Loser66
  • Loser66
Use induction to prove \(\sum_{k=1}^n \dfrac{1}{k^2}\leq 2-\dfrac{1}{n}\) for all positive integers n. Please, help
Mathematics
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SOLVED
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katieb
  • katieb
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Loser66
  • Loser66
HS: \(\sum_{k=1}^n \dfrac{1}{k^2}\leq 2 - \dfrac{1}{n}\) Induction step \(\sum_{k=1}^{n+1} \dfrac{1}{k^2}= \sum_{k=1}^n \dfrac{1}{k^2}+\dfrac{1}{(n+1)^2}\leq 2-\dfrac{1}{(n+1)^2}\)
Loser66
  • Loser66
oh, the last one is not ^2 \(\sum_{k=1}^{n+1} \dfrac{1}{k^2}= \sum_{k=1}^n \dfrac{1}{k^2}+\dfrac{1}{(n+1)^2}\leq 2-\dfrac{1}{(n+1)}\)
Loser66
  • Loser66
This is my attempt: the first part of the sum is \(\leq 2 - (1/n)\) , hence the LHS is \(\leq 2 -(1/n) + (1/n+1)^2\)

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Loser66
  • Loser66
So, we need prove \(2-\dfrac{1}{n} +\dfrac{1}{(n+1)^2}\leq 2- \dfrac{1}{n+1}\)
Loser66
  • Loser66
Then I stuck
ganeshie8
  • ganeshie8
clear the denominators and simplify
ganeshie8
  • ganeshie8
it simplifies nicely and you end up with \(1\ge 0\)
Loser66
  • Loser66
Yes, I did, but those are denominators, hence when I take reciprocal, the signs switch around.
Loser66
  • Loser66
oh, let me try again. :)
ganeshie8
  • ganeshie8
\(2-\dfrac{1}{n} +\dfrac{1}{(n+1)^2}\leq 2- \dfrac{1}{n+1}\) \(\iff\) \(\dfrac{1}{n} -\dfrac{1}{(n+1)^2}\ge \dfrac{1}{n+1}\) \(\iff\) \((n+1)^2-n\ge n(n+1)\)
Loser66
  • Loser66
Thanks a lot.
Loser66
  • Loser66
I got it. Thank you so much
ganeshie8
  • ganeshie8
np

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