Answered: "How to prove: ∑k=1∞k−12k(1+k)(1+2k)=loge8−2"

Yahir Haas

Answered question

2022-01-21

How to prove:

\sum_{k = 1}^{\infty} \frac{k - 1}{2 k (1 + k) (1 + 2 k)} = \log_{e} 8 - 2

sjkuzy5

Beginner2022-01-22Added 11 answers

Let me try. Note that the

R H S = 3 \ln 2 + 2 = 2 + 3 \sum_{k = 1}^{\infty} \frac{{(- 1)}^{k + 1}}{k} = 2 + 3 \sum_{k = 1}^{\infty} \frac{1}{2 k - 1} - \frac{1}{2 k}

We have

\sum_{k = 1}^{\infty} \frac{k - 1}{2 k (k + 1) (2 k + 1)} = \sum_{k = 1}^{\infty} \frac{2 k - (k + 1)}{2 k (k + 1) (2 k + 1)}

= \sum_{k = 1}^{\infty} (\frac{1}{(k + 1) (2 k + 1)} - \frac{1}{2 k (2 k + 1)})

= \sum_{k = 1}^{\infty} (\frac{2}{2 k + 1} - \frac{1}{k + 1} - \frac{1}{2 k} + \frac{1}{2 k + 1})

= 2 + 3 \sum_{k = 1}^{\infty} \frac{1}{2 k - 1} - \frac{1}{2 k}

= R H S

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