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Proof by induction (summation)

  • Thread starter marksyncm
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  • #1
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Homework Statement



Prove by induction that ##\sum\limits_{k=1}^{2n} \frac{1}{k(k+1)} = \frac{2n}{2n+1}##

2. The attempt at a solution

First I showed that it is true for ##n=1## and ##n=2##. Then, assuming it is true for all ##n##, I attempt to show that it is true for ##n+1##:

$$\sum\limits_{k=1}^{2(n+1)} \frac{1}{k(k+1)} = \frac{2(n+1)}{2(n+1)+1} = ... $$

My problem is what the right side (##...##) of the equation should be. I can't just say that it is equal to ##\frac{2n}{2n+1} + \frac{1}{(n+1)(n+2)}##, because the summation notation requires me to double my ##n## - I understand this to mean that each time we add 1 to ##n##, we are adding two additional terms to the summation total - one is ##2(n+1)-1## and the other is ##2(n+1)##. Based on this reasoning, I concluded that the above equation should be:

$$\sum\limits_{k=1}^{2(n+1)} \frac{1}{k(k+1)} = \frac{2(n+1)}{2(n+1)+1} = \frac{2n}{2n+1} + \frac{1}{(2n+1)(2n+2)} + \frac{1}{(2n+2)(2n+3)}$$

However, I was unable to reduce the right side to resemble the left side, and online algebra tools were of no help. Am I going about this wrong?
 
Last edited:

Answers and Replies

  • #2
RPinPA
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I believe this is the source of your problems:
Look at the expression $$\frac{2n}{2n+1}$$. Now replace ##n## by ##(n+1)##. That gives you $$\frac{2(n+1)}{2(n+1)+1} = \frac{2n+2}{2n+3}$$

You had ##(2n+2)/(2n+4)##.

If you're still uncertain what's going on, consider this. The general expression ##2n/(2n+1)## is an even divided by an odd. Your expression is an even divided by an even.
 
  • #3
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Sorry, this was actually a problem in my LaTeX - I have now corrected this. In my calculations, I had used ##\frac{2(n+1)}{2(n+1)+1}## and was unable to get the right side of the equation to resemble this, even using Mathway.
 
  • #4
RPinPA
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That doesn't line up with the rest of your calculations, as that denominator is not ##2(n+2) = 2n+4##. Look at the last line of your original question. What is the denominator on the left hand side?

You can indeed show that if you combine your three fractions over the common denominator ##(2n + 1)(2n + 2)(2n + 3)##, the numerator reduces to ##(2n + 1)(2n + 2)^2## and so the fraction reduces to ##(2n + 2)/(2n+3)##, the desired result.
 
  • #5
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That doesn't line up with the rest of your calculations, as that denominator is not ##2(n+2) = 2n+4##. Look at the last line of your original question. What is the denominator on the left hand side?
Apologies, I'm not sure I see which part you are referring to. I'm not sure I see where it still says ##2n+4## in a denominator in my original question?
 
  • #6
RPinPA
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Apologies, I'm not sure I see which part you are referring to. I'm not sure I see where it still says ##2n+4## in a denominator in my original question?
Ah, OK, it's been edited. I was still seeing ##2(n+2)## when I wrote that. So once more, your left hand side can be rewritten as (2n + 2)/(2n + 3) and it is indeed possible and not too messy to reduce the right-hand side to that form.
 
  • #7
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Ok, thank you - now that I know that my approach is correct, I can keep cracking on the reduction. Appreciated!
 

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