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Question Number 200976 by Blackpanther last updated on 27/Nov/23

Answered by Mathspace last updated on 28/Nov/23

$$\sum _{n=1}^{\mathrm{\infty }}\frac{cos\left(n\pi \right)}{ln3}=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{ln3}$$$${lim}_{n\to \mathrm{\infty }}{(-1)}^n\ne 0\Rightarrow thisserie$$$$isdivergente$$$$2)\sum _{n=1}^{\mathrm{\infty }}{n}^{-\frac{1}{4}}=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^{\frac{1}{4}}}$$$$n^{\frac{1}{4}}<n^{\frac{1}{2}}\Rightarrow \frac{1}{n^{\frac{1}{4}}}>\frac{1}{n^{\frac{1}{2}}}=\frac{1}{\sqrt{n}}$$$$but\mathrm{\Sigma }\frac{1}{\sqrt{n}}isdiv.\Rightarrow \mathrm{\Sigma }{n}^{-\frac{1}{4}}$$$$isdivergente$$$$3)\mathrm{\Sigma }\frac{1}{nlogn}$$$$letf\left(t\right)=\frac{1}{tlogt}(t⩾2)$$$$f^{{}^{\prime }}\left(t\right)=-\frac{logt+1}{{\left(tlogt\right)}^2}<0\Rightarrow fis$$$$decreasingso\mathrm{\Sigma }{u}_{n}and$$$${\int }_2^{\mathrm{\infty }}\frac{dt}{tlogt}havesamenature$$$$logt=u\Rightarrow {\int }_2^{\mathrm{\infty }}\frac{dt}{tlogt}$$$$={\int }_{log2}^{\mathrm{\infty }}\frac{e^{u}du}{e^{u}u}={\int }_{log2}^{\mathrm{\infty }}\frac{du}{u}$$$$={\left[logu\right]}_{log2}^{\mathrm{\infty }}=+\mathrm{\infty }\Rightarrow$$$$thisserieisdiv.$$$$$$

Answered by Mathspace last updated on 28/Nov/23

$$4)\mathrm{\Sigma }{(-1)}^n\frac{\left(2n\right)!}{{(n!)}^2}$$$$u_n={(-1)}^n\frac{\left(2n\right)!}{{(n!)}^2}$$$$\mid \frac{u_{n+1}}{u_n}\mid =\frac{(2n+2)!}{{(n+1)!)}^2}\times \frac{n{!}^2}{\left(2n\right)!}$$$$=\frac{(2n+2)(2n+1)}{{(n+1)}^2}\to 4>1$$$$\Rightarrow thisserieisdivergente$$

Answered by Mathspace last updated on 28/Nov/23

$$5)\sum _{n=1}^{\mathrm{\infty }}\frac{6n+6}{{(-1)}^n}=6\sum _{n=1}^{\mathrm{\infty }}(n+1){(-1)}^n(n+1=p)$$$$=6\sum _{p=2}^{\mathrm{\infty }}p{(-1)}^{p-1}$$$$={lim}_{n\to +\mathrm{\infty }}6\sum _{p=2}^{n}p{(-1)}^{p-1}$$$$wehave\sum _{p=2}^n{x}^p$$$$=\frac{1-x^{n+1}}{1-x}-1-x$$$$=\frac{1-x^{n+1}-(1-x)(1+x)}{1-x}$$$$=\frac{1-x^{n+1}-1+x^2}{1-x}=\frac{x^2-x^{n+1}}{1-x}$$$$and\sum _{p=2}^n{px}^{p-1}=\frac{d}{dx}\left(\frac{x^2-x^{n+1}}{1-x}\right)$$$$=\frac{(2x-(n+1)x^n)(1-x)+x^2-x^{n+1}}{{(1-x)}^2}$$$$=\frac{2x-2x^2-(n+1)x^n+(n+1)x^{n+1}+x^2-x^{n+1}}{{(1-x)}^2}$$$$=\frac{2x-x^2-{nx}^n+(n+1)x^{n+1}}{{(1-x)}^2}$$$$\sum _2^{n}p{(-1)}^{p-1}$$$$=\frac{-2-1-n{(-1)}^n+(n+1){(-1)}^{n+1}}{4}$$$$\to \mathrm{\infty }\Rightarrow thisserieisdivergente$$$$$$

Answered by Mathspace last updated on 28/Nov/23

$$6)\frac{lnn}{n}isdecreasing$$$${\int }_2^{\mathrm{\infty }}\frac{logx}{x}{=}_{}{\left[{log}^{2}x\right]}_2^{\mathrm{\infty }}-{\int }_2^{\mathrm{\infty }}\frac{logx}{x}dx$$$$\Rightarrow 2{\int }_2^{\mathrm{\infty }}\frac{logx}{x}dx=\frac{1}{2}{\left[{log}^{2}x\right]}_2^{\mathrm{\infty }}=\mathrm{\infty }$$$$\Rightarrow \mathrm{\Sigma }\frac{logn}{n}isdiv.$$$$$$

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