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Question Number 125282 by Mammadli last updated on 09/Dec/20

$$1.\underset{\mathit{n}=1}{\sum ^{\mathrm{\infty }}}\sqrt[\mathit{n}]{\frac{2\mathit{n}-1}{2\mathit{n}+1}}=?$$$$2.\underset{\mathit{n}\to \mathrm{\infty }}{\mathit{l}\mathit{i}\mathit{m}}\sqrt[\mathit{n}]{\frac{2\mathit{n}-1}{2\mathit{n}+1}}=?$$

Answered by mathmax by abdo last updated on 09/Dec/20

$$\mathrm{let}{\mathrm{u}}_{\mathrm{n}}={\left(\frac{2\mathrm{n}-1}{2\mathrm{n}+1}\right)}^{\frac{1}{\mathrm{n}}}\Rightarrow {\mathrm{u}}_{\mathrm{n}}={\left(\frac{2\mathrm{n}+1-2}{2\mathrm{n}+1}\right)}^{\frac{1}{\mathrm{n}}}={(1-\frac{2}{2\mathrm{n}+1})}^{\frac{1}{\mathrm{n}}}$$$$\sim 1-\frac{2}{\mathrm{n}(2\mathrm{n}+1)}\Rightarrow {\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{u}}_{\mathrm{n}}=1$$$$2)\mathrm{we}\mathrm{have}\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{}^{\mathrm{n}}\sqrt{\frac{2\mathrm{n}-1}{2\mathrm{n}+1}}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{u}}_{\mathrm{n}}\mathrm{but}{\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{u}}_{\mathrm{n}}\ne 0\Rightarrow$$$$\mathrm{\Sigma }{\mathrm{u}}_{\mathrm{n}}\mathrm{is}\mathrm{divergent}!$$

Answered by Dwaipayan Shikari last updated on 09/Dec/20

$$2.$$$$\underset{n\to \mathrm{\infty }}{\lim }{\left(\frac{2n-1}{2n+1}\right)}^{\frac{1}{n}}=\underset{n\to \mathrm{\infty }}{\lim }{(1-\frac{2}{2n+1})}^{\frac{1}{n}}=\underset{n\to \mathrm{\infty }}{\lim }(1-\frac{2}{n(2n+1)})=1$$

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