Proof by induction inequality

**deezy** · October 4th 2012, 05:06 PM

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

Proof: i) Show P(1) holds.

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

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

ii) Assume P(k) holds for some $k \epsilon \mathbb{N}$
That is, assume $\sum_{i = 1}^k \frac{1}{i^2} \le 2 - \frac{1}{k}$ .

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

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

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

**Prove It** · October 4th 2012, 05:19 PM

Originally Posted by deezy

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

Proof: i) Show P(1) holds.

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

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

ii) Assume P(k) holds for some $k \epsilon \mathbb{N}$
That is, assume $\sum_{i = 1}^k \frac{1}{i^2} \le 2 - \frac{1}{k}$ .

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

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

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

What you have done is fine, but you need to state why that inequality is true, the reason being that $\displaystyle \begin{align*} (k + 1)^2 > k + 1 \end{align*}$ for all $\displaystyle \begin{align*} k > 1 \end{align*}$ . Then you can say $\displaystyle \begin{align*} 2 - \frac{1}{k} + \frac{1}{k + 1} < 2 + \frac{1}{k + 1} \end{align*}$ .

**deezy** · October 4th 2012, 05:33 PM

But I'm not sure how I can use $2 + \frac{1}{k + 1}$ since it isn't less than $2 - \frac{1}{k + 1}$ .

**MathoMan** · October 5th 2012, 01:51 PM

$2-\frac{1}{n}+\frac{1}{(n+1)^2}=2-\left(\frac{1}{n}-\frac{1}{(n+1)^2} \right)=2-\frac{n^2+n+1}{n(n+1)^2}$

Now you just have to prove that $\frac{n^2+n+1}{n(n+1)^2}$ is greater than or equal to $\frac{1}{n+1}$ ,
because that would give you the result $2-\frac{n^2+n+1}{n(n+1)^2}\leq 2-\frac{1}{n+1}$ .

But needed inequality holds since
$\frac{n^2+n+1}{n(n+1)^2}& =\frac{n(n+1)+1}{n(n+1)^2}=\frac{n(n+1)}{n(n+1)^2} +\frac{1}{n(n+1)^2}= \\ &=\frac{1}{n+1}+\frac{1}{n(n+1)^2}\geq \frac{1}{n+1}$ .

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