$$ \sum_{n=1}^{999} \log_{10}\left(\frac{n+1}{n}\right) $$
Can anybody help me how to calculate this summation?
$$ \sum_{n=1}^{999} \log_{10}\left(\frac{n+1}{n}\right) $$
Can anybody help me how to calculate this summation?
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You were on the right track! Like Minh Nguyen Nhat, you have to take $\log(1000!)-\log(999!)$ as $\log(1000!/999!)$ to get the answer.
$\log(2/1)+\log(3/2)+\log(4/3)+$...$+\log(1000/999)=\log(1000!/999!)=\log(1000)=3$