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Question Number 25300 by SAGARSTARK last updated on 07/Dec/17

Commented by prakash jain last updated on 08/Dec/17

$$\mathrm{You}\mathrm{may}\mathrm{also}\mathrm{want}\mathrm{to}\mathrm{see}$$$$\mathrm{Q}2088.\mathrm{Different}\mathrm{but}\mathrm{similar}$$$$\mathrm{question}.$$

Commented by moxhix last updated on 08/Dec/17

$${\left(tanx\right)}^2=1(x=\pi /4)$$$${\left(tanx\right)}^2<1(0⩽x<\pi /4)$$$$\therefore \underset{n\to \mathrm{\infty }}{lim}{\left(tanx\right)}^{2n}=\{\begin{array}{l}1(x=\pi /4)\\ 0(0⩽x<\pi /4)\end{array}$$$$\text{Double subscripts: use braces to clarify}$$$$={\int }_0^{\pi /4}0dx=0$$$$$$$$a_n={\int }_0^{\pi /4}{tan}^{2n}xdx$$$$={\int }_0^{\pi /4}{tan}^{2n-2}x(\frac{1}{{cos}^{2}x}-1)dx$$$$={\int }_0^{\pi /4}{tan}^{2n-2}x\left(\frac{1}{{cos}^{2}x}\right)dx-a_{n-1}$$$$t=tanx,dt=\frac{1}{{cos}^{2}x}dx,x:[0,\pi /4]\Rightarrow t:[0,1]$$$$={\int }_0^1{t}^{2n-2}dt-a_{n-1}$$$$a_n=\frac{1}{2n-1}-a_{n-1}$$$$\therefore \frac{1}{2n-1}=a_n+a_{n-1}$$$$$$$$a_1={\int }_0^{\pi /4}{tan}^{2}xdx={[tanx-x]}_0^{\pi /4}=1-\frac{\pi }{4}$$$$$$$$\underset{k=1}{\sum ^n}\frac{{(-1)}^{k-1}}{2k-1}=1+\underset{k=2}{\sum ^n}{(-1)}^{k-1}(a_k+a_{k-1})$$$$=1+\{-(a_2+a_1)+(a_3+a_2)-...+{(-1)}^{n-1}(a_n+a_{n-1})[$$$$=1+(-a_1+{(-1)}^{n-1}{a}_n)$$$$\underset{(n\to \mathrm{\infty })}{\to }1-a_1+0=\frac{\pi }{4}$$

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