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Question Number 218850 by hardmath last updated on 16/Apr/25

Answered by MrGaster last updated on 17/Apr/25

$$T_n=\frac{1}{\ln \left(\underset{r=1}{\prod ^n}\frac{r^2}{1+r^2}\right)}=\frac{1}{\underset{r=1}{\sum ^n}\ln \left(\frac{r^2}{1+r^2}\right)}$$$$=\frac{1}{\underset{r=1}{\sum ^n}(2\ln r-\ln (1+r^2))}$$$$=\frac{1}{2\underset{r=1}{\sum ^n}\ln r-\underset{r=1}{\sum ^n}\ln (1+r^2)}$$$$=\frac{1}{2\ln (n!)-\underset{r=1}{\sum ^n}\ln (1+r^2)}$$$$S=\ln (1+\frac{1}{2}+\frac{1}{3}+\dots +\frac{1}{100})$$$$=\ln \left(\frac{100!}{\underset{r=1}{\prod ^{100}}r}\right)$$$$=\ln (100!)-\underset{r=1}{\sum ^{100}}\ln r$$$$T_n>\frac{4n}{{(n-1)}^2-n^2\ln 4}$$$$\frac{1}{2\ln (n!)-\underset{r=1}{\sum ^r}\ln (1+r^2)}>\frac{4n}{{(n-1)}^2-n^2\ln 4}$$$$2\ln (n!)-\underset{r=1}{\sum ^n}\ln (1+r^2)<\frac{{(n-1)}^2-n^2\ln 4}{4n}$$$$S<\frac{100}{99}\ln 50$$$$\ln \left(\frac{100!}{\underset{r=1}{\prod ^{100}}r}\right)<\frac{100}{99}\ln 50$$$$\ln (100!)-\underset{r=1}{\sum ^{100}}\ln r<\frac{100}{99}\ln 50$$$$T_n<\frac{2}{n(1-\ln 2)-{\cos }^{-1}\left(\frac{2n}{1+n^2}\right)}$$$$\frac{1}{2\ln (n!)-\underset{r=1}{\sum ^n}\ln (1+r^2)}<\frac{2}{n(1-\ln 2)-{\cos }^{-1}\left(\frac{2n}{1+n^2}\right)}$$$$2\ln (n!)-\underset{r=1}{\sum ^n}\ln (1+r^2)>\frac{n(1-\ln 2)-{\cos }^{-1}\left(\frac{2n}{1+n^2}\right)}{2}$$$$S+T_n>\frac{4n+100}{\frac{n}{100}+n\ln 4}$$$$\ln \left(\frac{100!}{\underset{r=1}{\prod ^{100}}r}\right)+\frac{1}{2\ln (n!)-\underset{r=1}{\sum ^n}\ln (1+r^2)}>\frac{4n+100}{\frac{n}{100}+n\ln 4}$$$$\ln (100!)-\underset{r=1}{\sum ^{100}}\ln r+\frac{1}{2\ln (n!)-\underset{r=1}{\sum ^n}\ln (1+r^2)}>\frac{4n+100}{\frac{n}{100}+n\ln 4}$$$$\mathbf{A}$$

Commented by hardmath last updated on 17/Apr/25

$$\mathrm{cool}\mathrm{my}\mathrm{dear}\mathrm{professor},\mathrm{thank}\mathrm{you}\mathrm{very}\mathrm{much}$$

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