Proof Using General principle of math induction

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Homework Help Overview

The problem involves proving the inequality 1 + 1/4 + 1/9 + ... + 1/n^2 ≤ 2 - 1/n for every positive integer n, utilizing the principle of mathematical induction.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to establish a base case for n=1 and then formulates an induction hypothesis. They express uncertainty about how to proceed from their induction step.
  • Some participants question the validity of simply replacing n with k+1 without demonstrating the necessary steps to connect the two forms of the inequality.
  • Another participant raises a concern about the interpretation of the series notation, clarifying that the ellipsis indicates continuation of the series rather than a finite sum.

Discussion Status

Contextual Notes

Participants note that the problem requires careful handling of the induction hypothesis and the proper interpretation of the series notation. There is an emphasis on the need for clarity in the steps taken during the proof.

kolley
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Homework Statement



prove that 1+1/4+1/9+...+1/n^2< or = 2-1/n for every positive integer n

Homework Equations





The Attempt at a Solution



proved it was correct for n=1, then replaced the n with k, changed it to k+1 to get:

1/(k+1)^2 < or = 2-1/(k+1)

don't know how to proceed
 
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kolley said:

Homework Statement



prove that 1+1/4+1/9+...+1/n^2< or = 2-1/n for every positive integer n

Homework Equations





The Attempt at a Solution



proved it was correct for n=1, then replaced the n with k, changed it to k+1 to get:

1/(k+1)^2 < or = 2-1/(k+1)

don't know how to proceed
Note: <= means "less than or equal to." You don't need to write < or =.
You're not supposed to show that 1/(k+1)^2 <= 2-1/(k+1). You need to show that 1+1/4+1/9+...+1/(k + 1)^2 <= 2 - 1/(k + 1).

Your induction hypothesis is 1+1/4+1/9+...+1/k^2 <= 2 - 1/k. How can you get from this statement to the one you want to prove?
 
Sorry, I left out part of mine. I had 49/36 +1/(k+1)^2 <= 2-1/(k+1)

since 1+1/4+1/9 is equal to 49/36, is this correct or am I still on the wrong track?
 
Your part where u show the base case is correct but u can't just replace k with k+1. otherwise it would be a tautology not a proof. you have to show you can get it into the form where 1+1/4+1/9+...+1/k+1/(k+1) <= 2 - 1/(k+1)

Here is a simple proof: Show by induction that

<br /> 1+2+3+...+n = \frac{n*(n+1)}{2}<br /> for n = 1 1*(1+1) = \frac{2}{2} = 1. so this is true for the base case.<br />

now using rules of algebra if we add to one side, we add to the other, so

<br /> 1+2+3+..+n+(n+1) = \frac{n*(n+1)}{2} + (n+1) = \frac{n^2+3n+2}{2} = \frac{(n+1)*(n+2)}{2}= \frac{(n+1)*((n+1)+1)}{2}<br />

So by induction, we are done.
 
Last edited:
kolley said:
Sorry, I left out part of mine. I had 49/36 +1/(k+1)^2 <= 2-1/(k+1)

since 1+1/4+1/9 is equal to 49/36, is this correct or am I still on the wrong track?
You're still on the wrong track.

1+1/4+1/9+...+1/n^2 does not mean 1 + 1/4 + 1/9 + 1/n^2. The ellipsis - the three dots -- means "continuing in the same fashion." IOW, it means 1 + 1/4 + 1/9 + 1/16 + 1/25 + ... and so on, up to 1/n^2 for whatever value n happens to be.
 

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