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

.......Advanced ... ... ... Calculus....... simplify ::: Ω_n =Σ_(k=1) ^(2n+1) log(1+tan(((kπ)/(4(2n+1))))) moreover , find the value of:: Ω= lim_(n→∞) (Ω_n /n) =???

$$\:.......\mathscr{A}{dvanced}\:...\:\:...\:\:...\:\mathscr{C}{alculus}....... \\ $$$$\:{simplify}\:::: \\ $$$$\:\Omega_{{n}} =\underset{{k}=\mathrm{1}} {\overset{\mathrm{2}{n}+\mathrm{1}} {\sum}}{log}\left(\mathrm{1}+{tan}\left(\frac{{k}\pi}{\mathrm{4}\left(\mathrm{2}{n}+\mathrm{1}\right)}\right)\right) \\ $$$$\:{moreover}\:,\:\:\:\:{find}\:{the}\:{value}\:{of}:: \\ $$$$\Omega=\:{lim}_{{n}\rightarrow\infty} \frac{\Omega_{{n}} }{{n}}\:=??? \\ $$

Answered by mindispower last updated on 02/Apr/21

=Σ_(k=0) ^(2n+1) log(1+tg(((kπ)/(4(2n+1))))=Σ_(k=0) ^(2n+1) log(1+tg((π/4)−((kπ)/(4(2n+1))))) ∴ln(1)=0,Σ_(l=m) ^p f(l)=Σf(p+m−l) =Σ_(k=0) ^(2n+1) log(1+((1−tg(((kπ)/(4(2n+1)))))/(1+tg(((kπ)/(4(2n+1))))))) =Σ_(k=0) ^(2n+1) log((2/(1+tg(((kπ)/(4(2n+1)))))))=Σ_(k=0) ^(2n+1) log(2)−^ Ω_n ⇒2Ω_n =(2n+2)log(2)⇒Ω_n =(n+1)log(2) (Ω_n /n)=(1+(1/n))log(2)⇒lim_(n→∞) (Ω_n /n)=log(2)

$$=\underset{{k}=\mathrm{0}} {\overset{\mathrm{2}{n}+\mathrm{1}} {\sum}}{log}\left(\mathrm{1}+{tg}\left(\frac{{k}\pi}{\mathrm{4}\left(\mathrm{2}{n}+\mathrm{1}\right)}\right)=\underset{{k}=\mathrm{0}} {\overset{\mathrm{2}{n}+\mathrm{1}} {\sum}}{log}\left(\mathrm{1}+{tg}\left(\frac{\pi}{\mathrm{4}}−\frac{{k}\pi}{\mathrm{4}\left(\mathrm{2}{n}+\mathrm{1}\right)}\right)\right)\right. \\ $$$$\therefore{ln}\left(\mathrm{1}\right)=\mathrm{0},\underset{{l}={m}} {\overset{{p}} {\sum}}{f}\left({l}\right)=\Sigma{f}\left({p}+{m}−{l}\right) \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{\mathrm{2}{n}+\mathrm{1}} {\sum}}{log}\left(\mathrm{1}+\frac{\mathrm{1}−{tg}\left(\frac{{k}\pi}{\mathrm{4}\left(\mathrm{2}{n}+\mathrm{1}\right)}\right)}{\mathrm{1}+{tg}\left(\frac{{k}\pi}{\mathrm{4}\left(\mathrm{2}{n}+\mathrm{1}\right)}\right)}\right) \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{\mathrm{2}{n}+\mathrm{1}} {\sum}}{log}\left(\frac{\mathrm{2}}{\mathrm{1}+{tg}\left(\frac{{k}\pi}{\mathrm{4}\left(\mathrm{2}{n}+\mathrm{1}\right)}\right)}\right)=\underset{{k}=\mathrm{0}} {\overset{\mathrm{2}{n}+\mathrm{1}} {\sum}}{log}\left(\mathrm{2}\right)−^{} \Omega_{{n}} \\ $$$$\Rightarrow\mathrm{2}\Omega_{{n}} =\left(\mathrm{2}{n}+\mathrm{2}\right){log}\left(\mathrm{2}\right)\Rightarrow\Omega_{{n}} =\left({n}+\mathrm{1}\right){log}\left(\mathrm{2}\right) \\ $$$$\frac{\Omega_{{n}} }{{n}}=\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right){log}\left(\mathrm{2}\right)\Rightarrow\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\Omega_{{n}} }{{n}}={log}\left(\mathrm{2}\right) \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 02/Apr/21

very nice mr power grateful...

$$\:{very}\:{nice}\:{mr}\:{power} \\ $$$$\:{grateful}... \\ $$

Answered by mnjuly1970 last updated on 02/Apr/21

Ω_n =Σ_(k=0) ^n log(1+tan(((kπ)/(4(2n+1)))))+Σ_(k=n+1) ^(2n+1) log(1+tan(((kπ)/(4(2n+1))))) =Σ_(k=0) ^n log(1+tan(((kπ)/(4(2n+1)))))+Σ_(k=0) ^n log(1+tan((((2n+1−k)π)/(4(2n+1))))) =Σ_(k=0) ^n log(1+tan(((kπ)/(4(2n+1)))))+Σ_(k=0) ^n log(1+((1−tan(((kπ)/(4(2n+1))))/(1+tan(((kπ)/(4(2n+1))))))) =(n+1)log(2) ....✓ ... (Ω_n /n)=((n+1)/n)log(2)⇒ lim_(n→∞) (Ω_n /n)=log(2) .....Ω=log(2).....✓ hint:Σ_(r=k) ^n a(r)=a(k)+a(k+1)+...+a(n) =Σ_(r=0) ^(n−k) a(n−r) ✓

$$\:\:\Omega_{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{log}\left(\mathrm{1}+{tan}\left(\frac{{k}\pi}{\mathrm{4}\left(\mathrm{2}{n}+\mathrm{1}\right)}\right)\right)+\underset{{k}={n}+\mathrm{1}} {\overset{\mathrm{2}{n}+\mathrm{1}} {\sum}}{log}\left(\mathrm{1}+{tan}\left(\frac{{k}\pi}{\mathrm{4}\left(\mathrm{2}{n}+\mathrm{1}\right)}\right)\right) \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{log}\left(\mathrm{1}+{tan}\left(\frac{{k}\pi}{\mathrm{4}\left(\mathrm{2}{n}+\mathrm{1}\right)}\right)\right)+\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{log}\left(\mathrm{1}+{tan}\left(\frac{\left(\mathrm{2}{n}+\mathrm{1}−{k}\right)\pi}{\mathrm{4}\left(\mathrm{2}{n}+\mathrm{1}\right)}\right)\right) \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{log}\left(\mathrm{1}+{tan}\left(\frac{{k}\pi}{\mathrm{4}\left(\mathrm{2}{n}+\mathrm{1}\right)}\right)\right)+\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{log}\left(\mathrm{1}+\frac{\mathrm{1}−{tan}\left(\frac{{k}\pi}{\mathrm{4}\left(\mathrm{2}{n}+\mathrm{1}\right.}\right)}{\mathrm{1}+{tan}\left(\frac{{k}\pi}{\mathrm{4}\left(\mathrm{2}{n}+\mathrm{1}\right)}\right)}\right) \\ $$$$=\left({n}+\mathrm{1}\right){log}\left(\mathrm{2}\right)\:....\checkmark \\ $$$$\:...\:\frac{\Omega_{{n}} }{{n}}=\frac{{n}+\mathrm{1}}{{n}}{log}\left(\mathrm{2}\right)\Rightarrow\:{lim}_{{n}\rightarrow\infty} \frac{\Omega_{{n}} }{{n}}={log}\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:.....\Omega={log}\left(\mathrm{2}\right).....\checkmark \\ $$$$\:\:\:{hint}:\underset{{r}={k}} {\overset{{n}} {\sum}}{a}\left({r}\right)={a}\left({k}\right)+{a}\left({k}+\mathrm{1}\right)+...+{a}\left({n}\right) \\ $$$$=\underset{{r}=\mathrm{0}} {\overset{{n}−{k}} {\sum}}{a}\left({n}−{r}\right)\:\:\checkmark \\ $$

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