Qual é a soma da série ?.... Pergunta de ideia deDaboiul

January 2020 1 56 Report

Qual é a soma da série $\sum_{n=2}^{\infty} ln(1- \frac{1}{n^2})$ ?

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aquiles1987 $\sum_{n=2}^{\infty} ln( \frac{ n^2 -1}{n^2}) = \sum_{n=2}^{\infty} ln( \frac{n-1}{n} ) +ln( \frac{n+1}{n} )\\
ln \frac{1}{2} + ln \frac{3}{2} +ln \frac{2}{3} +ln \frac{4}{3} +ln \frac{3}{4} +ln \frac{5}{4} ...... \\
S_n=ln \frac{1}{2} + ln \frac{3}{2} +ln \frac{2}{3} +ln \frac{4}{3} +ln \frac{3}{4} +ln \frac{5}{4} ......+ln( \frac{n-1}{n} ) +ln( \frac{n+1}{n} )\\
S_n=ln \frac{1}{2} +ln( \frac{3}{2} \frac{2}{3} \frac{4}{3} \frac{3}{4} \frac{4}{3} .. \frac{n-1}{n} )+ln \frac{n+1}{n} \\

$ $S_n=ln \frac{1}{2} +ln1+ln \frac{n+1}{n} \\ \lim_{n \to \infty} ln \frac{1}{2} +ln1+ln \frac{n+1}{n} =ln \frac{1}{2} +ln1+ln1=ln \frac{1}{2} +0+0=ln \frac{1}{2}$

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