0

I am to use mathematical induction to prove:

$$\sum_{i=1}^n\frac{1}{i^2}<2 - \frac{1}{n}$$

my base case is n = 3:

LHS: $\frac{1}{1}+\frac{1}{4}+\frac{1}{9}= \frac{49}{36}$

RHS: $2-\frac{1}{3}=\frac{5}{3}$

base case holds true.

assume $n = k$

$$2-\frac{1}{k}$$

now show $k+1$

$$2-\frac{1}{k+1}$$

Inductive Step:

$$2-\frac{1}{k}+ \frac{1}{{k+1}^2}$$

Now I am stuck I have no clue what math I should be doing from here. I thought about trying to FOIL the $\frac{1}{{k+1}^2}$ but that didn't really give me anything I could work with. Any tips? Or general pointers when doing this stuff? Thanks guys!

Did
  • 279,727
user126959
  • 21

1 Answers1

0

Hint: Your inductive step will work if you can show $$2-\frac1k +\frac1{(k+1)^2} \le 2-\frac1{(k+1)}$$ so consider whether $$\left( 2-\frac1{(k+1)}\right) - \left( 2-\frac1k +\frac1{(k+1)^2} \right)$$ is positive, zero or negative.

Henry
  • 157,058
  • So if that first part is true then the proof is true? Shouldn't the inductive step and the hypothesis be equal? – user126959 Mar 14 '14 at 09:29
  • 1
    @user126959 - you must show that $\sum_{i=1}^{k+1}$ $\frac{1}{i^2}$ $<$ $2 - \frac{1}{k+1}$ knowing that $\sum_{i=1}^k$ $\frac{1}{i^2}$ $<$ $2 - \frac{1}{k}$. You must use Daniel Fischer's suggestion, so that : $\sum_{i=1}^{k+1}$ $\frac{1}{i^2}$ $<$ $2 - \frac{1}{k+1}$ $=$ $\sum_{i=1}^k$ $\frac{1}{i^2}$ $+$ $\frac{1}{(k+1)^2}$ $<$ $2 - \frac{1}{k}$ $+$ $\frac{1}{(k+1)^2}$. But this is $<$ $2 - \frac{1}{k} + \frac{1}{(k+1)^2}$ $<$ $2 - \frac{1}{k} + \frac{1}{k(k+1)}$, and $2 - \frac{1}{k} + \frac{1}{k(k+1)}$ $=$ $2 - \frac{k}{k(k+1)}$, i.e. $2 - \frac{1}{k+1}$. – Mauro ALLEGRANZA Mar 14 '14 at 09:54