Question

\[1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + . . . + \frac{1}{n^2} < 2 - \frac{1}{n}\] for all n ≥ 2, n ∈ N 

 

Answer 1

Let P(n) be the given statement.
Thus, we have: 

\[P\left( n \right): 1 + \frac{1}{4} + \frac{1}{9} + . . . + \frac{1}{n^2} < 2 - \frac{1}{n}\]

\[\text{ Step} 1: P(2): \frac{1}{2^2} = \frac{1}{4} < 2 - \frac{1}{2}\]

\[\text{ Thus, } P\left( 2 \right) \text{ is true } . \left[ \text{ We have not taken n = 1 because it is not possible . We will start this function from n = 2 onwards . } \right]\]

\[\text{ Step} 2: \]

\[\text{ Let P } \left( m \right) \text{ be true .}  \]

\[\text{ Now } , \]

\[1 + \frac{1}{4} + \frac{1}{9} + . . . + \frac{1}{m^2} < 2 - \frac{1}{m}\]

\[\text{ We need to prove that P(m + 1) is true }  . \]

\[\text{ We know that P(m) is true }  . \]

\[\text{ Thus, we have: }  \]

\[1 + \frac{1}{4} + \frac{1}{9} + . . . + \frac{1}{m^2} < 2 - \frac{1}{m}\]

\[ \Rightarrow 1 + \frac{1}{4} + \frac{1}{9} + . . . + \frac{1}{m^2} + \frac{1}{\left( m + 1 \right)^2} < 2 - \frac{1}{m} + \frac{1}{\left( m + 1 \right)^2} \left[ \text{ Adding }  \frac{1}{(m + 1 )^2} to \text{ both sides }  \right]\] 

\[ \Rightarrow P\left( m + 1 \right) < 2 - \frac{1}{m + 1} \left[ \because \left( m + 1 \right)^2 > m + 1, \frac{1}{\left( m + 1 \right)^2} < \frac{1}{m + 1} \Rightarrow \frac{1}{m} - \frac{1}{\left( m + 1 \right)^2} < \frac{1}{m + 1} as m < m + 1 \right] \]

\[\text{ Thus, }  P\left( m + 1 \right) \text{ is true }  . \]

\[\text{ By principle of mathematical induction, P(n) is true for all n } \in N, n \geq 2 .\]