Prove that 1 +1/4 + 1/9 +1/16 +.....+ 1/n^2 < 2 - 1/n for all n > 2, n ϵ N. ​

Question

Prove that 1 + \(\frac{1}4\)+ \(\frac{1}9\) + \(\frac{1}{16}\)+.....+ \(\frac{1}{n^2}\) < 2 - \(\frac{1}n\) for all n > 2, n ϵ N.

Answer 1

Let the given statement be P(n) 

Thus,P(n) = 1 + \(\frac{1}4\)+ \(\frac{1}9\) + \(\frac{1}{16}\)+.....+ \(\frac{1}{n^2}\) < 2 - \(\frac{1}n\) for all n > 2, n ϵ N.

step1;

P(n); \(\frac{1}{2^2}\) = \(\frac{1}4\) < 2 - \(\frac{1}{2}\)

Thus, P(2) is true. 

Let, P(m) be true, 

Now,

step2; 1 + \(\frac{1}4\)+ \(\frac{1}9\) + \(\frac{1}{16}\)+.....+ \(\frac{1}{m^2}\) < 2 - \(\frac{1}m\)

Now, we need to prove that P(m+1) is true whenever P(m) is true.

we have 1 + \(\frac{1}4\)+ \(\frac{1}9\) + \(\frac{1}{16}\)+.....+ \(\frac{1}{m^2}\) < 2 - \(\frac{1}m\)

adding \(\frac{1}{(m+1)^2}\) on both sides

Thus, Pm+1 is true. By the principle of mathematical induction, P(n) is true for all n∈N, n ≥ 2.