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# LetFind the largest so that

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LetFind the largest  so that

Guest Jan 25, 2015

#1
+19207
+5

Let

Find the largest  so that

$$f(2)\cdot f(3)\cdot f(4)\cdots f(n-1)\cdot f(n) < 1.98.$$$$$\boxed{ \small{\text{ \prod \limits_{i=2}^{n} \frac{i^2}{i^2-1} < 1.98 }} } \qquad \dfrac{i^2}{i^2-1} = \left(\dfrac{i}{i-1} \right) * \left(\dfrac{i}{i+1} \right) \\ \\ \\ \small{\text{ \prod \limits_{i=2}^{n} \dfrac{i^2}{i^2-1} = \prod\limits_{i=2}^{n} \left(\dfrac{i}{i-1} \right) * \left(\dfrac{i}{i+1} \right) = \prod\limits_{i=2}^{n} \left(\dfrac{i}{i-1} \right) * \prod\limits_{i=2}^{n} \left(\dfrac{i}{i+1} \right) }} \\\\ \small{\text{ = \left( \frac{2}{2-1} \cdot \frac{3}{3-1} \cdot \frac{4}{4-1} \cdot \frac{5}{5-1} \dots \frac{n}{n-1} \right) \cdot \left( \frac{2}{2+1} \cdot \frac{3}{3+1} \cdot \frac{4}{4+1} \cdot \frac{5}{5+1} \dots \frac{n-1}{n} \cdot \frac{n}{n+1} \right) }} \\\\ \small{\text{ = \left( \frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \dots \frac{n}{n-1} \right) \cdot \left( \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5} \cdot \frac{5}{6} \dots \frac{n-1}{n} \cdot \frac{n}{n+1} \right) }} \\\\ \small{\text{ = \left( \frac{2}{1} \cdot \frac{{\not{3}} }{ {\not{2}} } \cdot \frac{{\not{4}} }{ {\not{3}} } \cdot \frac{{\not{5}} }{ {\not{4}} } \dots \frac{{\not{n}} }{ {\not{n-1}} } \right) \cdot \left( \frac{ {\not{2}} }{{\not{3}}} \cdot \frac{ {\not{3}} }{{\not{4}}} \cdot \frac{ {\not{4}} }{{\not{5}}} \cdot \frac{ {\not{5}} }{{\not{6}}} \dots \frac{ {\not{n-1}} }{{\not{n}} } \cdot \frac{n}{n+1} \right) }} \\\\ \small{\text{ = \left( \frac{2}{1} \right) \cdot \left( \frac{n}{n+1} \right) }} \\\\ \boxed{ \small{\text{ \left( \frac{2}{1} \right) \cdot \left( \frac{n}{n+1} \right) < 1.98 }} } \\\\ \small{\text{ \dfrac{n}{n+1} < \dfrac{1.98}{2} } \\\\ \small{\text{ \dfrac{n}{n+1} < 0.99 } \\\\ \small{\text{ n < 0.99 *(n+1) }} \\ \small{\text{ n < 0.99 *n+ 0.99 }} \\ \small{\text{ n - 0.99 *n < 0.99 }} \\ \small{\text{ n *(1 - 0.99) < 0.99 }} \\$$$

$$\\ \small{\text{  n * 0.01 < 0.99  }} \\ \small{\text{  n < \frac{ 0.99 } { 0.01 }  }} \\ \small{\text{  n < 99  }} \\ \boxed{ \small{\text{  n = 98  }} }$$

heureka  Jan 26, 2015
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#1
+19207
+5

Let

Find the largest  so that

$$f(2)\cdot f(3)\cdot f(4)\cdots f(n-1)\cdot f(n) < 1.98.$$$$$\boxed{ \small{\text{ \prod \limits_{i=2}^{n} \frac{i^2}{i^2-1} < 1.98 }} } \qquad \dfrac{i^2}{i^2-1} = \left(\dfrac{i}{i-1} \right) * \left(\dfrac{i}{i+1} \right) \\ \\ \\ \small{\text{ \prod \limits_{i=2}^{n} \dfrac{i^2}{i^2-1} = \prod\limits_{i=2}^{n} \left(\dfrac{i}{i-1} \right) * \left(\dfrac{i}{i+1} \right) = \prod\limits_{i=2}^{n} \left(\dfrac{i}{i-1} \right) * \prod\limits_{i=2}^{n} \left(\dfrac{i}{i+1} \right) }} \\\\ \small{\text{ = \left( \frac{2}{2-1} \cdot \frac{3}{3-1} \cdot \frac{4}{4-1} \cdot \frac{5}{5-1} \dots \frac{n}{n-1} \right) \cdot \left( \frac{2}{2+1} \cdot \frac{3}{3+1} \cdot \frac{4}{4+1} \cdot \frac{5}{5+1} \dots \frac{n-1}{n} \cdot \frac{n}{n+1} \right) }} \\\\ \small{\text{ = \left( \frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \dots \frac{n}{n-1} \right) \cdot \left( \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5} \cdot \frac{5}{6} \dots \frac{n-1}{n} \cdot \frac{n}{n+1} \right) }} \\\\ \small{\text{ = \left( \frac{2}{1} \cdot \frac{{\not{3}} }{ {\not{2}} } \cdot \frac{{\not{4}} }{ {\not{3}} } \cdot \frac{{\not{5}} }{ {\not{4}} } \dots \frac{{\not{n}} }{ {\not{n-1}} } \right) \cdot \left( \frac{ {\not{2}} }{{\not{3}}} \cdot \frac{ {\not{3}} }{{\not{4}}} \cdot \frac{ {\not{4}} }{{\not{5}}} \cdot \frac{ {\not{5}} }{{\not{6}}} \dots \frac{ {\not{n-1}} }{{\not{n}} } \cdot \frac{n}{n+1} \right) }} \\\\ \small{\text{ = \left( \frac{2}{1} \right) \cdot \left( \frac{n}{n+1} \right) }} \\\\ \boxed{ \small{\text{ \left( \frac{2}{1} \right) \cdot \left( \frac{n}{n+1} \right) < 1.98 }} } \\\\ \small{\text{ \dfrac{n}{n+1} < \dfrac{1.98}{2} } \\\\ \small{\text{ \dfrac{n}{n+1} < 0.99 } \\\\ \small{\text{ n < 0.99 *(n+1) }} \\ \small{\text{ n < 0.99 *n+ 0.99 }} \\ \small{\text{ n - 0.99 *n < 0.99 }} \\ \small{\text{ n *(1 - 0.99) < 0.99 }} \\$$$

$$\\ \small{\text{  n * 0.01 < 0.99  }} \\ \small{\text{  n < \frac{ 0.99 } { 0.01 }  }} \\ \small{\text{  n < 99  }} \\ \boxed{ \small{\text{  n = 98  }} }$$

heureka  Jan 26, 2015

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