# Another transposed harmonic alternating series

Calculus Level 5

$\large \sum_{k=1}^{c} \left(\dfrac{1}{2k}\right)-1+\sum_{k=c+1}^{2c} \left(\dfrac{1}{2k}\right)-\dfrac{1}{3}+\sum_{k=2c+1}^{3c}\left(\dfrac{1}{2k}\right)-\dfrac{1}{5}+\cdots=\dfrac{1}{2}\ln\frac{5}{4}$

In the previous series, the number $$c$$ is a positive integer and the sum of the series is the number $$\dfrac{1}{2}\ln\dfrac{5}{4}$$. Find the value of $$c$$ if there is any, otherwise, enter a 666 as your answer.

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