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Question Number 137397 by mnjuly1970 last updated on 02/Apr/21

$$.......\mathcal{A}dvanced.........\mathcal{C}alculus.......$$$$simplify:::$$$${\mathrm{\Omega }}_n=\underset{k=1}{\sum ^{2n+1}}log(1+tan\left(\frac{k\pi }{4(2n+1)}\right))$$$$moreover,findthevalueof::$$$$\mathrm{\Omega }={lim}_{n\to \mathrm{\infty }}\frac{{\mathrm{\Omega }}_n}{n}=???$$

Answered by mindispower last updated on 02/Apr/21

$$=\underset{k=0}{\sum ^{2n+1}}log(1+tg\left(\frac{k\pi }{4(2n+1)}\right)=\underset{k=0}{\sum ^{2n+1}}log(1+tg(\frac{\pi }{4}-\frac{k\pi }{4(2n+1)}))$$$$\therefore ln\left(1\right)=0,\underset{l=m}{\sum ^p}f\left(l\right)=\mathrm{\Sigma }f(p+m-l)$$$$=\underset{k=0}{\sum ^{2n+1}}log(1+\frac{1-tg\left(\frac{k\pi }{4(2n+1)}\right)}{1+tg\left(\frac{k\pi }{4(2n+1)}\right)})$$$$=\underset{k=0}{\sum ^{2n+1}}log\left(\frac{2}{1+tg\left(\frac{k\pi }{4(2n+1)}\right)}\right)=\underset{k=0}{\sum ^{2n+1}}log\left(2\right){-}^{}{\mathrm{\Omega }}_n$$$$\Rightarrow 2{\mathrm{\Omega }}_n=(2n+2)log\left(2\right)\Rightarrow {\mathrm{\Omega }}_n=(n+1)log\left(2\right)$$$$\frac{{\mathrm{\Omega }}_n}{n}=(1+\frac{1}{n})log\left(2\right)\Rightarrow \underset{n\to \mathrm{\infty }}{\lim }\frac{{\mathrm{\Omega }}_n}{n}=log\left(2\right)$$$$$$

Commented by mnjuly1970 last updated on 02/Apr/21

$$verynicemrpower$$$$grateful...$$

Answered by mnjuly1970 last updated on 02/Apr/21

$${\mathrm{\Omega }}_n=\underset{k=0}{\sum ^n}log(1+tan\left(\frac{k\pi }{4(2n+1)}\right))+\underset{k=n+1}{\sum ^{2n+1}}log(1+tan\left(\frac{k\pi }{4(2n+1)}\right))$$$$=\underset{k=0}{\sum ^n}log(1+tan\left(\frac{k\pi }{4(2n+1)}\right))+\underset{k=0}{\sum ^n}log(1+tan\left(\frac{(2n+1-k)\pi }{4(2n+1)}\right))$$$$=\underset{k=0}{\sum ^n}log(1+tan\left(\frac{k\pi }{4(2n+1)}\right))+\underset{k=0}{\sum ^n}log(1+\frac{1-tan\left(\frac{k\pi }{4(2n+1}\right)}{1+tan\left(\frac{k\pi }{4(2n+1)}\right)})$$$$=(n+1)log\left(2\right)....✓$$$$...\frac{{\mathrm{\Omega }}_n}{n}=\frac{n+1}{n}log\left(2\right)\Rightarrow {lim}_{n\to \mathrm{\infty }}\frac{{\mathrm{\Omega }}_n}{n}=log\left(2\right)$$$$.....\mathrm{\Omega }=log\left(2\right).....✓$$$$hint:\underset{r=k}{\sum ^n}a\left(r\right)=a\left(k\right)+a(k+1)+...+a\left(n\right)$$$$=\underset{r=0}{\sum ^{n-k}}a(n-r)✓$$

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