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Question Number 71759 by naka3546 last updated on 19/Oct/19

Commented by Prithwish sen last updated on 19/Oct/19

$$\mathbf{x}=\frac{1}{3}.\frac{1}{3}.\frac{2}{4}.\frac{3}{5}.........\frac{998}{1000}.\frac{999}{1001}=\frac{2\times 1001001}{3\mathrm{x}1000\times 1001}\approx 0.67$$$$\therefore 100\mathrm{x}\approx 67$$$$\mathbf{By}\mathbf{considering}$$$$\frac{{\mathbf{n}}^3-1}{{\mathbf{n}}^3+1}.\frac{{(\mathbf{n}+1)}^3-1}{{(\mathbf{n}+1)}^3-1}.....=\frac{(\mathbf{n}-1).\mathbf{n}.(\mathbf{n}+1).(\mathbf{n}+2)....}{(\mathbf{n}+1).({\mathbf{n}}^2-\mathbf{n}+1).(\mathbf{n}+2).(\mathbf{n}+3)....}$$$$\mathbf{please}\mathbf{check}.$$

Commented by Abdo msup. last updated on 20/Oct/19

$$X_n=\prod _{k=2}^n\frac{k^3-1}{k^3+1}\Rightarrow X_n=\prod _{k=2}^n\frac{k-1}{k+1}\times \frac{k^2+k+1}{k^2-k+1}$$$$=\prod _{k=2}^n\frac{k-1}{k+1}\times \prod _{k=2}^n\frac{k^2+k+1}{k^2-k+1}but$$$$\prod _{k=2}^n\frac{k-1}{k+1}=\frac{1}{3}\frac{2}{4}\frac{3}{5}.....\frac{n-1}{n+1}=2\frac{(n-1)!}{(n+1)!}$$$$=\frac{2(n-1)!}{(n+1)n(n-1)!}=\frac{2}{n^2+n}$$$$ln\left(X_n\right)=ln\left(\frac{2}{n^2+n}\right)+(\sum _{k=2}^{n}ln(k^2+k+1)-ln(k^2-k+1))$$$$letvk=ln(k^2+k+1)\Rightarrow v_{k-1}=ln({(k-1)}^2+k-1+1)$$$$=ln(k^2-2k+1+k)=ln(k^2-k+1)\Rightarrow$$$$\sum _{k=2}^n(ln(k^2+k+1)-ln(k^2-k+1))$$$$=\sum _{k=2}^n(v_k-v_{k-1})=v_2-v_1+v_3-v_2+...+v_n-v_{n-1}$$$$=v_n-v_1=ln(n^2+n+1)-ln\left(3\right)\Rightarrow$$$$ln\left(X_n\right)=ln\left(\frac{2}{n^2+n}\right)+ln\left(\frac{n^2+n+1}{3}\right)\Rightarrow$$$$X_n=\frac{2}{n^2+n}\times \frac{n^2+n+1}{3}=\frac{2}{3}\times \frac{n^2+n+1}{n^2+n}$$$$x=X_{{10}^3}=\frac{2}{3}\times \frac{{10}^6+{10}^3+1}{{10}^6+{10}^3}\Rightarrow$$$$100x=\frac{200}{3}(1+\frac{1}{{10}^6+{10}^3})=....$$$$$$

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