Sums of logarithms: \(\displaystyle{\sum_{{{k}={a}+{1}}}^{{b}}}{\log{{\left({1}+\frac{{1}}{{k}}\right)}}}\)

London Douglas

London Douglas

Answered question

2022-04-02

Sums of logarithms:
k=a+1blog(1+1k)

Answer & Explanation

Karsyn Wu

Karsyn Wu

Beginner2022-04-03Added 17 answers

Write
k=a+1blog(1+1k)=log(k=a+1bk+1k)
or
k=a+1blog(1+1k)=k=a+1blog(k+1)log(k)
Then you can work with a telescoping product or sum.

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