# Prove by Induction.

1. Jan 29, 2008

### PFStudent

1. The problem statement, all variables and given/known data
1.1.2

Write,

$${\frac{1}{1\cdot2}} + {\frac{1}{2\cdot3}} + {\frac{1}{3\cdot4}} + . . . + {\frac{1}{99\cdot100}}$$

as a fraction in lowest terms.

3. The attempt at a solution

Rewriting the problem as a summation,

$${{\sum_{i = 1}^{n}}{\frac{1}{i(i+1)}}} = {\frac{1}{1\cdot2}} + {\frac{1}{2\cdot3}} + {\frac{1}{3\cdot4}} + . . . + {\frac{1}{n\cdot(n+1)}}$$

Then considering the first few terms,

$${\frac{1}{1\cdot2}} = \frac{1}{2}$$

$${\frac{1}{1\cdot2}} + {\frac{1}{2\cdot3}} = \frac{2}{3}$$

$${\frac{1}{1\cdot2}} + {\frac{1}{2\cdot3}} + {\frac{1}{3\cdot4}} = \frac{3}{4}$$

This leads to the conjecture that,

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

How would I prove the above by induction?

-PFStudent

Last edited: Jan 29, 2008
2. Jan 29, 2008

### rock.freak667

Assume the statement is true for n=N, then prove true for n=N+1

EDIT:

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

3. Jan 29, 2008

### jdavel

the last term in your sum (where i = n) is going to be 1/n(n+1). if you take the sum one further (to i = n+1) what will the last term be now?

4. Jan 29, 2008

### PFStudent

Hey,

Thanks for the reply rock.freak667 and jdavel.

Assume,

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

is true for k. That is,

$${{\sum_{i = 1}^{k}}{\frac{1}{i(i+1)}}} = {\frac{k}{k+1}}$$

Then to prove it is true for k+1, consider the following,

$${{\sum_{i = 1}^{k+1}}{\frac{1}{i(i+1)}}} = {{\sum_{i = 1}^{k}}{\frac{1}{i(i+1)}}} + {\frac{1}{k+1((k+1)+1)}}$$

Which reduces to,

$${{\sum_{i = 1}^{k+1}}{\frac{1}{i(i+1)}}} = {\frac{k}{k+1}} + {\frac{1}{{k^2}+3k+1)}}$$

However, how do I know the above is true?

In other words how do I know if the proof by induction actually proved it?

Thanks,

-PFStudent

5. Jan 29, 2008

### rock.freak667

Well you basically want to get

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

basically the sum is the same but instead of "n" put k+1

6. Jan 29, 2008

### jdavel

you're missing a pair of parenetheses in your 3rd equation that's leading to an error in the last term of your 4th equation. what should that last term be?

7. Jan 29, 2008

### PFStudent

Hey,

Consider the following,

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

Assume it is true for k,

$${{\sum_{i = 1}^{k}}{\frac{1}{i(i+1)}}} = {\frac{k}{k+1}}$$

Now, if it is true for k+1 the result expected is,

$${{\sum_{i = 1}^{k+1}}{\frac{1}{i(i+1)}}} = {\frac{k+1}{(k+1)+1}}$$

$${{\sum_{i = 1}^{k+1}}{\frac{1}{i(i+1)}}} = {\frac{k+1}{k+2}}$$

Then to prove that, consider the following,

$${{\sum_{i = 1}^{k+1}}{\frac{1}{i(i+1)}}} = {{\sum_{i = 1}^{k}}{\frac{1}{i(i+1)}}} + {\frac{1}{(k+1)((k+1)+1)}}$$

$${{\sum_{i = 1}^{k+1}}{\frac{1}{i(i+1)}}} = {\left({\frac{k}{k+1}}\right)} + {\frac{1}{(k+1)(k+2)}}$$

$${{\sum_{i = 1}^{k+1}}{\frac{1}{i(i+1)}}} = {{\frac{k(k+2)}{(k+1)(k+2)}}} + {\frac{1}{(k+1)(k+2)}}$$

Which reduces to,

$${{\sum_{i = 1}^{k+1}}{\frac{1}{i(i+1)}}} = {\frac{(k+1)(k+1)}{(k+1)(k+2)}}$$

$${{\sum_{i = 1}^{k+1}}{\frac{1}{i(i+1)}}} = {\frac{k+1}{k+2}}$$

Proof.

Thanks,

-PFStudent

Last edited: Jan 29, 2008
8. Jan 30, 2008

### HallsofIvy

Staff Emeritus
Were you required to use induction?

Since
$$\frac{1}{n(n+1)}= \frac{1}{n}- \frac{1}{n+1}$$
that is a "telescoping" series and the sum is immediate.

9. Jan 30, 2008

### PFStudent

Hey,

Could I have proved,

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

by showing that it is a telescoping series? If so, how?

Thanks,

-PFStudent

10. Jan 30, 2008

### HallsofIvy

Staff Emeritus
$$\sum_{i=1}^{100} \frac{1}{i(i+1}= \frac{1}{1\cdot2}} + {\frac{1}{2\cdot3}} + {\frac{1}{3\cdot4}} + . . . + {\frac{1}{99\cdot100}$$
$$= (\frac{1}{1}- \frac{1}{2})+ (\frac{1}{2}-\frac{1}{3})+ (\frac{1}{3}-\frac{1}{4})+ \cdot\cdot\cdot+ (\frac{1}{99}- \frac{1}{100})$$
The last fraction in each pair cancels the first fraction in the next pair (except of course in the last pair). Every fraction except the first and last cancel so the sum is
$$\frac{1}{1}- \frac{1}{100}= \frac{100-1}{100}= \frac{99}{100}$$

More generally,
$$\sum_{i= 1}^n \frac{1}{i(i+1)}= \frac{1}{1}- \frac{1}{n+1}= \frac{n+1-1}{n+1}= \frac{n}{n+1}$$

11. Feb 4, 2008

### PFStudent

Hey,

Thanks HallsofIvy for showing how it could have been proved by showing it was a telescoping sum.

However, how would you know that the individual sums can be rewritten as the sum or difference of two fractions. In other words, how did you figure out the following,

$$\sum_{i=1}^{100} \frac{1}{i(i+1}= \frac{1}{1\cdot2}} + {\frac{1}{2\cdot3}} + {\frac{1}{3\cdot4}} + . . . + {\frac{1}{99\cdot100} = \left(\frac{1}{1}-\frac{1}{2}\right) + \left(\frac{1}{2}-\frac{1}{3}\right)+ \left(\frac{1}{3}-\frac{1}{4}\right) + \cdot\cdot\cdot + \left(\frac{1}{99}- \frac{1}{100}\right)$$

That is, how did you figure out that,

$${{\sum_{i = 1}^{n}}{\frac{1}{i(i+1)}}} = {\left({{\frac{A_{1}}{1}}-{\frac{B_{1}}{2}}}\right)} + {\left({{\frac{A_{2}}{2}}-{\frac{B_{2}}{3}}}\right)} +. . . + {\left({{\frac{A_{n}}{n}}-{\frac{B_{n}}{n+1}}}\right)}$$

Additionally, is finding the above the same as using the technique of partial fractions?

Also, what is the general way of finding $$A$$ and $$C$$ for the following (where: $$B, D, E, and{{.}}F$$; are all given),

$${\frac{E}{F}} = {{{\frac{A}{B}}\pm{\frac{C}{D}}}}$$

Where it can be shown that,

$${E} = {AD \pm BC}$$

$${F} = {BD}$$

Which is the same as,

$${\frac{E}{F}} = {\frac{AD \pm BC}{BD}}$$

Thanks,

-PFStudent

Last edited: Feb 4, 2008
12. Feb 6, 2008

### PFStudent

Hey,

I was looking over this and realized that the following,

$${{\sum_{i = 1}^{n}}{\frac{1}{i(i+1)}}} = {\left({{\frac{A_{1}}{1}}-{\frac{B_{1}}{2}}}\right)} + {\left({{\frac{A_{2}}{2}}-{\frac{B_{2}}{3}}}\right)} +. . . + {\left({{\frac{A_{n}}{n}}-{\frac{B_{n}}{n+1}}}\right)}$$

Was better written as,

$${{\sum_{i = 1}^{n}}{\frac{1}{i(i+1)}}} = {\left({{\frac{C_{1}}{1}}-{\frac{C_{2}}{2}}}\right)} + {\left({{\frac{C_{2}}{2}}-{\frac{C_{3}}{3}}}\right)} + {\cdot} {\cdot} {\cdot} + {\left({{\frac{C_{n-1}}{n-1}}-{\frac{C_{n}}{n}}}\right)} + {\left({{\frac{C_{n}}{n}}-{\frac{C_{n+1}}{n+1}}}\right)}$$

But, I'm still having some trouble figuring how would one decipher what the constants "C" must be.

Thanks,

-PFStudent

13. Feb 6, 2008

### rock.freak667

$$\sum_{n=1} ^{N} \frac{1}{n(n+1)} \equiv \sum_{n=1} ^{N} (\frac{1}{n}-\frac{1}{n+1})$$

Use partial fractions on 1/n(n+1)

$$\sum_{n=1} ^{N} \frac{1}{n}-\frac{1}{n+1}$$

then input n=1,2,3,...,N-2,N-1,N. then add them all up.

14. Feb 6, 2008

### PFStudent

Hey,

I see that however, how would you have figured out what the constants for the partial fractions of,

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

would be?

In other words, how would you have solved the following,

$${{\sum_{i = 1}^{n}}{\frac{1}{i(i+1)}}} = {\left({{\frac{C_{1}}{1}}-{\frac{C_{2}}{2}}}\right)} + {\left({{\frac{C_{2}}{2}}-{\frac{C_{3}}{3}}}\right)} + {\cdot} {\cdot} {\cdot} + {\left({{\frac{C_{n-1}}{n-1}}-{\frac{C_{n}}{n}}}\right)} + {\left({{\frac{C_{n}}{n}}-{\frac{C_{n+1}}{n+1}}}\right)}$$

for: $$C_{1}, C_{2}, C_{3},..., C_{n-1}, C_{n}$$?

That is where I am stuck.

Thanks,

-PFStudent

15. Feb 6, 2008

### rock.freak667

I don't think you need to do all of that.

16. Feb 11, 2008

### PFStudent

Hey,

Thanks for the reply, rock.freak667. Your right I did not need to prove that it is a telescoping sum.

However, I still would like to know how would one have realized that,

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

is a telescoping sum?

Further after the above realization one would have had to solve for the constants: $$C_{1}, C_{2}, C_{3},..., C_{n-1}, C_{n}$$; for the following,

$${{\sum_{i = 1}^{n}}{\frac{1}{i(i+1)}}} = {\left({{\frac{C_{1}}{1}}-{\frac{C_{2}}{2}}}\right)} + {\left({{\frac{C_{2}}{2}}-{\frac{C_{3}}{3}}}\right)} + {\cdot} {\cdot} {\cdot} + {\left({{\frac{C_{n-1}}{n-1}}-{\frac{C_{n}}{n}}}\right)} + {\left({{\frac{C_{n}}{n}}-{\frac{C_{n+1}}{n+1}}}\right)}$$

And that is, what I wanted to know. How do you do that?

Thanks,

-PFStudent

Last edited: Feb 11, 2008