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?](https://www.tinkutara.com/question/Q134227.png)
$$\mathrm{I}\:\:\mathrm{struck}\:\mathrm{upon}\:\mathrm{this} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{16}^{{n}} =\frac{−\mathrm{1}}{\mathrm{15}} \\ $$$$\int_{\mathrm{0}} ^{\:\pi} \mathrm{cos}^{\mathrm{4}} \:{x}\:\mathrm{sin}\:{x}\:{dx}=\frac{−\mathrm{1}}{\mathrm{5}}\left[\mathrm{cos}^{\mathrm{5}} \:{x}\right]_{\mathrm{0}} ^{\pi} =\mathrm{0}.\mathrm{4} \\ $$$$\mathrm{in}\:\mathrm{another}\:\mathrm{way} \\ $$$${I}=\int^{\:} \mathrm{cos}^{\mathrm{4}} \:{x}\:\mathrm{sin}\:{x}\:{dx} \\ $$$${I}^{\:} =\mathrm{cos}^{\mathrm{4}} \:\left({x}\right)\:\left(−\mathrm{cos}\:{x}\right)−\int\mathrm{4cos}^{\mathrm{3}} \:{x}\left(−\mathrm{sin}\:{x}\right)\left(−\mathrm{cos}\:{x}\right){dx} \\ $$$${I}^{\:} =\mathrm{cos}^{\mathrm{5}} \:{x}−\mathrm{4}\int\mathrm{cos}^{\mathrm{4}} \:{x}\:\mathrm{sin}\:{xdx} \\ $$$${I}^{\:} =\mathrm{cos}^{\mathrm{5}} \:{x}−\mathrm{4}{I} \\ $$$${I}=−\mathrm{cos}^{\mathrm{5}} \:{x}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{4}\right)^{{n}} +{C} \\ $$$$\left[{I}\right]_{\mathrm{0}} ^{\pi} =\left[\left(−\left(−\mathrm{1}\right)\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{4}\right)^{{n}} \right)−\left(−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{4}\right)^{{n}} \right)\right] \\ $$$$=\mathrm{2}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{4}\right)^{{n}} =\mathrm{2}\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{16}^{{n}} −\mathrm{4}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{16}^{{n}} \right) \\ $$$$=\mathrm{2}\left(−\mathrm{3}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{16}^{{n}} \right)=−\mathrm{6}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{16}^{{n}} \\ $$$$−\mathrm{6}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{16}^{{n}} =\mathrm{0}.\mathrm{4}\Rightarrow\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{16}^{{n}} =\frac{−\mathrm{0}.\mathrm{4}}{\mathrm{6}}=\frac{−\mathrm{1}}{\mathrm{15}} \\ $$$$\mathrm{did}\:\mathrm{I}\:\mathrm{do}\:\mathrm{something}\:\mathrm{wrong}? \\ $$
Commented by Dwaipayan Shikari last updated on 01/Mar/21
$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{16}^{{n}} \rightarrow\:{Diverges}\rightarrow\infty \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{16}^{{n}} }=\frac{\mathrm{1}}{\mathrm{15}} \\ $$