Series Convergence Verification

In summary: You're just looking at the last term in each expression. The question is - what happens when you simplify this?
  • #1
Jbreezy
582
0

Homework Statement


Verify that the infinite series converges


Homework Equations



[itex]\Big(\sum_{n=1}^\infty\frac{1}{n(n+1)}\Big)[/itex]

The Attempt at a Solution


So I did just that and I got
1/n -1/(n+1)

So I thought to take the limit of this as n goes to infinity I got 0 as the limit. My book took the limit as n goes to infinity of 1 - 1/(n+1) which they say is equal to 1. Which I don't disagree with I don't know how they got from what I got ( and them) 1/n -1/(n+1) to 1 - 1/(n+1)
For a look at what they did see http://www.calcchat.com/book/Calculus-ETF-5e/

You have to put in the section and question though. It is section 9.2 question 29

Have a look I'm lost thanks
 
Physics news on Phys.org
  • #2
Do you have to actually find the limit or just show the convergence? Consider the comparison test if is the latter. If it is the former, they used partial sums to obtain ##1 - 1/(n+1)## from which they extracted the limit.
 
  • #3
CAF123 said:
Do you have to actually find the limit or just show the convergence? Consider the comparison test if is the latter. If it is the former, they used partial sums to obtain ##1 - 1/(n+1)## from which they extracted the limit.

They found the limit. When you do partial fractions you get 1/n -1/(n+1)
Which they get too. But they say that 1/n -1/(n+1) = ##1 - 1/(n+1)## Which I don't understand.


The step between those two.
 
  • #4
Jbreezy said:
They found the limit. When you do partial fractions you get 1/n -1/(n+1)
Which they get too. But they say that 1/n -1/(n+1) = ##1 - 1/(n+1)## Which I don't understand.

No, they are summing together ##n## partial sums to obtain ##S_n = 1 - \frac{1}{n+1}##. They then take the limit of partial sums (as the number of partial of sums tends to infinity) which is equivalent to summation you are looking for.
 
  • #5
$$S_n =\frac 1 {1\cdot 2} + \frac 1 {2\cdot 3} + \frac 1 {3\cdot 4} +\dots+ \frac 1 {n \cdot (n+1)}$$ $$
=(\frac 1 1 - \frac 1 2)+(\frac 1 2 - \frac 1 3)+(\frac 1 3 - \frac 1 4)+\dots +(\frac 1 n - \frac 1 {n+1})$$What happens to that second expression when you simplify it?
 
Last edited by a moderator:
  • #6
LCKurtz said:
$$S_n =\frac 1 {1\cdot 2} + \frac 1 {2\cdot 3} + \frac 1 {3\cdot 4} +\dots+ \frac 1 {n \cdot (n+1)}$$ $$
=(\frac 1 1 - \frac 1 2)+(\frac 1 2 - \frac 1 3)+(\frac 1 3 - \frac 1 4)+\dots +(\frac 1 n - \frac 1 {n+1})$$What happens to that second expression when you simplify it?

The second expression is just the first expression when you simplify it.

I'm confused because I don't understand where this came from 1- (1/(n+1))
Why do they take the limit of that and not the limit of 1/n - 1/(n+1)
This is my question. I wanted to take the limit of the latter. Why must you do the former and what is the reason they did it this way.
 
Last edited by a moderator:
  • #7
That's not what LCKurtz asked. What happens to the 2nd expression when you simplify it?
 
  • #8
Mark44 said:
That's not what LCKurtz asked. What happens to the 2nd expression when you simplify it?

I just answered that. I said 1/n(n+1) = 1/n -(1/n+1) Put a common denominator and the n's in the numerator cancel.
 
  • #9
Don't just go backwards. What happens when you remove the parentheses?
 
  • #10
Jbreezy said:
I just answered that. I said 1/n(n+1) = 1/n -(1/n+1) Put a common denominator and the n's in the numerator cancel.

You're not looking at the entire expressions - you're just looking at the last term in each expression. The question is - what happens when you simplify this?
$$ (\frac 1 1 - \frac 1 2)+(\frac 1 2 - \frac 1 3)+(\frac 1 3 - \frac 1 4)+\dots +(\frac 1 n - \frac 1 {n+1})$$
 
Last edited:
  • #11
You get 1/2 +...1/n -1/(n+1)
 
  • #12
Jbreezy said:
You get 1/2 +...1/n -1/(n+1)
I don't. How did you get it?
 
  • #13
Mark44 said:
You're not looking at the entire expressions - you're just looking at the last term in each expression. The question is - what happens when you simplify this?
$$ (\frac 1 1 - \frac 1 2)+(\frac 1 2 - \frac 1 3)+(\frac 1 3 - \frac 1 2)+\dots +(\frac 1 n - \frac 1 {n+1})$$

$$ (\frac 1 1 - \frac 1 2)+(\frac 1 2 - \frac 1 3)+(\frac 1 3 - \frac 1 2)+\dots +(\frac 1 n - \frac 1 {n+1})$$
Two 1/2 dissapear
= $$ (\frac 1 1 - \frac 1 3)+(\frac 1 3 - \frac 1 2)+\dots +(\frac 1 n - \frac 1 {n+1})$$

Two 1/3 dissapear

=$$ (\frac 1 1 - \frac 1 2)+\dots +(\frac 1 n - \frac 1 {n+1})$$

$$(\frac 2 2 - \frac 1 2 + \dots +(\frac 1 n - \frac 1 {n+1})$$

So you get what I said right?
 
  • #14
There was a mistake in what LCKurtz wrote, that I didn't notice when I copied and pasted it. It should have been as below. The corrected term is 1/3 - 1/4
Mark44 said:
You're not looking at the entire expressions - you're just looking at the last term in each expression. The question is - what happens when you simplify this?
$$ (\frac 1 1 - \frac 1 2)+(\frac 1 2 - \frac 1 3)+(\frac 1 3 - \frac 1 4)+\dots +(\frac 1 n - \frac 1 {n+1})$$
 
  • #15
3/4 + 1/n - (1/n+1)

right? How does this help?
 
  • #16
Jbreezy said:
3/4 + 1/n - (1/n+1)

right?
No.
$$ \frac 1 1 - \frac 1 2+\frac 1 2 - \frac 1 3+\frac 1 3 - \frac 1 4+\dots +\frac 1 n - \frac 1 {n+1}$$

I have removed all of the parentheses. When you simplify the above, what are you left with?
 
  • #17
The1/2 goes with the -1/2 and the 1/3 goes with the -1/3 leaving the 1/1 and the -1/4 combining those is 3/4 and you have 1/n -1/n+1 left like I said before. How is that wrong?
 
  • #18
You aren't getting the pattern. Here it is with some more terms.
$$(\frac 1 1 - \frac 1 2)+(\frac 1 2 - \frac 1 3)+(\frac 1 3 - \frac 1 4)+(\frac 1 4 - \frac 1 5) + (\frac 15 - \frac 1 6) + \dots +(\frac 1 n - \frac 1 {n+1})$$

This really isn't complicated. Possibly you are missing the significance of "..." which means "continues in the same fashion."
 
  • #19
Yeah I feel like this is getting away from my original question. I get the pattern it just keeps going that is that 1/n - 1/n+1 is telling me in general it continues this way. I just want to know why they take the limit of 1 - 1/(n+1)

and not the limit of 1/n - 1/n+1
 
  • #20
I don't think you get the pattern. What's the term that comes before (1/n - 1/(n + 1))?
 
  • #21
1/(n-1) - 1/n
 
  • #22
Yes, so it looks like this:
$$(\frac 1 1 - \frac 1 2)+(\frac 1 2 - \frac 1 3)+(\frac 1 3 - \frac 1 4)+(\frac 1 4 - \frac 1 5) + (\frac 1 5 - \frac 1 6) + \dots +(\frac 1 {n-1} - \frac 1 n) + (\frac 1 n - \frac 1 {n+1})$$

So when you simplify, will 1/(n - 1) be left? When you're all done, there should be only two terms left. What are they?

This is what is called a "telescoping" series, for an obvious reason.
 
  • #23
Mark44 said:
Yes, so it looks like this:
$$(\frac 1 1 - \frac 1 2)+(\frac 1 2 - \frac 1 3)+(\frac 1 3 - \frac 1 4)+(\frac 1 4 - \frac 1 5) + (\frac 1 5 - \frac 1 6) + \dots +(\frac 1 {n-1} - \frac 1 n) + (\frac 1 n - \frac 1 {n+1})$$

So when you simplify, will 1/(n - 1) be left? When you're all done, there should be only two terms left. What are they?

This is what is called a "telescoping" series, for an obvious reason.

$$(\frac 1 {n-1} - \frac 1 {n+1})$$

So, I'm still not sure why they took the limit of $$(1-\frac 1 {n+1})$$
 
  • #24
Jbreezy said:
$$(\frac 1 {n-1} - \frac 1 {n+1})$$
No.
Jbreezy said:
So, I'm still not sure why they took the limit of $$(1-\frac 1 {n+1})$$
Let me make things very simple. How about this?
$$(\frac 1 1 - \frac 1 2)+(\frac 1 2 - \frac 1 3)+(\frac 1 3 - \frac 1 4)+(\frac 1 4 - \frac 1 5) + (\frac 1 5 - \frac 1 6)$$

What does the above simplify to?

How about now?

$$(\frac 1 1 - \frac 1 2)+(\frac 1 2 - \frac 1 3)+ \dots + (\frac 1 5 - \frac 1 6)$$
 
  • #25
5/6 for both
 
  • #26
Mark44 said:
Yes, so it looks like this:
$$(\frac 1 1 - \frac 1 2)+(\frac 1 2 - \frac 1 3)+(\frac 1 3 - \frac 1 4)+(\frac 1 4 - \frac 1 5) + (\frac 1 5 - \frac 1 6) + \dots +(\frac 1 {n-1} - \frac 1 n) + (\frac 1 n - \frac 1 {n+1})$$

So when you simplify, will 1/(n - 1) be left? When you're all done, there should be only two terms left. What are they?

This is what is called a "telescoping" series, for an obvious reason.

There are not two terms left if you simplify this expression.
$$(\frac 1 1 - \frac 1 2)+(\frac 1 2 - \frac 1 3)+(\frac 1 3 - \frac 1 4)+(\frac 1 4 - \frac 1 5) + (\frac 1 5 - \frac 1 6) + \dots +(\frac 1 {n-1} - \frac 1 n) + (\frac 1 n - \frac 1 {n+1})$$

$$ = (\frac 1 1 - \frac 1 6) + \dots +(\frac 1 {n-1} - \frac 1 n) + (\frac 1 n - \frac 1 {n+1})$$

$$ = (\frac 5 6 +\dots + (\frac 1{n-1} - \ \frac 1 {n+1})$$

Three terms.
 
  • #27
or two

combining the last

$$ \frac 5 6 + \dots + \frac 2 {n^2 -1} $$
 
  • #28
Let's take the case where n =10:$$
(\frac 1 1 - \frac 1 2)+(\frac 1 2 - \frac 1 3)+(\frac 1 3 - \frac 1 4)+(\frac 1 4 - \frac 1 5) + (\frac 1 5 - \frac 1 6) + (\frac 1 6 - \frac 1 7)+ (\frac 1 7 - \frac 1 8)+ (\frac 1 8 - \frac 1 9)+ (\frac 1 9 - \frac 1 {10})+ (\frac 1 {10} - \frac 1 {11})$$where I haven't left out any terms. Do you see that the only terms that don't cancel are the ##1## at the beginning and the ##\frac 1 {11}## at the end, so you are left with ##1-\frac 1 {11}##?

Now to save writing, the above might have been written$$
(\frac 1 1 - \frac 1 2)+(\frac 1 2 - \frac 1 3)+\cdots + (\frac 1 {10} - \frac 1 {11})$$The dots indicate that the pattern continues for the missing terms. But the missing terms still cancel.

It's the same idea for$$
(\frac 1 1 - \frac 1 2)+(\frac 1 2 - \frac 1 3)+\cdots + (\frac 1 {n} - \frac 1 {n+1})$$
 
  • #29
Yeah I see is that not what I have just written above?
 
  • #30
Jbreezy said:
Yeah I see is that not what I have just written above?
No, the 1/(n-1) term should cancel too.
Consider this: Let ##S_n## denote the ##n^{th}## partial sum. Then $$S_1 = \left(1 - \frac{1}{2}\right),$$ $$S_2 = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right),$$ and $$S_3 = \left(1-\frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right)$$

Simplify ##S_2## and ##S_3## and hopefully you see a pattern. Then write for general ##S_n##.
 
  • #31
Got it thanks to all.
 

1. What is series convergence verification?

Series convergence verification is a method used in mathematics to determine whether a series, which is a sum of infinitely many terms, will have a finite sum or will diverge to infinity. It involves applying various tests and criteria to the terms of the series to determine its behavior.

2. Why is series convergence verification important?

Series convergence verification is important because it helps us determine the validity and accuracy of mathematical equations and models. It also allows us to manipulate and simplify complex series to make them more manageable for further analysis and applications.

3. What are some common tests used for series convergence verification?

Some common tests used for series convergence verification include the ratio test, the comparison test, the integral test, and the limit comparison test. These tests involve evaluating the behavior of the terms of the series and comparing them to known series with known convergence properties.

4. How do you know if a series converges or diverges?

If the series passes one of the convergence tests, it is considered to be convergent and will have a finite sum. If the series fails all of the convergence tests, it is considered to be divergent and will have an infinite sum. It is important to note that some series may not pass any of the tests, in which case further analysis and techniques may be needed to determine its behavior.

5. Can series convergence verification be used for all types of series?

No, series convergence verification is not applicable to all types of series. It is most commonly used for infinite series with real-valued terms, but may also be used for power series, geometric series, and other specific types of series. It is important to understand the properties and behavior of the series before applying any convergence tests.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
93
  • Calculus and Beyond Homework Help
Replies
3
Views
358
  • Calculus and Beyond Homework Help
Replies
2
Views
660
  • Calculus and Beyond Homework Help
Replies
1
Views
94
  • Calculus and Beyond Homework Help
Replies
2
Views
734
  • Calculus and Beyond Homework Help
Replies
5
Views
941
  • Calculus and Beyond Homework Help
Replies
5
Views
436
  • Calculus and Beyond Homework Help
Replies
1
Views
285
  • Calculus and Beyond Homework Help
Replies
4
Views
97
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
Back
Top