+26397 Let
Find the largest 
so that
$f(2)⋅f(3)⋅f(4)⋯f(n−1)⋅f(n)<1.98.$
\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{\textcolor[rgb]{1,0,0}{\not{3}} }{ \textcolor[rgb]{0,0,1}{\not{2}} } \cdot \frac{\textcolor[rgb]{1,0,0}{\not{4}} }{ \textcolor[rgb]{0,0,1}{\not{3}} } \cdot \frac{\textcolor[rgb]{1,0,0}{\not{5}} }{ \textcolor[rgb]{0,0,1}{\not{4}} } \dots \frac{\textcolor[rgb]{1,0,0}{\not{n}} }{ \textcolor[rgb]{0,0,1}{\not{n-1}} } \right) \cdot \left( \frac{ \textcolor[rgb]{0,0,1}{\not{2}} }{\textcolor[rgb]{1,0,0}{\not{3}}} \cdot \frac{ \textcolor[rgb]{0,0,1}{\not{3}} }{\textcolor[rgb]{1,0,0}{\not{4}}} \cdot \frac{ \textcolor[rgb]{0,0,1}{\not{4}} }{\textcolor[rgb]{1,0,0}{\not{5}}} \cdot \frac{ \textcolor[rgb]{0,0,1}{\not{5}} }{\textcolor[rgb]{1,0,0}{\not{6}}} \dots \frac{ \textcolor[rgb]{0,0,1}{\not{n-1}} }{\textcolor[rgb]{1,0,0}{\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 $ }} }
+26397 Let
Find the largest so that
$f(2)⋅f(3)⋅f(4)⋯f(n−1)⋅f(n)<1.98.$
\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{\textcolor[rgb]{1,0,0}{\not{3}} }{ \textcolor[rgb]{0,0,1}{\not{2}} } \cdot \frac{\textcolor[rgb]{1,0,0}{\not{4}} }{ \textcolor[rgb]{0,0,1}{\not{3}} } \cdot \frac{\textcolor[rgb]{1,0,0}{\not{5}} }{ \textcolor[rgb]{0,0,1}{\not{4}} } \dots \frac{\textcolor[rgb]{1,0,0}{\not{n}} }{ \textcolor[rgb]{0,0,1}{\not{n-1}} } \right) \cdot \left( \frac{ \textcolor[rgb]{0,0,1}{\not{2}} }{\textcolor[rgb]{1,0,0}{\not{3}}} \cdot \frac{ \textcolor[rgb]{0,0,1}{\not{3}} }{\textcolor[rgb]{1,0,0}{\not{4}}} \cdot \frac{ \textcolor[rgb]{0,0,1}{\not{4}} }{\textcolor[rgb]{1,0,0}{\not{5}}} \cdot \frac{ \textcolor[rgb]{0,0,1}{\not{5}} }{\textcolor[rgb]{1,0,0}{\not{6}}} \dots \frac{ \textcolor[rgb]{0,0,1}{\not{n-1}} }{\textcolor[rgb]{1,0,0}{\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 $ }} }
