I}_2+mathcal{I}_3-mathcal{I}_4) and (displaystyle -dfrac{15}{16}zeta(5)) gives

    [egin{align*} sum^infty_{n=1}frac{(-1)^nH_n}{n^4} =&3mathcal{S}-6{ m Li}_5left(dfrac{1}{2} ight)-frac{31}{16}zeta(5)-3{ m Li}_4left(dfrac{1}{2} ight)ln{2}+frac{pi^4}{240}ln{2}\&-frac{3}{2}zeta(3)ln^2{2}+frac{pi^2}{3}zeta(3)+frac{pi^2}{12}ln^3{2}-frac{3}{40}ln^5{2} end{align*}]

    But consider (displaystyle f(z)=frac{picsc(pi z)(gamma+psi_0(-z))}{z^4}). At the positive integers,

    [sum^infty_{n=1}{ m Res}(f,n)=sum^infty_{n=1}operatorname*{Res}_{z=n}left[frac{(-1)^n}{z^4(z-n)^2}+frac{(-1)^nH_n}{z^4(z-n)} ight]=sum^infty_{n=1}frac{(-1)^nH_n}{n^4}+frac{15}{4}zeta(5) ]

    At (z=0),

    [egin{align*} { m Res}(f,0) &=[z^3]left(frac{1}{z}+frac{pi^2}{6}z+frac{7pi^4}{360}z^3 ight)left(frac{1}{z}-frac{pi^2}{6}z-zeta(3)z^2-frac{pi^4}{90}z^3-zeta(5)z^4 ight)\ &=-zeta(5)-frac{pi^2}{6}zeta(3) end{align*}]

    At the negative integers,

    [egin{align*} sum^infty_{n=1}{ m Res}(f,-n) &=sum^infty_{n=1}frac{(-1)^nH_n}{n^4}+frac{15}{16}zeta(5) end{align*}]

    Since the sum of the residues is zero,

    [sum^infty_{n=1}frac{(-1)^nH_n}{n^4}=-frac{59}{32}zeta(5)+frac{pi^2}{12}zeta(3) ]

    Hence,

    [egin{align*} -frac{59}{32}zeta(5)+frac{pi^2}{12}zeta(3) =&3mathcal{S}-6{ m Li}_5left(dfrac{1}{2} ight)-frac{31}{16}zeta(5)-3{ m Li}_4left(dfrac{1}{2} ight)ln{2}+frac{pi^4}{240}ln{2}\&-frac{3}{2}zeta(3)ln^2{2}+frac{pi^2}{3}zeta(3)+frac{pi^2}{12}ln^3{2}-frac{3}{40}ln^5{2} end{align*}]

    This implies that

    [oxed{egin{align*} {sum^infty_{n=1}frac{H_n}{n^42^n}} {=}&color{blue}{2{ m Li}_5left(dfrac{1}{2} ight)+frac{1}{32}zeta(5)+{ m Li}_4left(dfrac{1}{2} ight)ln{2}-frac{pi^4}{720}ln{2}+frac{1}{2}zeta(3)ln^2{2}}\&color{blue}{-frac{pi^2}{12}zeta(3)-frac{pi^2}{36}ln^3{2}+frac{1}{40}ln^5{2}} end{align*}}]

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  • 原文地址:https://www.cnblogs.com/Renascence-5/p/5436454.html
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