Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 106033 by mohammad17 last updated on 02/Aug/20

Answered by mathmax by abdo last updated on 02/Aug/20

$$\mathrm{let}\mathrm{I}=\int \frac{1-{\mathrm{x}}^2}{{\mathrm{x}}^2(1+{\mathrm{x}}^2)}\mathrm{dx}\mathrm{let}\mathrm{decompose}\mathrm{F}\left(\mathrm{x}\right)=\frac{1-{\mathrm{x}}^2}{{\mathrm{x}}^2(1+{\mathrm{x}}^2)}\Rightarrow$$$$\mathrm{F}\left(\mathrm{x}\right)=(1-{\mathrm{x}}^2)\{\frac{1}{{\mathrm{x}}^2}-\frac{1}{1+{\mathrm{x}}^2}\}=\frac{1-{\mathrm{x}}^2}{{\mathrm{x}}^2}-\frac{1-{\mathrm{x}}^2}{1+{\mathrm{x}}^2}$$$$=\frac{1}{{\mathrm{x}}^2}-1+\frac{{\mathrm{x}}^2-1}{{\mathrm{x}}^2+1}=\frac{1}{{\mathrm{x}}^2}-1+\frac{{\mathrm{x}}^2+1-2}{{\mathrm{x}}^2+1}=\frac{1}{{\mathrm{x}}^2}-\frac{2}{{\mathrm{x}}^2+1}\Rightarrow$$$$\mathrm{I}=\int (\frac{1}{{\mathrm{x}}^2}-\frac{2}{{\mathrm{x}}^2+1})\mathrm{dx}=-\frac{1}{\mathrm{x}}-2\arctan \left(\mathrm{x}\right)+\mathrm{C}$$

Answered by Dwaipayan Shikari last updated on 02/Aug/20

$$f\left(x\right)=cosx$$$$f^{\prime }\left(x\right)=-sinx$$$$f^{″}\left(x\right)=-cosx$$$$f^{‴}\left(x\right)=sinx$$$$f^{\left(iv\right)}\left(x\right)=cosx$$$$f^{\left(v\right)}\left(x\right)=-sinx$$$$f^{\left(vi\right)}\left(x\right)=-cosx$$$$f^{\left(vii\right)}\left(x\right)=sinx$$$$......$$$$f\left(x\right)=f\left(0\right)+f^{\prime }\left(0\right)x+\frac{f^{″}\left(0\right)}{2!}{x}^2+\frac{f^{‴}\left(0\right)}{3!}{x}^3+\frac{f^{\left(iv\right)}\left(0\right)}{4!}{x}^4+\frac{f^{\left(v\right)}}{5!}{x}^5+....$$$$f\left(x\right)=1+0-\frac{x^2}{2!}+0+\frac{x^4}{4!}+...$$$$cosx=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+..$$$$$$

Commented by mohammad17 last updated on 02/Aug/20

$$thankyousir$$

Answered by Dwaipayan Shikari last updated on 02/Aug/20

$$2)\int \frac{d\theta }{1+cos\theta }$$$$=2\int \frac{dt}{1+\frac{1-t^2}{1+t^2}}.\frac{1}{1+t^2}(tan\frac{\theta }{2}=t)$$$$=2\int \frac{dt}{2}=t+C=tan\frac{\theta }{2}+C$$$$$$$$\mathrm{Another}\mathrm{way}$$$$\int \frac{d\theta }{1+cos\theta }=\int \frac{d\theta }{2{cos}^2\frac{\theta }{2}}=\frac{1}{2}\int {sec}^2\frac{\theta }{2}d\theta =tan\frac{\theta }{2}+C$$

Commented by mohammad17 last updated on 02/Aug/20

$$thankyousircanyouhelpmeinnumber\left(4\right)intheintigral$$

Answered by Dwaipayan Shikari last updated on 02/Aug/20

$$3)\int log(1+x^2)dx$$$$I\left(a\right)=\int log(1+{ax}^2)dx$$$${\mathrm{I}}^{\prime }\left(a\right)=\int \frac{x^2}{1+{ax}^2}dx$$$${\mathrm{I}}^{\prime }\left(a\right)=\frac{1}{a}\int \frac{1+{ax}^2}{1+{ax}^2}-\frac{1}{1+{ax}^2}dx$$$${\mathrm{I}}^{\prime }\left(a\right)=\frac{1}{a}\int 1-\frac{1}{a^2}\int \frac{1}{\frac{1}{a}+x^2}dx$$$${\mathrm{I}}^{\prime }\left(a\right)=\frac{x}{a}-\frac{1}{a^2}.\sqrt{a}{tan}^{-1}\left(\sqrt{a}x\right)+C$$$${\mathrm{I}}^{\prime }\left(a\right)=\frac{x}{a}-a^{-\frac{3}{2}}{tan}^{-1}\left(\sqrt{a}x\right)+C$$$$I\left(a\right)=\int \frac{x}{a}-a^{\frac{-3}{2}}{tan}^{-1}\left(\sqrt{a}x\right)+Cda$$$$I\left(a\right)=xlog\left(a\right)-a^{-\frac{3}{2}}{tan}^{-1}\left(u\right)da\{\sqrt{a}x=u,\frac{x}{2\sqrt{a}}=\frac{du}{da}\}$$$$I\left(a\right)=xlog\left(a\right)-\frac{1}{x}\frac{2x}{2a\sqrt{a}}\int {tan}^{-1}\left(u\right)da+Ca$$$$I\left(a\right)=xlog\left(a\right)-\frac{2}{x}\int \frac{1}{a}{tan}^{-1}\left(u\right)du+Ca$$$$I\left(a\right)=xlog\left(a\right)-\frac{2}{x^3}\int u^2{tan}^{-1}udu+Ca$$$$I\left(a\right)=xlog\left(a\right)-\frac{2}{x^3}{tan}^{-1}u.\frac{u^3}{3}+\frac{1}{3}\int \frac{u^3}{1+u^2}+Ca$$$$I\left(a\right)=xlog\left(a\right)-\frac{u^3}{3}.\frac{2}{x^3}{tan}^{-1}\left(\sqrt{a}x\right)+\frac{1}{3}\int \frac{u^3+u}{1+u^2}-\frac{u}{1+u^2}+Ca$$$$I\left(a\right)=xlog\left(a\right)-\frac{2u^3}{3x^3}.{tan}^{-1}\left(\sqrt{a}x\right)+\frac{1}{6}{u}^2-\frac{1}{6}log(1+u^2)+Ca$$$$I\left(a\right)=xlog\left(a\right)-\frac{2a^{\frac{3}{2}}{x}^3}{3x^3}{tan}^{-1}\left(\sqrt{a}x\right)+\frac{1}{6}{ax}^2-\frac{1}{6}log(1+{ax}^2)+Ca$$$$I\left(1\right)=-\frac{2}{3}{tan}^{-1}\left(x\right)+\frac{1}{6}{x}^2-\frac{1}{6}log(1+x^2)+C$$$$\mathrm{another}\mathrm{way}$$$$\int log(1+x^2)=xlog(1+x^2)-\int \frac{2x^2}{1+x^2}$$$$=xlog(1+x^2)-2x+2{tan}^{-1}x+C$$

Commented by mohammad17 last updated on 02/Aug/20

$$thankyousir$$

Answered by Dwaipayan Shikari last updated on 02/Aug/20

$$4)\int \frac{1-x^2}{x^2(1+x^2)}dx$$$$\int \frac{1}{x^2(1+x^2)}-\frac{1}{1+x^2}$$$$=\int \frac{1}{x^2}-\frac{1}{1+x^2}-{tan}^{-1}x$$$$=-\frac{1}{x}-2{tan}^{-1}x+C$$

Answered by bemath last updated on 02/Aug/20

$$\left(4\right)\int \frac{(1-{\mathrm{x}}^2)}{{\mathrm{x}}^2(1+{\mathrm{x}}^2)}\mathrm{dx}=\int \frac{(1+{\mathrm{x}}^2)-{2\mathrm{x}}^2}{{\mathrm{x}}^2(1+{\mathrm{x}}^2)}\mathrm{dx}$$$$=\int \frac{\mathrm{dx}}{{\mathrm{x}}^2}-2\int \frac{\mathrm{dx}}{1+{\mathrm{x}}^2}$$$$=-\frac{1}{\mathrm{x}}-{2\tan }^{-1}\left(\mathrm{x}\right)+\mathrm{c}$$

Commented by mohammad17 last updated on 02/Aug/20

$$thankyousir$$

Answered by mathmax by abdo last updated on 02/Aug/20

$${\mathrm{q}}_1)\mathrm{nature}\mathrm{of}\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{(1-\frac{2\mathrm{a}}{\mathrm{n}})}^{\mathrm{n}}\mathrm{we}\mathrm{have}{(1-\frac{2\mathrm{a}}{\mathrm{n}})}^{\mathrm{n}}={\mathrm{e}}^{\mathrm{nln}(1-\frac{2\mathrm{a}}{\mathrm{n}})}$$$${\ln }^{{}^{\prime }}(1-\mathrm{u})=\frac{-1}{1-\mathrm{u}}=-\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{u}}^{\mathrm{n}}\Rightarrow \ln (1-\mathrm{u})=-\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{\mathrm{u}}^{\mathrm{n}+1}}{\mathrm{n}+1}$$$$=-\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{u}}^{\mathrm{n}}}{\mathrm{n}}\Rightarrow \ln (1-\mathrm{u})=-\mathrm{u}-\frac{{\mathrm{u}}^2}{2}+\mathrm{o}\left({\mathrm{u}}^3\right)\Rightarrow$$$$\ln (1-\frac{2\mathrm{a}}{\mathrm{n}})=-\frac{2\mathrm{a}}{\mathrm{n}}-\frac{{4\mathrm{a}}^2}{{\mathrm{n}}^2}+\mathrm{o}\left(\frac{1}{{\mathrm{n}}^3}\right)\Rightarrow \mathrm{nln}(1-\frac{2\mathrm{a}}{\mathrm{n}})=-2\mathrm{a}-\frac{{4\mathrm{a}}^2}{\mathrm{n}}+\mathrm{o}\left(\frac{1}{{\mathrm{n}}^2}\right)\Rightarrow$$$${\mathrm{e}}^{\mathrm{nln}(1-\frac{2\mathrm{a}}{\mathrm{n}})}={\mathrm{e}}^{-2\mathrm{a}-\frac{{4\mathrm{a}}^2}{\mathrm{n}}+\mathrm{o}\left(\frac{1}{{\mathrm{n}}^2}\right)}\sim {\mathrm{e}}^{-2\mathrm{a}}{\mathrm{e}}^{-\frac{{4\mathrm{a}}^2}{\mathrm{n}}}\sim {\mathrm{e}}^{-2\mathrm{a}}\{1-\frac{{4\mathrm{a}}^2}{\mathrm{n}}\}$$$$\mathrm{\Sigma }\frac{{4\mathrm{a}}^2}{\mathrm{n}}\mathrm{diverges}\mathrm{if}\mathrm{a}\ne 0\Rightarrow \mathrm{\Sigma }{\mathrm{u}}_{\mathrm{n}}\mathrm{diverges}$$$$\mathrm{\Sigma }{\mathrm{u}}_{\mathrm{n}}\mathrm{diverges}\mathrm{if}\mathrm{a}=0$$

Commented by mohammad17 last updated on 02/Aug/20

$$thankyousir$$

Commented by abdomathmax last updated on 02/Aug/20

$$\mathrm{you}\mathrm{are}\mathrm{welcome}$$

Answered by bemath last updated on 02/Aug/20

$$\mathrm{Q}3.\left(2\right)\int \frac{\mathrm{d}\theta }{1+\cos \theta }=\int \frac{1-\cos \theta }{\sin {}^2\theta }\mathrm{d}\theta$$$$=\int ({\csc }^2\theta -\cot \theta \csc \theta )\mathrm{d}\theta$$$$=-\cot \theta +\csc \theta +\mathrm{c}$$

Commented by mohammad17 last updated on 02/Aug/20

$$thankyousir$$

Answered by mathmax by abdo last updated on 02/Aug/20

$$\mathrm{nature}\mathrm{of}\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{(1+\mathrm{n})}^{\frac{1}{\mathrm{lnn}}}\mathrm{let}{\mathrm{u}}_{\mathrm{n}}={(1+\mathrm{n})}^{\frac{1}{\mathrm{lnn}}}$$$$\Rightarrow {\mathrm{u}}_{\mathrm{n}}={\mathrm{e}}^{\frac{1}{\ln \left(\mathrm{n}\right)}\ln (1+\mathrm{n})}={\mathrm{e}}^{\frac{\ln \left(\mathrm{n}\right)+\ln (1+\frac{1}{\mathrm{n}})}{\mathrm{lnn}}}={\mathrm{e}}^{1+\frac{\ln (1+\frac{1}{\mathrm{n}})}{\ln \left(\mathrm{n}\right)}}$$$$\sim \mathrm{e}\times {\mathrm{e}}^{\frac{1}{\mathrm{nln}\left(\mathrm{n}\right)}}\Rightarrow {\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{u}}_{\mathrm{n}}=\mathrm{e}\Rightarrow \mathrm{\Sigma }{\mathrm{u}}_{\mathrm{n}}\mathrm{diverges}$$

Commented by mohammad17 last updated on 02/Aug/20

$$thankyousir$$

Answered by mathmax by abdo last updated on 02/Aug/20

$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{cosx}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=\mathrm{o}}^{\mathrm{\infty }}\frac{{\mathrm{f}}^{\left(\mathrm{n}\right)}\left(0\right)}{\mathrm{n}!}{\mathrm{x}}^{\mathrm{n}}\mathrm{but}{\mathrm{f}}^{\left(\mathrm{n}\right)}\left(\mathrm{x}\right)=\cos (\mathrm{x}+\frac{\mathrm{n}\pi }{2})\Rightarrow$$$${\mathrm{f}}^{\left(\mathrm{n}\right)}\left(\mathrm{n}\right)=\cos \left(\frac{\mathrm{n}\pi }{2}\right)\mathrm{n}=2\mathrm{p}\Rightarrow {\mathrm{f}}^{\left(\mathrm{n}\right)}\left(\mathrm{x}\right)={(-1)}^{\mathrm{p}}$$$$\mathrm{n}=2\mathrm{p}+1\Rightarrow {\mathrm{f}}^{\left(\mathrm{n}\right)}\left(\mathrm{x}\right)=\cos \left(\frac{(2\mathrm{p}+1)\pi }{2}\right)=\cos (\mathrm{p}\pi +\frac{\pi }{2})=0\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}{\mathrm{x}}^{2\mathrm{n}}}{\left(2\mathrm{n}\right)!}=1-\frac{{\mathrm{x}}^2}{2!}+\frac{{\mathrm{x}}^4}{4!}+...$$

Commented by mohammad17 last updated on 02/Aug/20

$$thankyousir$$

Answered by mathmax by abdo last updated on 02/Aug/20

$$\mathrm{S}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}{(\mathrm{x}-4)}^{\mathrm{n}}}{{(\mathrm{n}+1)}^2}\mathrm{let}{\mathrm{U}}_{\mathrm{n}}=\frac{{(-1)}^{\mathrm{n}}{(\mathrm{x}-4)}^{\mathrm{n}}}{{(\mathrm{n}+1)}^2}\mathrm{we}\mathrm{have}$$$$\mid \frac{{\mathrm{U}}_{\mathrm{n}+1}}{{\mathrm{U}}_{\mathrm{n}}}\mid =\mid \frac{{(-1)}^{\mathrm{n}+1}{(\mathrm{x}-4)}^{\mathrm{n}+1}}{{(\mathrm{n}+2)}^2}\times \frac{{(\mathrm{n}+1)}^2}{{(-1)}^{\mathrm{n}}{(\mathrm{x}-4)}^{\mathrm{n}}}\mid$$$$={\left(\frac{\mathrm{n}+1}{\mathrm{n}+2}\right)}^2\mid \mathrm{x}-4\mid \to \mid \mathrm{x}-4\mid (\mathrm{n}\to \mathrm{\infty })\mathrm{so}$$$$\mathrm{if}\mid \mathrm{x}-4\mid <1\mathrm{the}\mathrm{serie}\mathrm{converge}(-1<\mathrm{x}-4<1\Rightarrow 3<\mathrm{x}<5)$$$$\mathrm{if}\mid \mathrm{x}-4\mid >1\mathrm{the}\mathrm{serie}\mathrm{diverges}$$$$\mathrm{x}=4\Rightarrow \mathrm{S}=0$$

Commented by mohammad17 last updated on 02/Aug/20

$$thankyousir$$

Answered by mathmax by abdo last updated on 02/Aug/20

$$\mathrm{convergence}\mathrm{of}\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{(2\mathrm{n}+1){(\mathrm{x}-2)}^{\mathrm{n}}}{{\mathrm{n}}^3}\mathrm{let}{\mathrm{v}}_{\mathrm{n}}=\frac{(2\mathrm{n}+1){(\mathrm{x}-2)}^{\mathrm{n}}}{{\mathrm{n}}^3}$$$$\mathrm{we}\mathrm{have}\mid \frac{{\mathrm{v}}_{\mathrm{n}+1}}{{\mathrm{v}}_{\mathrm{n}}}\mid =\mid \frac{(2\mathrm{n}+3){(\mathrm{x}-2)}^{\mathrm{n}+1}}{{(\mathrm{n}+1)}^3}\times \frac{{\mathrm{n}}^3}{(2\mathrm{n}+1){(\mathrm{x}-2)}^{\mathrm{n}}}\mid$$$$={\left(\frac{\mathrm{n}}{\mathrm{n}+1}\right)}^3\frac{2\mathrm{n}+3}{2\mathrm{n}+1}\times \mid \mathrm{x}-2\mid \to \mid \mathrm{x}-2\mid \mathrm{so}$$$$\mathrm{if}\mid \mathrm{x}-2\mid <1\Rightarrow \mathrm{\Sigma }{\mathrm{v}}_{\mathrm{n}}\mathrm{converge}(-1<\mathrm{x}-2<1\Rightarrow 1<\mathrm{x}<3\Rightarrow {\mathrm{D}}_{\mathrm{c}}=]1,3\left[\right)$$$$\mathrm{if}\mid \mathrm{x}-2\mid >1\Rightarrow \mathrm{\Sigma }{\mathrm{v}}_{\mathrm{n}}\mathrm{diverges}$$$$\mathrm{x}=2\Rightarrow \mathrm{\Sigma }{\mathrm{v}}_{\mathrm{n}}=0$$$$$$$$$$

Commented by mohammad17 last updated on 02/Aug/20

$$thankyousir$$

Answered by mathmax by abdo last updated on 02/Aug/20

$$\int \frac{\mathrm{d}\theta }{1+\cos \theta }{=}_{\tan \left(\frac{\theta }{2}\right)=\mathrm{t}}\int \frac{2\mathrm{dt}}{(1+{\mathrm{t}}^2)(1+\frac{1-{\mathrm{t}}^2}{1+{\mathrm{t}}^2})}=\int \frac{2\mathrm{dt}}{1+{\mathrm{t}}^2+1-{\mathrm{t}}^2}$$$$=\int \mathrm{dt}=\tan \left(\frac{\theta }{2}\right)+\mathrm{C}$$$$\int \ln (1+{\mathrm{x}}^2)\mathrm{dx}{=}_{\mathrm{by}\mathrm{parts}}\mathrm{xln}(1+{\mathrm{x}}^2)-\int \mathrm{x}.\frac{2\mathrm{x}}{1+{\mathrm{x}}^2}\mathrm{dx}$$$$=\mathrm{xln}(1+{\mathrm{x}}^2)-2\int \frac{{\mathrm{x}}^2+1-1}{1+{\mathrm{x}}^2}\mathrm{dx}$$$$=\mathrm{xln}(1+{\mathrm{x}}^2)-2\mathrm{x}+2\mathrm{arctanx}+\mathrm{C}$$$$$$

Commented by mohammad17 last updated on 02/Aug/20

$$thankyousir$$

Answered by mathmax by abdo last updated on 02/Aug/20

$$\mathrm{A}=\int \frac{\mathrm{cotan}\left(\mathrm{lnx}\right)}{\mathrm{x}}\mathrm{dx}\mathrm{let}\mathrm{lnx}=\mathrm{t}\Rightarrow \mathrm{A}=\int \frac{\mathrm{cotan}\left(\mathrm{t}\right)}{{\mathrm{e}}^{\mathrm{t}}}{\mathrm{e}}^{\mathrm{t}}\mathrm{dt}$$$$=\int \frac{\mathrm{cost}}{\mathrm{sint}}\mathrm{dt}=\ln \mid \sin \left(\mathrm{t}\right)\mid +\mathrm{C}=\ln \mid \sin \left(\mathrm{lnx}\right)\mid +\mathrm{C}$$

Answered by mathmax by abdo last updated on 02/Aug/20

$${\mathrm{q}}_5)\mathrm{for}\mathrm{n}⩾1\frac{1}{{\mathrm{n}}^2\sqrt{\mathrm{n}}}⩽\frac{1}{{\mathrm{n}}^2}\mathrm{but}\mathrm{\Sigma }\frac{1}{{\mathrm{n}}^2}\mathrm{converges}\Rightarrow \mathrm{\Sigma }{\mathrm{u}}_{\mathrm{n}}\mathrm{converges}$$$$\mathrm{we}\mathrm{have}\mathrm{\forall }\mathrm{n}⩾1\frac{{\sin }^2\left(\mathrm{n}\right)}{2^{\mathrm{n}}}⩽\frac{1}{2^{\mathrm{n}}}\mathrm{and}\mathrm{\Sigma }\frac{1}{2^{\mathrm{n}}}\mathrm{converges}\Rightarrow \mathrm{\Sigma }{\mathrm{u}}_{\mathrm{n}}\mathrm{convergs}$$$$\mathrm{for}\mathrm{x}⩾2\mathrm{the}\mathrm{function}\phi \left(\mathrm{x}\right)=\frac{1}{\mathrm{x}\sqrt{\mathrm{lnx}}}\mathrm{is}\mathrm{decreazing}\Rightarrow$$$$\sum _{\mathrm{n}=2}^{\mathrm{\infty }}\frac{1}{\mathrm{n}\sqrt{\mathrm{lnn}}}\mathrm{and}{\int }_2^{\mathrm{\infty }}\frac{\mathrm{dx}}{\mathrm{x}\sqrt{\mathrm{lnx}}}\mathrm{have}\mathrm{tbe}\mathrm{same}\mathrm{nature}$$$$\mathrm{changement}\sqrt{\mathrm{lnx}}=\mathrm{t}\mathrm{give}\mathrm{lnx}={\mathrm{t}}^2\Rightarrow \mathrm{x}={\mathrm{e}}^{{\mathrm{t}}^2}\Rightarrow$$$${\int }_2^{\mathrm{\infty }}\frac{\mathrm{dx}}{\mathrm{x}\sqrt{\mathrm{lnx}}}={\int }_{\sqrt{\ln 2}}^{+\mathrm{\infty }}\frac{2\mathrm{t}{\mathrm{e}}^{{\mathrm{t}}^2}}{{\mathrm{e}}^{{\mathrm{t}}^2}\mathrm{t}}\mathrm{dt}={\int }_{\sqrt{\ln 2}}^{+\mathrm{\infty }}2\mathrm{dt}=+\mathrm{\infty }\Rightarrow \mathrm{this}\mathrm{serie}\mathrm{diverges}..$$

Commented by mohammad17 last updated on 02/Aug/20

$$thankyousir$$

Commented by abdomathmax last updated on 02/Aug/20

$$\mathrm{you}\mathrm{are}\mathrm{welcome}\mathrm{sir}$$

Answered by mathmax by abdo last updated on 02/Aug/20

$$\sum _{\mathrm{n}=5}^{\mathrm{\infty }}{\left(\frac{\mathrm{e}}{\mathrm{n}}\right)}^{\mathrm{n}-1}=\sum _{\mathrm{n}=4}^{\mathrm{\infty }}{\left(\frac{\mathrm{e}}{\mathrm{n}+1}\right)}^{\mathrm{n}}\mathrm{let}{\mathrm{v}}_{\mathrm{n}}={\left(\frac{\mathrm{e}}{\mathrm{n}+1}\right)}^{\mathrm{n}}\mathrm{we}\mathrm{see}{\mathrm{v}}_{\mathrm{n}}>0\Rightarrow$$$$\Rightarrow \frac{{\mathrm{v}}_{\mathrm{n}+1}}{{\mathrm{v}}_{\mathrm{n}}}=\frac{{\mathrm{e}}^{\mathrm{n}+1}}{{(\mathrm{n}+2)}^{\mathrm{n}+1}}\times \frac{{(\mathrm{n}+1)}^{\mathrm{n}}}{{\mathrm{e}}^{\mathrm{n}}}=\mathrm{e}\times \frac{{(\mathrm{n}+1)}^{\mathrm{n}}}{{(\mathrm{n}+2)}^{\mathrm{n}}(\mathrm{n}+2)}$$$$=\frac{\mathrm{e}}{\mathrm{n}+2}{\left(\frac{\mathrm{n}+1}{\mathrm{n}+2}\right)}^{\mathrm{n}}=\frac{\mathrm{e}}{\mathrm{n}+2}{\left(\frac{\mathrm{n}+2-1}{\mathrm{n}+2}\right)}^{\mathrm{n}}=\frac{\mathrm{e}}{\mathrm{n}+2}{(1-\frac{1}{\mathrm{n}+2})}^{\mathrm{n}}$$$$=\frac{\mathrm{e}}{\mathrm{n}+2}{\mathrm{e}}^{\mathrm{nln}(1-\frac{1}{\mathrm{n}+2})}\sim \frac{\mathrm{e}}{\mathrm{n}+2}{\mathrm{e}}^{-\frac{\mathrm{n}}{\mathrm{n}+2}}\Rightarrow 0\Rightarrow \mathrm{\Sigma }{\mathrm{v}}_{\mathrm{n}}\mathrm{converges}$$

Commented by mohammad17 last updated on 02/Aug/20

$$thankyousir$$

Commented by abdomathmax last updated on 02/Aug/20

$$\mathrm{you}\mathrm{are}\mathrm{welcome}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com