易证明1/m={1/m-1/(m+n)}+{1/(m+n)-1/(m+2n)}+....=
∑n/(m+in)(m+in+n)[i=0→∞]
那么(1/n)*∑1/m =∑∑1/(m+in)(m+in+n)∑分别m=1→n,m=1→n,i=0→∞
=∑1/L*(L+n) l=1→∞
可知i取 0 1 2 ....及m=1 2 ....n,那么可以的到L取到1 到无穷
下面写的求和都是1到无穷
S=∑{1/n*(∑1/m)}^2=∑{∑1/L*(L+n)}^2=
∑1/i*L(L+n)(i+n)[L,i,n]=∑1/i*L(L+n)(i+n)[L<i]+∑1/i*L(L+n)(i+n)[L=i]+∑1/i*L(L+n)(i+n)[L>i]=
∑1/(L+n)^2*L^2 +2∑1/i*L(L+n)(i+n)[L<i]
=∑1/i^2*L^2 [l<i]+2∑1/L(L+m)(L+n)(L+m+n)
=1/2{∑1/i^2*L^2 - ∑1/i^2*L^2 [i=L]+2 ∑{1/Lmn(L+m+n) -
1/mn(L+m)(L+n) }=1/2{(∑1/L^2)^2-∑1/L^4}+1/∑Lmn(L+m+n)-2S
利用∑1/L^2=pi^2/6 ∑1/L^4=pi^4/90
我们利用上面结论可以得到1/2(3S-pi^4/120)=∑1/Lmn(L+m+n)
=∑(0→1)∫x^(L+m+n-1)/Lmn dx
=(0→1)∫(∑x^k/k)^3*(dx/x)=(0→1)∫{ln(1-x)}^3*(dx/x)
=(0→1)∫(lnx)^3*dx/(1-x) =-∑(0→1)∫x^(n-1)*(lnx)^3dx
=∑6/n^4=pi^4/15
所以可以得到S=17pi^4/360
∑n/(m+in)(m+in+n)[i=0→∞]
那么(1/n)*∑1/m =∑∑1/(m+in)(m+in+n)∑分别m=1→n,m=1→n,i=0→∞
=∑1/L*(L+n) l=1→∞
可知i取 0 1 2 ....及m=1 2 ....n,那么可以的到L取到1 到无穷
下面写的求和都是1到无穷
S=∑{1/n*(∑1/m)}^2=∑{∑1/L*(L+n)}^2=
∑1/i*L(L+n)(i+n)[L,i,n]=∑1/i*L(L+n)(i+n)[L<i]+∑1/i*L(L+n)(i+n)[L=i]+∑1/i*L(L+n)(i+n)[L>i]=
∑1/(L+n)^2*L^2 +2∑1/i*L(L+n)(i+n)[L<i]
=∑1/i^2*L^2 [l<i]+2∑1/L(L+m)(L+n)(L+m+n)
=1/2{∑1/i^2*L^2 - ∑1/i^2*L^2 [i=L]+2 ∑{1/Lmn(L+m+n) -
1/mn(L+m)(L+n) }=1/2{(∑1/L^2)^2-∑1/L^4}+1/∑Lmn(L+m+n)-2S
利用∑1/L^2=pi^2/6 ∑1/L^4=pi^4/90
我们利用上面结论可以得到1/2(3S-pi^4/120)=∑1/Lmn(L+m+n)
=∑(0→1)∫x^(L+m+n-1)/Lmn dx
=(0→1)∫(∑x^k/k)^3*(dx/x)=(0→1)∫{ln(1-x)}^3*(dx/x)
=(0→1)∫(lnx)^3*dx/(1-x) =-∑(0→1)∫x^(n-1)*(lnx)^3dx
=∑6/n^4=pi^4/15
所以可以得到S=17pi^4/360