利用
[sum_{j=1}^{n}frac{left ( -1
ight )^{j-1}}{j}=ln 2+left ( -1
ight )^{n-1}int_{0}^{1}frac{x^{n}}{1+x}\, mathrm{d}x
]
我们有
[egin{align*}
sum_{k=1}^{infty }frac{left ( -1
ight )^{k}}{k}sum_{j=1}^{2k}frac{left ( -1
ight )^{j}}{j}&=ln 2sum_{k=1}^{infty }frac{left ( -1
ight )^{k}}{k}-sum_{k=1}^{infty }frac{left ( -1
ight )^{k}}{k}int_{0}^{1}frac{x^{2k}}{1+x}\, mathrm{d}x\
&=ln^22+int_{0}^{1}frac{1}{1+x}sum_{k=1}^{infty }frac{left ( -x^{2}
ight )^{k}}{k}\, mathrm{d}x\
&=ln^22-int_{0}^{1}frac{lnleft ( 1+x^{2}
ight )}{1+x}\, mathrm{d}x\
&=ln^22+frac{pi ^{2}}{48}-frac{3}{4}ln^22\
&=frac{pi ^{2}}{48}+frac{1}{4}ln^22
end{align*}]