Mathematical induction using inequalities

Question

I just started with induction and i can solve cases where you need to prove that the left hand side is equal with the right hand side. But now i am trying to do this one:

Prove that $$ 1+1/4+1/9+1/16+ ... +1/(n^2) < 2-(1/n) $$ for n > 1

Base n = 2: $$ 1/4 < 2-(1/2) <=> 1/4 < 1/2 $$ Assume true for n = k:

$$ 1+1/4+1/9+1/16+ ... +1/(k^2) < 2-(1/k) $$

Show true for n = k + 1:

$$ 1+1/4+1/9+1/16+ ... + 1/(k^2) + 1/(k+1)^2 < 2-(1/(k+1)) $$

Im stuck here, usually when i prove expressions with equalities i just rewrite $$ 1+1/4+1/9+1/16+ ... + 1/(k^2) $$

in the above to

$$ 2-(1/(k)) $$

But now im not sure how to continue..

Answer 1

Do the same as usual, i.e. substitution just instead of equality use an inequality ;-)

To be more specific, just take all what is known in one bracket: $$\underbrace{1+1/4+1/9+...+1/k^2}_{<2-1/k}+1/(k+1)^2$$ and substitute, using "<" $$1+1/4+1/9+...+1/k^2+1/(k+1)^2<2-1/k+1/(k+1)^2$$

What is left, is to prove that: $$2-1/k+1/(k+1)^2\leq2-1/(k+1).$$ Hope you can do it!

Then, combining both would give you the desired outcome.

Answer 2

You can do the same thing with inequalities

$1 + ..... + 1/k^2 < 2-1/k$

So $1 + ...... + 1/k^2 + 1/(k+1)^2 < 2- 1/k + 1/(k+1)^2$

Now it just a matter of showing $2 - 1/k + 1/(k+1)^2 \le 2 - 1/(k + 1)^2$

which is just a matter of showing $1/k - 1/(k+1)^2 \ge 1/(k+1)$

which is ... well lets see...:

$1/k - 1/(k+1)^2 = ((k+1)^2-k)/k(k+1)^2 = (k^2 + k + 1)/k(k+1)^2 = k(k + 1 + 1/k)/k(k+1)^2 > k(k+1)/k(k+1)^ = 1/(k+1)$

So it seems to be true.