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Question Number 134227 by Raxreedoroid last updated on 01/Mar/21

I  struck upon this  Σ_(n=0) ^∞ 16^n =((−1)/(15))  ∫_0 ^( π) cos^4  x sin x dx=((−1)/5)[cos^5  x]_0 ^π =0.4  in another way  I=∫^  cos^4  x sin x dx  I^  =cos^4  (x) (−cos x)−∫4cos^3  x(−sin x)(−cos x)dx  I^  =cos^5  x−4∫cos^4  x sin xdx  I^  =cos^5  x−4I  I=−cos^5  xΣ_(n=0) ^∞ (−4)^n +C  [I]_0 ^π =[(−(−1)Σ_(n=0) ^∞ (−4)^n )−(−Σ_(n=0) ^∞ (−4)^n )]  =2Σ_(n=0) ^∞ (−4)^n =2(Σ_(n=0) ^∞ 16^n −4Σ_(n=0) ^∞ 16^n )  =2(−3Σ_(n=0) ^∞ 16^n )=−6Σ_(n=0) ^∞ 16^n   −6Σ_(n=0) ^∞ 16^n =0.4⇒Σ_(n=0) ^∞ 16^n =((−0.4)/6)=((−1)/(15))  did I do something wrong?

$$\mathrm{I}\:\:\mathrm{struck}\:\mathrm{upon}\:\mathrm{this} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{16}^{{n}} =\frac{−\mathrm{1}}{\mathrm{15}} \\ $$$$\int_{\mathrm{0}} ^{\:\pi} \mathrm{cos}^{\mathrm{4}} \:{x}\:\mathrm{sin}\:{x}\:{dx}=\frac{−\mathrm{1}}{\mathrm{5}}\left[\mathrm{cos}^{\mathrm{5}} \:{x}\right]_{\mathrm{0}} ^{\pi} =\mathrm{0}.\mathrm{4} \\ $$$$\mathrm{in}\:\mathrm{another}\:\mathrm{way} \\ $$$${I}=\int^{\:} \mathrm{cos}^{\mathrm{4}} \:{x}\:\mathrm{sin}\:{x}\:{dx} \\ $$$${I}^{\:} =\mathrm{cos}^{\mathrm{4}} \:\left({x}\right)\:\left(−\mathrm{cos}\:{x}\right)−\int\mathrm{4cos}^{\mathrm{3}} \:{x}\left(−\mathrm{sin}\:{x}\right)\left(−\mathrm{cos}\:{x}\right){dx} \\ $$$${I}^{\:} =\mathrm{cos}^{\mathrm{5}} \:{x}−\mathrm{4}\int\mathrm{cos}^{\mathrm{4}} \:{x}\:\mathrm{sin}\:{xdx} \\ $$$${I}^{\:} =\mathrm{cos}^{\mathrm{5}} \:{x}−\mathrm{4}{I} \\ $$$${I}=−\mathrm{cos}^{\mathrm{5}} \:{x}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{4}\right)^{{n}} +{C} \\ $$$$\left[{I}\right]_{\mathrm{0}} ^{\pi} =\left[\left(−\left(−\mathrm{1}\right)\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{4}\right)^{{n}} \right)−\left(−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{4}\right)^{{n}} \right)\right] \\ $$$$=\mathrm{2}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{4}\right)^{{n}} =\mathrm{2}\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{16}^{{n}} −\mathrm{4}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{16}^{{n}} \right) \\ $$$$=\mathrm{2}\left(−\mathrm{3}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{16}^{{n}} \right)=−\mathrm{6}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{16}^{{n}} \\ $$$$−\mathrm{6}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{16}^{{n}} =\mathrm{0}.\mathrm{4}\Rightarrow\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{16}^{{n}} =\frac{−\mathrm{0}.\mathrm{4}}{\mathrm{6}}=\frac{−\mathrm{1}}{\mathrm{15}} \\ $$$$\mathrm{did}\:\mathrm{I}\:\mathrm{do}\:\mathrm{something}\:\mathrm{wrong}? \\ $$

Commented by Dwaipayan Shikari last updated on 01/Mar/21

Σ_(n=0) ^∞ 16^n → Diverges→∞  Σ_(n=0) ^∞ (1/(16^n ))=(1/(15))

$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{16}^{{n}} \rightarrow\:{Diverges}\rightarrow\infty \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{16}^{{n}} }=\frac{\mathrm{1}}{\mathrm{15}} \\ $$

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