Prove \sum_{k=1}^\infty\frac{(-1)^k}{k}\sum_{j=1}^{2k}\frac{(-1)^j}{j}=\frac{\pi^2}{48}+\frac{1}{4}\ln^2 2

Zeenat Horn

Zeenat Horn

Answered question

2022-02-24

Prove k=1(1)kkj=12k(1)jj=π248+14ln22

Answer & Explanation

Kathryn Duggan

Kathryn Duggan

Beginner2022-02-25Added 7 answers

Use
j=1n(1)j1j=ln2+(1)n101xn1+xdx
we get
k=1(1)kkj=12k(1)jj=ln2k=1(1)kkk=1(1)kk01x2k1+xdx
=ln22+0111+xk=1(x2)kkdx=ln2201ln(1+x2)1+xdx
letf(t)=01ln(1+tx2)1+xdx
then
f(t)=01x2(1+x)(1+tx2)dx
=1t+101x11+tx2dx+1t+101dxx+1
=1t+1[12tln(1+tx2)1tarctan(tx)+ln(x+1)]01
Tye Rhodes

Tye Rhodes

Beginner2022-02-26Added 4 answers

Note that
j=12k(1)jj=j=1k12jj=1k12j1=12j=1k1j(j=12k1jj=1k12j)=j=1k1jj=12k1j
=HkH2k
Then,
k=1(1)kkj=12k(1)jj=k=1(1)kk(HkH2k)
=k=1(1)kkHk2k=1i2k2kH2k
=f(1)2R(f(i)) where f(x)=k=1xkkHk
However,
f(x)=k=1xkHk01tk1dt=011tk=1(xt)kHkdt=011t[ln(1xt)1xt]dt

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?