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Question Number 55906 by gunawan last updated on 06/Mar/19

$$\underset{n\to \mathrm{\infty }}{\lim }\stackrel{n}{\sum _{k=1}}\frac{8n^2}{n^4+1}=..$$

Answered by tanmay.chaudhury50@gmail.com last updated on 06/Mar/19

$$T_n=\frac{8n^2}{1+n^4}=\frac{4\times 2n^2}{1+{\left(n^2\right)}^2}=\frac{4\times \left(2tann\theta \right)}{1+{\left(tann\theta \right)}^2}=4sin\left(2n\theta \right)$$$$T_1=4sin\left(2\times 1\times \theta \right)\to (here{n}^2=tann\theta ]$$$$T_2=4sin\left(2\times 2\times \theta \right)$$$$...$$$$...$$$$now$$$$S=4[sin2\theta +sin4\theta +sin6\theta +...+sin2n\theta ]$$$$2sin2\theta sin\theta =cos\theta -cos3\theta$$$$2sin4\theta sin\theta =cos3\theta -cos5\theta$$$$2sin6\theta sin\theta =cos5\theta -cos7\theta$$$$....$$$$....$$$$2sin2n\theta sin\theta =cos(2n-1)\theta -cos(2n+1)\theta$$$$addthem$$$$2sin\theta [sin2\theta +sin4\theta +...+sin2n\theta ]=cos\theta -cos(2n+1)\theta$$$$2sin\theta \times \frac{S}{4}=2sin(n+1)\theta sinn\theta$$$$S=\frac{4sin(n+1)\theta sin\left(n\theta \right)}{sin\theta }$$$$[$$$$nowtann\theta =n^2\to sinn\theta =\frac{n^2}{\sqrt{1+n^4}}$$$$cosn\theta =\frac{1}{\sqrt{1+n^4}}$$$$sin\theta =\frac{1}{\sqrt{2}}cos\theta =\frac{1}{\sqrt{2}}$$$$S=\frac{4[sinn\theta cos\theta +cosn\theta sin\theta ]sinn\theta }{sin\theta }$$$$S=\frac{4[\frac{n^2}{\sqrt{1+n^4}}\times \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{1+n^4}}\times \frac{1}{\sqrt{2}}]\times \frac{n^2}{\sqrt{1+n^4}}}{\frac{1}{\sqrt{2}}}$$$$S=4[\frac{n^2}{n^2\sqrt{\frac{1}{n^4}+1}}+\frac{1}{\sqrt{1+n^4}}]\times \frac{n^2}{n^2\sqrt{\frac{1}{n^4}+1}}$$$$$$$$whenn\to \mathrm{\infty }$$$$S=4[\frac{1}{\sqrt{0+1}}+0]\times \frac{1}{\sqrt{0+1}}$$$$soansweris4$$$$plscheckisitcorrect...ihavedoubtalso$$$$becos{T}_{1}itselfis4...$$$$sohowScanbe4...$$$$plsplscheckthestepsmistakeifany...$$

Commented by mr W last updated on 06/Mar/19

$$logicisnotcorrectsir.$$$$youtake\theta asconstant,butwhenyou$$$$assume{n}^2=\tan \left(n\theta \right)youmean\theta$$$$isnotconstant,since\theta =\frac{{\tan }^{-1}{n}^2}{n},i.e.$$$$\theta isdependentfromn.$$$$thereforeyoucannotassume$$$$n^2=\tan \left(n\theta \right)$$

Commented by tanmay.chaudhury50@gmail.com last updated on 06/Mar/19

$$thankyousir...$$

Commented by mr W last updated on 06/Mar/19

$$doesithelp:$$$$T_n=\frac{8}{n^2+\frac{1}{n^2}}=\frac{8}{{(n+\frac{1}{n})}^2-2}=2\sqrt{2}(\frac{1}{n+\frac{1}{n}-\sqrt{2}}-\frac{1}{n+\frac{1}{n}+\sqrt{2}})$$

Commented by tanmay.chaudhury50@gmail.com last updated on 06/Mar/19

$$sirletmeuseyourhintpls...$$

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