# Inductive proof of summation formula

livcon

## Homework Statement

Prove by induction the following summation formula:
$$\frac{1}{1\1*2} + \frac{1}{2*3} + ... + \frac{1}{n(n+1)} = 1 - \frac{1}{n+1}$$
$$n \geq 1$$

-

## The Attempt at a Solution

Inductive step:
1. $$\frac{1}{1*2} + \frac{1}{2*3} + ... + \frac{1}{n(n+1)} + \frac{1}{(n+1)((n+1)+1)} = 1 - \frac{1}{(n+1)+1}$$

2. $$\frac{1}{1*2} + \frac{1}{2*3} + ... + \frac{1}{n(n+1)} + \frac{1}{(n+1)((n+1)+1)} = 1 - \frac{1}{n+1} + \frac{1}{(n+1)(n+2)}$$

To prove this (1) should equal (2) (right?), but I don't see how to manipulate (2) to make it equal to (1)

whs
Common denominator!

Staff Emeritus
Gold Member
but I don't see how to manipulate (2) to make it equal to (1)
You don't need to. All you need to do is to show these two sums of fractions are equal.

livcon
That's right, but in this case I've actually been staring at the problem for a while, so I'm really interested in how to do the algebra in (2). Thanks

Staff Emeritus
Gold Member
As a warmup exercise, prove the following:

$$\frac{1}{3} + \frac{1}{6} = \frac{3}{2} - 1$$

emyt
That's right, but in this case I've actually been staring at the problem for a while, so I'm really interested in how to do the algebra in (2). Thanks

see how 1 - 1/n+1 looks first

livcon
I appreciate the help, but I'm afraid I'm still blind here, proving the equivalence in the warmup exercise is easy Hurkyl, but I just don't see it in my problem.

Staff Emeritus
Gold Member
I appreciate the help, but I'm afraid I'm still blind here, proving the equivalence in the warmup exercise is easy Hurkyl, but I just don't see it in my problem.
Isn't proving
$$1 - \frac{1}{(n+1)+1} = 1 - \frac{1}{n+1} + \frac{1}{(n+1)(n+2)}$$​
exactly the same kind of problem, though? Why can't you do the same thing you did for the warmup problem?

livcon
I guess it's because in the warmup exercise I could easily add and subtract the fractions because it was all constants, but in the one involving the variable n I got stuck. So I would probably have an epiphany if someone would just write it out. This is just a matter of me not seeing the obvious steps in this particular problem.

$$1- \frac{1}{n+1}+ \frac{1}{(n+1)(n+2)}$$
The text passage contains a mathematical question in the form of a proof by induction for a summation formula. The question is about proving the formula:
$$\frac{1}{1*2} + \frac{1}{2*3} + ... + \frac{1}{n(n+1)} = 1 - \frac{1}{n+1}$$
for \( n \geq 1 \). The passage also includes an attempt to solve the problem and a discussion about how to manipulate the algebraic expressions to complete the proof.
single_question_with_answer

the question is clearly stated, and the text includes an attempt to solve it, which can be considered a partial answer. however, the complete algebraic manipulation to finalize the proof is not provided, so it is not a fully resolved answer.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# Inductive proof of summation formula

livcon

## Homework Statement

Prove by induction the following summation formula:
$$\frac{1}{1\1*2} + \frac{1}{2*3} + ... + \frac{1}{n(n+1)} = 1 - \frac{1}{n+1}$$
$$n \geq 1$$

-

## The Attempt at a Solution

Inductive step:
1. $$\frac{1}{1*2} + \frac{1}{2*3} + ... + \frac{1}{n(n+1)} + \frac{1}{(n+1)((n+1)+1)} = 1 - \frac{1}{(n+1)+1}$$

2. $$\frac{1}{1*2} + \frac{1}{2*3} + ... + \frac{1}{n(n+1)} + \frac{1}{(n+1)((n+1)+1)} = 1 - \frac{1}{n+1} + \frac{1}{(n+1)(n+2)}$$

To prove this (1) should equal (2) (right?), but I don't see how to manipulate (2) to make it equal to (1)

whs
Common denominator!

Staff Emeritus
Gold Member
but I don't see how to manipulate (2) to make it equal to (1)
You don't need to. All you need to do is to show these two sums of fractions are equal.

livcon
That's right, but in this case I've actually been staring at the problem for a while, so I'm really interested in how to do the algebra in (2). Thanks

Staff Emeritus
Gold Member
As a warmup exercise, prove the following:

$$\frac{1}{3} + \frac{1}{6} = \frac{3}{2} - 1$$

emyt
That's right, but in this case I've actually been staring at the problem for a while, so I'm really interested in how to do the algebra in (2). Thanks

see how 1 - 1/n+1 looks first

livcon
I appreciate the help, but I'm afraid I'm still blind here, proving the equivalence in the warmup exercise is easy Hurkyl, but I just don't see it in my problem.

Staff Emeritus
Gold Member
I appreciate the help, but I'm afraid I'm still blind here, proving the equivalence in the warmup exercise is easy Hurkyl, but I just don't see it in my problem.
Isn't proving
$$1 - \frac{1}{(n+1)+1} = 1 - \frac{1}{n+1} + \frac{1}{(n+1)(n+2)}$$​
exactly the same kind of problem, though? Why can't you do the same thing you did for the warmup problem?

livcon
I guess it's because in the warmup exercise I could easily add and subtract the fractions because it was all constants, but in the one involving the variable n I got stuck. So I would probably have an epiphany if someone would just write it out. This is just a matter of me not seeing the obvious steps in this particular problem.

$$1- \frac{1}{n+1}+ \frac{1}{(n+1)(n+2)}$$
user
The text passage contains a mathematical question in the form of a proof by induction for a summation formula. The question is about proving the formula:
$$\frac{1}{1*2} + \frac{1}{2*3} + ... + \frac{1}{n(n+1)} = 1 - \frac{1}{n+1}$$
for \( n \geq 1 \). The passage also includes an attempt to solve the problem and a discussion about how to manipulate the algebraic expressions to complete the proof.

TYPE: SINGLE_QUESTION_WITH_ANSWER

The question is clearly stated, and the text includes an attempt to solve it, which can be considered a partial answer. However, the complete algebraic manipulation to finalize the proof is not provided, so it is not a fully resolved answer.