Math Question and Answers


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 177099 by mokys last updated on 30/Sep/22

	Answered by TheHoneyCat last updated on 04/Oct/22

	$taking the \log of your product$ $L = \underset{n = 0}{\sum^{\infty}} x^{n} (\ln (1 + x^{n + 1}) - \ln (1 + x^{n}))$ $= \frac{1}{x} \underset{n = 0}{\sum^{\infty}} x^{n + 1} \ln (1 + x^{n + 1}) - \underset{n = 0}{\sum^{\infty}} \ln (1 + x^{n})$ $= \frac{1}{x} (\underset{n = 0}{\sum^{\infty}} x^{n} \ln (1 + x^{n}) - \ln 2) - \underset{n = 0}{\sum^{\infty}} \ln (1 + x^{n})$ $= \frac{- \ln 2}{x} + (\frac{1}{x} - 1) \underset{n = 0}{\sum^{\infty}} x^{n} \ln (1 + x^{n})$ $so now we just need to take care of that series .$ $0 < \ln (1 + x^{n}) < \ln 2$ $so$ $0 > (\frac{1}{x} - 1) Σ \dots > \frac{1 - x}{x} \frac{1}{x - 1} \ln 2 = \frac{- \ln 2}{x}$ $so this limit is indeed wel defined .$ $so going back to the exponential the limit is in$ $[1 / 4, 1 / 2]$ $calling l the excat value of the series, your$ $result is 1 / 2^{l}$ $I gave up getting the exact result .$ $But {it}^{'} s a classic one . You can find it in$ $calculus books .$ $I fear it has no explicit value thought .$ $If my memory is correct the result is$ $1 / 2^{α} where alpha is a famous constant$ $that appears in the asympotic study of \log$ $but my meroy is rarely correct .$ $; -)$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum