## Loser66 one year ago Use induction to prove $$\sum_{k=1}^n \dfrac{1}{k^2}\leq 2-\dfrac{1}{n}$$ for all positive integers n. Please, help

1. Loser66

HS: $$\sum_{k=1}^n \dfrac{1}{k^2}\leq 2 - \dfrac{1}{n}$$ Induction step $$\sum_{k=1}^{n+1} \dfrac{1}{k^2}= \sum_{k=1}^n \dfrac{1}{k^2}+\dfrac{1}{(n+1)^2}\leq 2-\dfrac{1}{(n+1)^2}$$

2. Loser66

oh, the last one is not ^2 $$\sum_{k=1}^{n+1} \dfrac{1}{k^2}= \sum_{k=1}^n \dfrac{1}{k^2}+\dfrac{1}{(n+1)^2}\leq 2-\dfrac{1}{(n+1)}$$

3. Loser66

This is my attempt: the first part of the sum is $$\leq 2 - (1/n)$$ , hence the LHS is $$\leq 2 -(1/n) + (1/n+1)^2$$

4. Loser66

So, we need prove $$2-\dfrac{1}{n} +\dfrac{1}{(n+1)^2}\leq 2- \dfrac{1}{n+1}$$

5. Loser66

Then I stuck

6. ganeshie8

clear the denominators and simplify

7. ganeshie8

it simplifies nicely and you end up with $$1\ge 0$$

8. Loser66

Yes, I did, but those are denominators, hence when I take reciprocal, the signs switch around.

9. Loser66

oh, let me try again. :)

10. ganeshie8

$$2-\dfrac{1}{n} +\dfrac{1}{(n+1)^2}\leq 2- \dfrac{1}{n+1}$$ $$\iff$$ $$\dfrac{1}{n} -\dfrac{1}{(n+1)^2}\ge \dfrac{1}{n+1}$$ $$\iff$$ $$(n+1)^2-n\ge n(n+1)$$

11. Loser66

Thanks a lot.

12. Loser66

I got it. Thank you so much

13. ganeshie8

np