[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*}}]