# Difficulties in a proof by mathematical induction (2) [duplicate]

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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## marked as duplicate by Kannappan Sampath, The Chaz 2.0, Henry, Henning Makholm, t.b.Apr 22 '12 at 1:04

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. –  user21436 Apr 22 '12 at 0:14
@KannappanSampath That is too advanced. It's not a duplicate of that one. –  Mob Apr 22 '12 at 0:18
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 to your last question. –  user21436 Apr 22 '12 at 0:21
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) –  Adam Rubinson Apr 22 '12 at 0:23
I think you made some arithmetic error. Maybe you can post your train of thought? –  sdcvvc Apr 22 '12 at 0:24

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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Thank You. . . . . –  Mob Apr 22 '12 at 1:21

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 to a question of which this question is a duplicate... –  The Chaz 2.0 Apr 22 '12 at 0:49
@TheChaz This whole question seems to be a duplicate, you are right. –  Argon Apr 22 '12 at 0:50
@TheChaz And the linked answer is precisely telescopic induction. –  Bill Dubuque Apr 22 '12 at 1:23

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. –  Bill Dubuque Apr 22 '12 at 0:44