1. ## Induction (involving inequality)

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

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

solution:
Instead of proving $1+ \frac{1}{2^2} + \frac{1}{3^2} + ... + \frac{1}{n^2} \leq 2, n>0$
we will prove something stronger:

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

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

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

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

$1+ \frac{1}{2^2} + \frac{1}{3^2} + ... + \frac{1}{n^2} + \frac{1}{(n+1)^2} \leq 2 - \frac{1}{n} +\frac{1}{(n+1)^2}
\leq 2 - \frac{1}{n+1}$
, so P(n+1) holds.

Questions:

1. How could someone think to prove
$1+ \frac{1}{2^2} + \frac{1}{3^2} + ... + \frac{1}{n^2} \leq 2 - \frac{1}{n}$ (ie. adding $- \frac{1}{n}$ )?

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

3. In a similar problem, having to prove
$\frac{8}{1*3} + \frac{8}{5*7} + ... + \frac{8}{(4n+1)*(4n+3)} \leq 4$

we shall prove
$\frac{8}{1*3} + \frac{8}{5*7} + ... + \frac{8}{(4n+1)*(4n+3)} \leq 4 - \frac{1}{n+1}$
Why choose $\frac{1}{n+1}$ and then add to both sides $\frac{8}{[(4(n+1)+1]*[4(n+1)+3]}$?

Thanks.

2. Originally Posted by drthea
1. How could someone think to prove
$1+ \frac{1}{2^2} + \frac{1}{3^2} + ... + \frac{1}{n^2} \leq 2 - \frac{1}{n}$ (ie. adding $- \frac{1}{n}$ )?
The problem here is that if you tried to prove, $1+ \tfrac{1}{2^2}+ ... + \tfrac{1}{n^2} \leq 2$ then by adding $\tfrac{1}{(n+1)^2}$ to both sides (which becomes necessary because of induction) we get, $1+...+\tfrac{1}{(n+1)^2} \leq 2 + \tfrac{1}{(n+1)^2}$, but we cannot continue any longer because $2 + \tfrac{1}{(n+1)^2} > 2$ and induction breaks down here.

2. Why should we add to both sides $\frac{1}{(n+1)^2}$ ?
That was already addressed on top. It is necessary because of induction.

3. In a similar problem, having to prove
$\frac{8}{1*3} + \frac{8}{5*7} + ... + \frac{8}{(4n+1)*(4n+3)} \leq 4$

we shall prove
$\frac{8}{1*3} + \frac{8}{5*7} + ... + \frac{8}{(4n+1)*(4n+3)} \leq 4 - \frac{1}{n+1}$
Why choose $\frac{1}{n+1}$ and then add to both sides $\frac{8}{[(4(n+1)+1]*[4(n+1)+3]}$?
Over here we might use $\tfrac{1}{n+1}$ instead of $\tfrac{1}{n}$ because we want to show this holds for $n\geq 0$. In order for this statement to make any sense we need to realize that $\tfrac{1}{n}$ is a problem for then we have a zero denominator. There may be other reasons but this is the first one that I thought of.