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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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    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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    @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