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Simple mathematical induction

  1. Jan 23, 2013 #1
    Prove that:

    1 + 1/√2 +...+ 1/√n < 2√n

    Work:

    So I've done the base case of n = 1, and I've set up the Indutive hypothesis as assuming n=k as 1 + 1/√2 +...+ 1/√k < 2√k

    and for the inductive step:

    1 + 1/√2 +...+ 1/√k + 1/√(k+1) < 2√k + 1/√(k+1)

    now I'm having trouble trying to manipulate 2√k + 1/√(k+1) to get 2√(k+1) alone. I tried a conjugate of 1/[2 √(k+1) + 2√k] [2 √(k+1) - 2√k / [2 √(k+1) - 2√k].......the closest thing I got was 2√(k+1) / 4.

    ANy suggestions?
     
  2. jcsd
  3. Jan 23, 2013 #2

    jbunniii

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    Here is a hint which should be useful:
    $$4(k^2 + k) < 4k^2 + 4k + 1 = (2k + 1)^2$$
    Taking the square root of both sides, which is OK because both sides are positive, we get
    $$2\sqrt{k^2 + k} < 2k + 1$$
    See if you can manipulate the right hand side of your inequality into a form where you can apply this.
     
    Last edited: Jan 23, 2013
  4. Jan 23, 2013 #3

    Ray Vickson

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    So, you want to show [tex] 2 \sqrt{k+1} - 2 \sqrt{k} \geq \frac{1}{\sqrt{k+1}}.[/tex] Re-write the LHJS using the type of trick you applied above.
     
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