Question

$\sum _{k=1}^{9}\log \left( \dfrac {k} {k+1}\right) $  işleminin sonucu ?

Answer 1

$\displaystyle\sum_{k=1}^{9} \log\left(\dfrac{k}{k+1}\right)=log(\dfrac{1}{2})+log(\dfrac{2}{3})+log(\dfrac{3}{4})+....log(\dfrac{9}{10})=\log\left(\dfrac{1.2.3.4.......9}{2.3.4.5.6....10}\right)=\boxed{\boxed{\boxed{\boxed{-1}}}}$

Answer 2

$$\sum_{k=1}^{n}\log \left(\frac{k}{k+1}\right)$$$$=$$$$\sum_{k=1}^{n} (\log k-\log (k+1))$$

$$=$$

$$(\log 1-\log 2)+(\log 2-\log 3)+\ldots +(\log (n-1)-\log n)$$

$$=$$

$$\log 1-\log n$$

$$=$$

$$\log \frac1n$$

$$=$$

$$-\log n$$

Comments

teşekkürler hocam :)