# Show that $\sum_{i=1}^n \frac{1}{i^2} \le 2 - \frac{1}{n}$

So I am able to calculate the given problem and prove $P(K) \implies P(k + 1)$; it's been sometime since I did proofs and I perform my steps I get what Wolfram Alpha shows as an alternate solution.

Any help is greatly appreciated

The problem is the following:

Show that

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

What I have:

$P(1)$: $$\sum_{i=1}^n \frac{1}{i^2} \le 2 - \frac{1}{n}$$

Replace n with 1

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

Conclusion $$1 \le 1$$

Prove: $$\sum_{i=1}^{k+1} \frac{1}{i^2} \le 2 - \frac{1}{k+1}$$

P(K) Assume $$\sum_{i=1}^k \frac{1}{i^2} \le 2 - \frac{1}{k}$$

$P(K) \implies P(k + 1)$

Performed Steps:

Working the LHS to match RHS

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

Edit: Fixed error on regrouping

$$2 - \left[\frac{1}{k} - \frac{1}{(k+1)^2}\right]$$

Work the fractions

$$2 - \left[\frac{1}{k} \frac{(k+1)^2}{(k+1)^2} - \frac{1}{(k+1)^2} \frac{k}{k} \right]$$

$$2 - \left[\frac{(k+1)^2 - k}{k(k+1)^2} \right]$$

$$2 - \left[\frac{k^2 + 2k + 1 - k}{k(k+1)^2} \right]$$

$$2 - \left[\frac{k^2 + k + 1}{k(k+1)^2} \right]$$

$$2 - \left[\frac{k(k+1) + 1}{k(k+1)^2} \right]$$

$$2 - \left[\frac{k(k+1)}{k(k+1)^2} + \frac{1}{k(k+1)^2} \right]$$

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

EDIT: I fixed my mistake of my regrouping and signs; had completely missed the regrouping.

This is the final step I got to. I am hung on where to go from here. The answers given have been really helpful and I'm happy with them. I'd just like to know the mistake I made or next step I should take.

$$\sum_{i=1}^{k+1} \frac{1}{i^2} \le 2 - \frac{1}{k+1}$$

Thanks for the help

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Illegible please try to use Latex. –  Sami Ben Romdhane Jan 14 at 18:54
Yeah I am trying to fix it, not sure what happened –  CoffeeMuncher Jan 14 at 18:55
I have edited it. Please confirm that this is correct. It seems the \over command does not work with MathJax. –  Cameron Williams Jan 14 at 18:58
Thanks Cameron looks correct. –  CoffeeMuncher Jan 14 at 18:59
Maybe a duplicate of math.stackexchange.com/questions/351166/… –  Josué Tonelli-Cueto Jan 14 at 19:23

$$\sum_{i=1}^n\frac1{i^2}=1+\sum_{i=2}^n\frac1{i^2}\leqslant1+\sum_{i=2}^n\frac1{i(i-1)}=1+\sum_{i=2}^n\left(\frac1{i-1}-\frac1i\right)=\ldots$$

Edit: (About the Edit to the question 2014-01-14 21:25:21)

I'd just like to know the mistake I made or next step I should take.

None, neither mistake nor next step. Actually, what you did yields the result since you proved that $$\sum_{i=1}^{k+1}\frac1{i^2}\leqslant 2 - \frac{1}{(k+1)} - \frac{1}{k(k+1)^2},$$ which implies $P(k+1)$ since $$2 - \frac{1}{(k+1)} - \frac{1}{k(k+1)^2}\leqslant2 - \frac{1}{(k+1)}.$$ Well done.

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(+1) This is the solution I tried to reall –  no identity Jan 14 at 20:25
Thanks! I was very set on getting the solution my professor gave instead of reviewing my own solution –  CoffeeMuncher Jan 15 at 4:56

You want to prove that $$P(k):\qquad \sum_{i=1}^k{1\over i^2}\leq 2-{1\over k}$$ implies $$P(k+1):\qquad \sum_{i=1}^{k+1}{1\over i^2}\leq 2-{1\over k+1}\ .$$ Therefore we have to prove that $$\left(2-{1\over k}\right)+{1\over (k+1)^2}\leq 2-{1\over k+1}\ ,$$ which is the same as $${1\over (k+1)^2}\leq {1\over k}-{1\over k+1}\quad\left(={1\over k(k+1)}\right)\ .$$ But this is obvious.

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You can approximate the sum by an integral to obtain the inequality $$\sum_{i=1}^n\frac1{i^2}=1+\sum_{i=2}^n\frac1{i^2}\le1+\int_1^n\frac1{x^2}\mathrm dx=2-\frac1n.$$

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