# Prove by Induction.

by PFStudent
Tags: induction, prove
 P: 171 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
 HW Helper P: 6,207 Assume the statement is true for n=N, then prove true for n=N+1 EDIT: Your formula should be $$\sum_{i=1} ^{n} \frac{1}{i(i+1)}=\frac{n}{n+1}$$
 P: 618 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?
P: 171

## Prove by Induction.

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
HW Helper
P: 6,207
 Quote by PFStudent In other words how do I know if the proof by induction actually proved it?

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
 P: 618 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?
 P: 171 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
 Math Emeritus Sci Advisor Thanks PF Gold P: 38,706 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.
P: 171
Hey,

 Quote by HallsofIvy 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.
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
 Math Emeritus Sci Advisor Thanks PF Gold P: 38,706 $$\sum_{i=1}^{100} \frac{1}{i(i+1}= \frac{1}{1\cdot2}} + {\frac{1}{2\cdot3}} + {\frac{1}{3\cdot4}} + . . . + {\frac{1}{99\cdot100}$$ $$= (\frac{1}{1}- \frac{1}{2})+ (\frac{1}{2}-\frac{1}{3})+ (\frac{1}{3}-\frac{1}{4})+ \cdot\cdot\cdot+ (\frac{1}{99}- \frac{1}{100})$$ 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, $$\sum_{i= 1}^n \frac{1}{i(i+1)}= \frac{1}{1}- \frac{1}{n+1}= \frac{n+1-1}{n+1}= \frac{n}{n+1}$$
P: 171
Hey,

 Quote by HallsofIvy $$\sum_{i=1}^{100} \frac{1}{i(i+1}= \frac{1}{1\cdot2}} + {\frac{1}{2\cdot3}} + {\frac{1}{3\cdot4}} + . . . + {\frac{1}{99\cdot100} = (\frac{1}{1}- \frac{1}{2})+ (\frac{1}{2}-\frac{1}{3})+ (\frac{1}{3}-\frac{1}{4})+ \cdot\cdot\cdot+ (\frac{1}{99}- \frac{1}{100})$$ 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, $$\sum_{i= 1}^n \frac{1}{i(i+1)}= \frac{1}{1}- \frac{1}{n+1}= \frac{n+1-1}{n+1}= \frac{n}{n+1}$$
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
 P: 171 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
 HW Helper P: 6,207 $$\sum_{n=1} ^{N} \frac{1}{n(n+1)} \equiv \sum_{n=1} ^{N} (\frac{1}{n}-\frac{1}{n+1})$$ 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.
P: 171
Hey,

 Quote by 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.
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
 HW Helper P: 6,207 I don't think you need to do all of that.
P: 171
Hey,

 Quote by rock.freak667 I don't think you need to do all of that.
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

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