Difficulties in a proof by mathematical induction (2)

Question

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.

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. — , Apr 22 '12 at 00:14
@KannappanSampath That is too advanced. It's not a duplicate of that one. — Mob, Apr 22 '12 at 00:18
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. — , Apr 22 '12 at 00:21
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 00:23
I think you made some arithmetic error. Maybe you can post your train of thought? — sdcvvc, Apr 22 '12 at 00:24

score 2 · Accepted Answer · edited Apr 13 '17 at 12:19

2

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

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

edited Apr 13 '17 at 12:19

Community

1

answered Apr 22 '12 at 00:28

Bill Dubuque

272,048

Argon · Answer 2 · 2012-04-22T00:53:29.027

2

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.

edited Apr 22 '12 at 00:53

answered Apr 22 '12 at 00:42

Argon

25,303

1

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 00:49
@TheChaz This whole question seems to be a duplicate, you are right. – Argon Apr 22 '12 at 00:50
1

@TheChaz And the linked answer is precisely telescopic induction. – Bill Dubuque Apr 22 '12 at 01:23

score 1 · Answer 3 · answered Apr 22 '12 at 00:36

1

First show it is true for $n=1$

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

answered Apr 22 '12 at 00:36

Henry

157,058

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 00:44

Difficulties in a proof by mathematical induction (2)

3 Answers3