Putnam 2017/B4

#1

Να υπολογιστεί το άθροισμα

$\displaystyle \sum_{k=0}^{\infty}\left(3\cdot\frac{\ln(4k+2)}{4k+2}-\frac{\ln(4k+3)}{4k+3}-\frac{\ln(4k+4)}{4k+4}-\frac{\ln(4k+5)}{4k+5}\right)$

dement · #2

Γράφουμε

$\displaystyle \sum_{k=0}^{\infty} \left( \frac{3 \ln (4k+2)}{4k+2} - \frac{\ln (4k+3)}{4k+3} - \frac{\ln (4k+4)}{4k+4} - \frac{\ln (4k+5)}{4k+5} \right) =$

$\displaystyle = \sum_{k=0}^{\infty} \left( \frac{\ln (4k+2)}{4k+2} - \frac{\ln (4k+3)}{4k+3} + \frac{\ln (4k+4)}{4k+4} - \frac{\ln (4k+5)}{4k+5} \right) + 2\sum_{k=0}^{\infty} \left( \frac{\ln (4k+2)}{4k+2} - \frac{\ln (4k+4)}{4k+4} \right) =$

$\displaystyle = \sum_{k=2}^{\infty} \frac{(-1)^k \ln k}{k} + 2 \sum_{k=1}^{\infty} \frac{(-1)^{k+1} \ln (2k)}{2k} = \ln 2 + \sum_{k=2}^{\infty} (-1)^k \left( \frac{\ln k}{k} - \frac{\ln(2k)}{k} \right) =$

$\displaystyle = \ln 2 - \ln 2 \sum_{k=2}^{\infty} \frac{(-1)^k}{k} = \ln 2 - \ln 2 (1 - \ln 2) = (\ln 2)^2$

Putnam 2017/B4

Putnam 2017/B4

Re: Putnam 2017/B4

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