Hello,

I am trying to understand the solution to a problem

but I can't understand it. Could you explain it, please?

Here is the problem:

Prove by induction that

$\displaystyle 1+ \frac{1}{2^2} + \frac{1}{3^2} + ... + \frac{1}{n^2} \leq 2, n>0$

solution:

Instead of proving $\displaystyle 1+ \frac{1}{2^2} + \frac{1}{3^2} + ... + \frac{1}{n^2} \leq 2, n>0$

we will prove something stronger:

$\displaystyle 1+ \frac{1}{2^2} + \frac{1}{3^2} + ... + \frac{1}{n^2} \leq 2 - \frac{1}{n} , n>0$

Basis: $\displaystyle n=1: 1 \leq 1$, holds

Inductive step: The equation holds for n, so $\displaystyle 1+ \frac{1}{2^2} + \frac{1}{3^2} + ... + \frac{1}{n^2} \leq 2 - \frac{1}{n} , n>0$.

By adding to both sides $\displaystyle \frac{1}{(n+1)^2}$ we have:

$\displaystyle 1+ \frac{1}{2^2} + \frac{1}{3^2} + ... + \frac{1}{n^2} + \frac{1}{(n+1)^2} \leq 2 - \frac{1}{n} +\frac{1}{(n+1)^2}

\leq 2 - \frac{1}{n+1}$ , so P(n+1) holds.

Questions:

1. How could someone think to prove

$\displaystyle 1+ \frac{1}{2^2} + \frac{1}{3^2} + ... + \frac{1}{n^2} \leq 2 - \frac{1}{n} $ (ie. adding $\displaystyle - \frac{1}{n}$ )?

2. Why should we add to both sides $\displaystyle \frac{1}{(n+1)^2}$ ?

3. In a similar problem, having to prove

$\displaystyle \frac{8}{1*3} + \frac{8}{5*7} + ... + \frac{8}{(4n+1)*(4n+3)} \leq 4 $

we shall prove

$\displaystyle \frac{8}{1*3} + \frac{8}{5*7} + ... + \frac{8}{(4n+1)*(4n+3)} \leq 4 - \frac{1}{n+1} $

Why choose $\displaystyle \frac{1}{n+1}$ and then add to both sides $\displaystyle \frac{8}{[(4(n+1)+1]*[4(n+1)+3]}$?

Thanks.