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Possible Duplicate:
proof by induction: n/(n+1)

Continuing from here, I got a splendid answer that helped a lot. I'm tackling one now, but I've run into problems.

Prove by mathematical induction that

$$ \sum_{r=1}^n \frac{1}{r(r+1)} = \frac{n}{n+1} $$

Since I'm not all that conversant with Tex. . . I got to . . .

$$ \frac{2}{n(n+2)} $$

How I got there is through fractional addition, expanding then factoring. I'm stuck now. Thanks. I would prefer simple answers.

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    Did you read through Bill's answer there about Telescopy and its usefulness? This one does not require induction if you knew the technique of telescopy.2012-04-22
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    @KannappanSampath That is too advanced. It's not a duplicate of that one.2012-04-22
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    That is not too advanced at all. In fact, you have written what the OP wrote out in a long form. The 3 answers there are, explain what you'd do in the form of hints. They should suffice if you mock what Asaf wrote out more explicitly in [his answer](http://math.stackexchange.com/a/134922/21436) to your last question.2012-04-22
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    I don't know where you got 2/(n(n+2)) from. Inductive step should be: Assume the sum from r=1 to n does equal n/(n+1). Then the sum from r=1 to n+1 equals (sum from r=1 to n) + 1/((n+1)(n+2)) = n/(n+1) + 1/((n+1)(n+2)) = ... (try and fill in the rest)2012-04-22
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    I think you made some arithmetic error. Maybe you can post your train of thought?2012-04-22
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    Thanks @AdamRubinson I'm currently on it.2012-04-22
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    Ahh! Found the error. Would get back to Adam and @sdcvvc.2012-04-22
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    Yeah. Got to (n+1)/(n+2)2012-04-22
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    Thanks everyone who contributed positively.2012-04-22

3 Answers 3

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Hint $\ $ By telescopic induction the proof reduces to the elementary school calculation

$$\rm F(n)\:\! =\:\! \frac{n}{n+1}\ \ \Rightarrow\ \ F(n) - F(n-1) \!\:=\!\: \frac{1}{n(n+1)}$$

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Say we want to prove $$\sum_{r=1}^n f(r)=g(n)$$

Induction can be used to prove this equality by confirming the following two steps:

$f(1)=g(1)$

$g(n+1)-g(n)=f(n+1)$


In this case $f(r)=\frac{1}{r(r+1)}$ and $g(n)=\frac{n}{n+1}$.

$f(1)=g(1)=1/2$ so the first step is true. To check the second step:

$$g(n+1)-g(n)=\frac{n+1}{n+2}-\frac{n}{n+1} = \frac{1}{(n+1)(n+2)} = f(n+1)$$

The second step is true, thus confirming the equality.

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    This is essentially a duplicate of [an answer](http://math.stackexchange.com/a/11842/7850) to a question of which *this* question is a duplicate...2012-04-22
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    @TheChaz This whole question seems to be a duplicate, you are right.2012-04-22
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    @TheChaz And the linked answer is precisely [telescopic induction.](http://math.stackexchange.com/a/134925/242)2012-04-22
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First show it is true for $n=1$

Then assume it is true up to $n-1$ and show that $$\sum_{r=1}^n \frac{1}{r(r+1)} = \frac{1}{n(n+1)} + \sum_{r=1}^{n-1} \frac{1}{r(r+1)} = \frac{1}{n(n+1)} + \frac{n-1}{n}= \frac{n}{n+1}.$$

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    Note your RHS is $\rm\:\dfrac{1}{n(n+1)} + F(n-1) = F(n)\:$ i.e. it's *telescopy*, as in my answer.2012-04-22