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Question Number 116451 by bemath last updated on 04/Oct/20

$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{\sqrt[\mathrm{n}]{(\mathrm{n}+1)(\mathrm{n}+2)(\mathrm{n}+3)...(\mathrm{n}+\mathrm{n})}}{\mathrm{n}}=?$$

Commented by Olaf last updated on 04/Oct/20

$$\mathbf{Sorry}{\mathbf{I}}^{\prime }\mathbf{m}\mathbf{wrong}!$$$$\mathrm{I}\mathbf{calculated}\mathbf{another}\mathbf{expression}!$$$$!??!{\mathbf{I}}^{\prime }\mathbf{m}\mathbf{tired}!$$

Answered by Olaf last updated on 04/Oct/20

$$u_n={}^n\sqrt{\frac{\underset{k=1}{\prod ^n}(n+k)}{n}}$$$$v_n=\ln u_n$$$$v_n=\frac{1}{n}(\ln \underset{k=1}{\prod ^n}(n+k)-\ln n)$$$$v_n=\frac{1}{n}(\ln \underset{k=1}{\prod ^n}n(1+\frac{k}{n})-\ln n)$$$$v_n=\frac{1}{n}(\ln n^n\underset{k=1}{\prod ^n}(1+\frac{k}{n})-\ln n)$$$$v_n=\frac{1}{n}(n\ln n+\ln \underset{k=1}{\prod ^n}(1+\frac{k}{n})-\ln n)$$$$v_n=\frac{1}{n}(n\ln n+\underset{k=1}{\sum ^n}\ln (1+\frac{k}{n})-\ln n)$$$$\ln (1+\frac{k}{n})\underset{\mathrm{\infty }}{\sim }\frac{k}{n}$$$$v_n\underset{\mathrm{\infty }}{\sim }\frac{1}{n}(n\ln n+\underset{k=1}{\sum ^n}\frac{k}{n}-\ln n)$$$$v_n\underset{\mathrm{\infty }}{\sim }\frac{1}{n}(n\ln n+\frac{1}{n}.\frac{n(n+1)}{2}-\ln n)$$$$v_n\underset{\mathrm{\infty }}{\sim }\ln n+\frac{n+1}{2n}-\frac{\ln n}{n}=+\mathrm{\infty }+\frac{1}{2}-0$$$$\underset{n\to \mathrm{\infty }}{\lim }{v}_n=+\mathrm{\infty }$$$$\underset{n\to \mathrm{\infty }}{\lim }{u}_n=\underset{n\to \mathrm{\infty }}{\lim }{e}^{v_n}=+\mathrm{\infty }$$$$$$$$$$

Commented by bemath last updated on 04/Oct/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Answered by Dwaipayan Shikari last updated on 04/Oct/20

$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\sqrt[\mathrm{n}]{(1+\frac{1}{\mathrm{n}})(1+\frac{2}{\mathrm{n}})(1+\frac{3}{\mathrm{n}})..}=\mathrm{y}$$$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\log (1+\frac{\mathrm{k}}{\mathrm{n}})=\mathrm{logy}$$$${\int }_0^1\log (1+\mathrm{x})\mathrm{dx}=\mathrm{logy}$$$${\left[\mathrm{xlog}(1+\mathrm{x})\right]}_0^1-{\int }_0^1\frac{\mathrm{x}}{1+\mathrm{x}}=\mathrm{logy}$$$$\log \left(2\right)-1+{\left[\log (1+\mathrm{x})\right]}_0^1=\mathrm{logy}$$$$2\log \left(2\right)-1=\mathrm{logy}$$$$\mathrm{y}=\frac{4}{\mathrm{e}}$$

Answered by mnjuly1970 last updated on 04/Oct/20

$$solution$$$$l={lim}_{n\to \mathrm{\infty }}\frac{\sqrt[n]{(n+1)(n+2)...(n+n)}}{n}$$$$l={lim}_{n\to \mathrm{\infty }}\sqrt[n]{(1+\frac{1}{n})(1+\frac{2}{n})...(1+\frac{n}{n})}$$$$log\left(l\right)={lim}_{n\to \mathrm{\infty }}\frac{1}{n}\underset{k=}{\sum ^n}log(1+\frac{k}{n})$$$$={\int }_0^{1}log(1+x)dx={\int }_1^{2}log\left(t\right)dt$$$$={[tlog\left(t\right)-t]}_1^2=2log\left(2\right)-2+1$$$$=log\left(4\right)-log\left(e\right)=log\left(\frac{4}{e}\right)$$$$\therefore l:=\frac{4}{e}✓✓$$$$...m.n.1970...$$$$$$$$$$

Commented by bemath last updated on 04/Oct/20

$$\mathrm{yes}...\mathrm{thank}\mathrm{you}$$

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