Question

Use induction to prove
\(\sum_{k=1}^n \dfrac{1}{k^2}\leq 2-\dfrac{1}{n}\) for all positive integers n.
Please, help

Answer 1

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

Induction step

\(\sum_{k=1}^{n+1} \dfrac{1}{k^2}= \sum_{k=1}^n \dfrac{1}{k^2}+\dfrac{1}{(n+1)^2}\leq 2-\dfrac{1}{(n+1)^2}\)

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

Then I stuck

Answer 6

clear the denominators and simplify

Answer 7

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

Answer 8

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

Answer 9

oh, let me try again. :)

Answer 10

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

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

\((n+1)^2-n\ge n(n+1)\)

Answer 11

Thanks a lot.

Answer 12

I got it. Thank you so much

Answer 13

np