How to prove: \sum_{k=1}^\infty\frac{k-1}{2k(1+k)(1+2k)}=\log_e 8-2

Yahir Haas

Yahir Haas

Answered question

2022-01-21

How to prove:
k=1k12k(1+k)(1+2k)=loge82

Answer & Explanation

sjkuzy5

sjkuzy5

Beginner2022-01-22Added 11 answers

Let me try. Note that the
RHS=3ln2+2=2+3k=1(1)k+1k=2+3k=112k112k
We have
k=1k12k(k+1)(2k+1)=k=12k(k+1)2k(k+1)(2k+1)
=k=1(1(k+1)(2k+1)12k(2k+1))
=k=1(22k+11k+112k+12k+1)
=2+3k=112k112k
=RHS

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