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**deezy** Prove $\displaystyle \sum_{i=1}^n \frac{1}{i^2} \le 2 - \frac{1}{n}$ for all $\displaystyle n \epsilon \mathbb{N}$

__Proof__: i) Show P(1) holds.

LHS = $\displaystyle \sum_{i=1}^1 \frac{1}{i^2} = \frac{1}{1} = 1$

RHS = $\displaystyle 2 - \frac{1}{1} = 2 - 1 = 1 $

LHS $\displaystyle \le $ RHS, thus, P(1) holds.

ii) Assume P(k) holds for some $\displaystyle k \epsilon \mathbb{N}$

That is, assume $\displaystyle \sum_{i = 1}^k \frac{1}{i^2} \le 2 - \frac{1}{k}$.

To show: P(k + 1): $\displaystyle \sum_{i=1}^{k+1} \frac{1}{i^2} \le 2 - \frac{1}{k+1}$

Then $\displaystyle \sum_{i=1}^{k+1} \frac{1}{i^2} = \left( \sum_{i=1}^{k} \frac{1}{i^2} \right) + \frac{1}{(k+1)^2} \le 2 - \frac{1}{k} + \frac{1}{(k+1)^2}$

Not sure what to do next. In my next step, I said $\displaystyle 2 - \frac{1}{k} + \frac{1}{(k+1)^2} \le 2 - \frac{1}{k} + \frac{1}{k+1}$. But I'm not sure if this is the right path to take.