# Solve the following

Question:

$1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\ldots+\frac{1}{n^{2}}<2-\frac{1}{n}$ for all $n \geq 2, n \in N$

Solution:

Let P(n) be the given statement.

Thus, we have:

$P(n): 1+\frac{1}{4}+\frac{1}{9}+\ldots+\frac{1}{n^{2}}<2-\frac{1}{n}$

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

Thus, $P(2)$ is true.

[We have not taken $n=1$ because it is not possible. We will start this function from $n=2$ onwards.]

Step 2 :

Let $P(m)$ be true.

Now,

$1+\frac{1}{4}+\frac{1}{9}+\ldots+\frac{1}{m^{2}}<2-\frac{1}{m}$

We need to prove that $P(m+1)$ is true.

We know that $P(m)$ is true.

Thus, we have :

$1+\frac{1}{4}+\frac{1}{9}+\ldots+\frac{1}{m^{2}}<2-\frac{1}{m}$

$\Rightarrow 1+\frac{1}{4}+\frac{1}{9}+\ldots+\frac{1}{m^{2}}+\frac{1}{(m+1)^{2}}<2-\frac{1}{m}+\frac{1}{(m+1)^{2}} \quad$ Adding $\frac{1}{(m+1)^{2}}$ to both sides

$\Rightarrow P(m+1)<2-\frac{1}{m+1}$

$\left[\because(m+1)^{2}>m+1, \frac{1}{(m+1)^{2}}<\frac{1}{m+1} \Rightarrow \frac{1}{m}-\frac{1}{(m+1)^{2}}<\frac{1}{m+1}\right.$ as $\left.m Thus,$P(m+1)$is true. By principle of$m$athematical$i$nduction,$P(n)$is true for all$n \in N, n \geq 2\$.