Question Number 25300 by SAGARSTARK last updated on 07/Dec/17
Commented by prakash jain last updated on 08/Dec/17
$$\mathrm{You}\:\mathrm{may}\:\mathrm{also}\:\mathrm{want}\:\mathrm{to}\:\mathrm{see} \\ $$$$\mathrm{Q2088}.\:\mathrm{Different}\:\mathrm{but}\:\mathrm{similar} \\ $$$$\mathrm{question}. \\ $$
Commented by moxhix last updated on 08/Dec/17
![(tanx)^2 =1 (x=π/4) (tanx)^2 <1 (0≤x<π/4) ∴lim_(n→∞) (tanx)^(2n) = { ((1 (x=π/4))),((0 (0≤x<π/4))) :} lim_(n→∞) a_n =∫_0 ^(π/4) lim_(n→∞) (tanx)^(2n) dx (∵tanx∈uniformly continuous) =∫_0 ^(π/4) 0dx=0 a_n =∫_0 ^(π/4) tan^(2n) x dx =∫_0 ^(π/4) tan^(2n−2) x((1/(cos^2 x))−1) dx =∫_0 ^(π/4) tan^(2n−2) x((1/(cos^2 x))) dx− a_(n−1) t=tanx, dt=(1/(cos^2 x))dx, x:[0,π/4]⇒t:[0,1] =∫_0 ^1 t^(2n−2) dt−a_(n−1) a_n =(1/(2n−1))−a_(n−1) ∴(1/(2n−1))=a_n +a_(n−1) a_1 =∫_0 ^(π/4) tan^2 x dx=[tanx−x]_0 ^(π/4) =1−(π/4) Σ_(k=1) ^n (((−1)^(k−1) )/(2k−1))=1+Σ_(k=2) ^n (−1)^(k−1) (a_k +a_(k−1) ) =1+{−(a_2 +a_1 )+(a_3 +a_2 )−...+(−1)^(n−1) (a_n +a_(n−1) )[ =1+(−a_1 +(−1)^(n−1) a_n ) →_((n→∞)) 1−a_1 +0=(π/4)](Q25330.png)
$$\left({tanx}\right)^{\mathrm{2}} =\mathrm{1}\:\left({x}=\pi/\mathrm{4}\right) \\ $$$$\left({tanx}\right)^{\mathrm{2}} <\mathrm{1}\:\left(\mathrm{0}\leqslant{x}<\pi/\mathrm{4}\right) \\ $$$$\therefore\underset{{n}\rightarrow\infty} {{lim}}\left({tanx}\right)^{\mathrm{2}{n}} =\begin{cases}{\mathrm{1}\:\left({x}=\pi/\mathrm{4}\right)}\\{\mathrm{0}\:\left(\mathrm{0}\leqslant{x}<\pi/\mathrm{4}\right)}\end{cases} \\ $$$$\underset{{n}\rightarrow\infty} {{lim}a}_{{n}} =\int_{\mathrm{0}} ^{\pi/\mathrm{4}} \underset{{n}\rightarrow\infty} {{lim}}\left({tanx}\right)^{\mathrm{2}{n}} {dx}\:\:\left(\because{tanx}\in{uniformly}\:{continuous}\right) \\ $$$$\:\:\:\:=\int_{\mathrm{0}} ^{\pi/\mathrm{4}} \mathrm{0}{dx}=\mathrm{0} \\ $$$$ \\ $$$${a}_{{n}} =\int_{\mathrm{0}} ^{\pi/\mathrm{4}} {tan}^{\mathrm{2}{n}} {x}\:{dx} \\ $$$$\:\:\:\:\:=\int_{\mathrm{0}} ^{\pi/\mathrm{4}} {tan}^{\mathrm{2}{n}−\mathrm{2}} {x}\left(\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {x}}−\mathrm{1}\right)\:{dx} \\ $$$$\:\:\:\:\:=\int_{\mathrm{0}} ^{\pi/\mathrm{4}} {tan}^{\mathrm{2}{n}−\mathrm{2}} {x}\left(\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {x}}\right)\:{dx}−\:{a}_{{n}−\mathrm{1}} \\ $$$${t}={tanx},\:{dt}=\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {x}}{dx},\:{x}:\left[\mathrm{0},\pi/\mathrm{4}\right]\Rightarrow{t}:\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\:\:\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\mathrm{2}{n}−\mathrm{2}} {dt}−{a}_{{n}−\mathrm{1}} \\ $$$${a}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}−{a}_{{n}−\mathrm{1}} \\ $$$$\therefore\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}={a}_{{n}} +{a}_{{n}−\mathrm{1}} \\ $$$$ \\ $$$${a}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\pi/\mathrm{4}} {tan}^{\mathrm{2}} {x}\:{dx}=\left[{tanx}−{x}\right]_{\mathrm{0}} ^{\pi/\mathrm{4}} =\mathrm{1}−\frac{\pi}{\mathrm{4}} \\ $$$$ \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{\mathrm{2}{k}−\mathrm{1}}=\mathrm{1}+\underset{{k}=\mathrm{2}} {\overset{{n}} {\sum}}\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({a}_{{k}} +{a}_{{k}−\mathrm{1}} \right) \\ $$$$\:\:\:\:=\mathrm{1}+\left\{−\left({a}_{\mathrm{2}} +{a}_{\mathrm{1}} \right)+\left({a}_{\mathrm{3}} +{a}_{\mathrm{2}} \right)−...+\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({a}_{{n}} +{a}_{{n}−\mathrm{1}} \right)\left[\right.\right. \\ $$$$\:\:\:\:=\mathrm{1}+\left(−{a}_{\mathrm{1}} +\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {a}_{{n}} \right) \\ $$$$\underset{\left({n}\rightarrow\infty\right)} {\rightarrow}\mathrm{1}−{a}_{\mathrm{1}} +\mathrm{0}=\frac{\pi}{\mathrm{4}} \\ $$