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Question Number 104174 by Dwaipayan Shikari last updated on 19/Jul/20

\underset{n = 1}{\prod^{\infty}} {(\frac{n}{n + 1})}^{2}

Answered by OlafThorendsen last updated on 19/Jul/20

S_{n} = \underset{k = 1}{\prod^{n}} {(\frac{k}{k + 1})}^{2}

S_{n} = {(\frac{1}{2})}^{2} {(\frac{2}{3})}^{2} {(\frac{3}{4})}^{2} \dots {(\frac{n}{n + 1})}^{2}

S_{n} = \frac{1}{{(n + 1)}^{2}}

Double subscripts: use braces to clarify

Answered by mathmax by abdo last updated on 19/Jul/20

let S_{n} = \prod_{k = 1}^{n} \frac{k^{2}}{{(k + 1)}^{2}} \Rightarrow \ln (S_{n}) = \sum_{k = 1}^{n} \ln (\frac{k^{2}}{{(k + 1)}^{2}}) = 2 \sum_{k = 1}^{n} {\ln (k) - \ln (k + 1)}

= 2 {\ln 1 - \ln (2) + \ln (2) - \ln (3) + \dots + \ln (n) - \ln (n + 1)}

= - 2 \ln (n + 1) \to - \infty \Rightarrow \lim_{n \to + \infty} \ln (S_{n}) = 0 \Rightarrow \lim_{n \to + \infty} S_{n} = 0