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Question Number 144261 by ZiYangLee last updated on 23/Jun/21

$$\mathrm{Find}\mathrm{the}\mathrm{value}\mathrm{of}\underset{n\to \mathrm{\infty }}{\lim }\underset{k=n}{\sum ^{2n}}\frac{{(-1)}^k}{k}.$$

Commented by Dwaipayan Shikari last updated on 23/Jun/21

$$\underset{n\to \mathrm{\infty }}{\lim }\underset{k=n}{\sum ^{2n}}\frac{{(-1)}^k}{k}$$$$=\underset{n\to \mathrm{\infty }}{\lim }\frac{{(-1)}^n}{n}-\frac{{(-1)}^{n+1}}{n+1}+\frac{{(-1)}^{n+2}}{n+2}+...\frac{1}{n+n}$$$$=\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n}\underset{k=1}{\sum ^n}\frac{{(-1)}^{n+k-1}}{1+\frac{k}{n}}=\frac{1}{n}\mathrm{\Sigma }\frac{e^{\pi i(n+k-1)}}{1+\frac{k}{n}}$$$$$$$$Ifitwas$$$$\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n}\underset{k=1}{\sum ^n}\frac{1}{1+\frac{k}{n}}={\int }_0^1\frac{1}{1+x}dx=\ln \left(2\right)$$

Answered by mathmax by abdo last updated on 24/Jun/21

$$\mathrm{p}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=\mathrm{n}}^{2\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}}\Rightarrow {\mathrm{P}}^{{}^{\prime }}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=\mathrm{n}}^{2\mathrm{n}}{(-1)}^{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}-1}$$$$=\frac{1}{\mathrm{x}}\sum _{\mathrm{k}=\mathrm{n}}^{2\mathrm{n}}{(-\mathrm{x})}^{\mathrm{k}}{=}_{\mathrm{k}-\mathrm{n}=\mathrm{j}}\frac{1}{\mathrm{x}}\sum _{\mathrm{j}=0}^{\mathrm{n}}{(-\mathrm{x})}^{\mathrm{n}+\mathrm{j}}$$$$=\frac{{(-\mathrm{x})}^{\mathrm{n}}}{\mathrm{x}}\sum _{\mathrm{j}=0}^{\mathrm{n}}{(-\mathrm{x})}^{\mathrm{j}}={(-1)}^{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}-1}\times \frac{1-{(-\mathrm{x})}^{\mathrm{n}+1}}{1+\mathrm{x}}$$$$=\frac{{(-1)}^{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}-1}}{1+\mathrm{x}}(1+{(-1)}^{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}+1})$$$$=\frac{{(-1)}^{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}-1}+{\mathrm{x}}^{2\mathrm{n}}}{1+\mathrm{x}}\Rightarrow \mathrm{p}\left(\mathrm{x}\right)={\int }_0^{\mathrm{x}}\frac{{(-1)}^{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}-1}+{\mathrm{x}}^{2\mathrm{n}}}{1+\mathrm{x}}\mathrm{dx}+\mathrm{C}$$$$\mathrm{p}\left(\mathrm{o}\right)=0\Rightarrow \mathrm{c}=0\Rightarrow \mathrm{p}\left(\mathrm{x}\right)={\int }_0^{\mathrm{x}}\frac{{(-1)}^{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}-1}+{\mathrm{x}}^{2\mathrm{n}}}{1+\mathrm{x}}\mathrm{dx}\mathrm{and}$$$$\sum _{\mathrm{k}=\mathrm{n}}^{2\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}}=\mathrm{p}\left(1\right)={\int }_0^1\frac{{(-1)}^{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}-1}+{\mathrm{x}}^{2\mathrm{n}}}{1+\mathrm{x}}\mathrm{dx}$$$$\mid \mathrm{p}\left(1\right)\mid ⩽{\int }_0^1{\mathrm{x}}^{\mathrm{n}-1}\mathrm{dx}+{\int }_0^1{\mathrm{x}}^{2\mathrm{n}}\mathrm{dx}=\frac{1}{\mathrm{n}}+\frac{1}{2\mathrm{n}+1}\to 0(\mathrm{n}\to +\mathrm{\infty })\Rightarrow$$$${\lim }_{\mathrm{n}\to \mathrm{\infty }}\sum _{\mathrm{k}=\mathrm{n}}^{2\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}}=0$$

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