Series Convergence Verification

[Jbreezy]

Homework Statement

Verify that the infinite series converges

Homework Equations

\Big(\sum_{n=1}^\infty\frac{1}{n(n+1)}\Big)

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

 

[CAF123]

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.

 

[Jbreezy]

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.

 

[CAF123]

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.

 

[LCKurtz]

$$S_n =\frac 1 {1\cdot 2} + \frac 1 {2\cdot 3} + \frac 1 {3\cdot 4} +\dots+ \frac 1 {n \cdot (n+1)}$$ $$
=(\frac 1 1 - \frac 1 2)+(\frac 1 2 - \frac 1 3)+(\frac 1 3 - \frac 1 4)+\dots +(\frac 1 n - \frac 1 {n+1})$$What happens to that second expression when you simplify it?

 

[Jbreezy]

$$S_n =\frac 1 {1\cdot 2} + \frac 1 {2\cdot 3} + \frac 1 {3\cdot 4} +\dots+ \frac 1 {n \cdot (n+1)}$$ $$
=(\frac 1 1 - \frac 1 2)+(\frac 1 2 - \frac 1 3)+(\frac 1 3 - \frac 1 4)+\dots +(\frac 1 n - \frac 1 {n+1})$$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.

 

[Mark44]

That's not what LCKurtz asked. What happens to the 2nd expression when you simplify it?

 

[Jbreezy]

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.

 

[LCKurtz]

Don't just go backwards. What happens when you remove the parentheses?

 

[Mark44]

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?
$$ (\frac 1 1 - \frac 1 2)+(\frac 1 2 - \frac 1 3)+(\frac 1 3 - \frac 1 4)+\dots +(\frac 1 n - \frac 1 {n+1})$$

 

[Jbreezy]

You get 1/2 +...1/n -1/(n+1)

 

[Mark44]

You get 1/2 +...1/n -1/(n+1)

I don't. How did you get it?

 

[Jbreezy]

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?

 

[Mark44]

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

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?
$$ (\frac 1 1 - \frac 1 2)+(\frac 1 2 - \frac 1 3)+(\frac 1 3 - \frac 1 4)+\dots +(\frac 1 n - \frac 1 {n+1})$$

 

[Jbreezy]

3/4 + 1/n - (1/n+1)

right? How does this help?

 

[Mark44]

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?

 

[Jbreezy]

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?

 

[Mark44]

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."

 

[Jbreezy]

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

 

[Mark44]

I don't think you get the pattern. What's the term that comes before (1/n - 1/(n + 1))?

 

[Jbreezy]

1/(n-1) - 1/n

 

[Mark44]

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.

 

[Jbreezy]

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

 

[Mark44]

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

No.

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?

$$(\frac 1 1 - \frac 1 2)+(\frac 1 2 - \frac 1 3)+ \dots + (\frac 1 5 - \frac 1 6)$$

 

[Jbreezy]

5/6 for both

 

[Jbreezy]

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.

 

[Jbreezy]

or two

combining the last

$$ \frac 5 6 + \dots + \frac 2 {n^2 -1} $$

 

[LCKurtz]

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$$
(\frac 1 1 - \frac 1 2)+(\frac 1 2 - \frac 1 3)+\cdots + (\frac 1 {n} - \frac 1 {n+1})$$

 

[Jbreezy]

Yeah I see is that not what I have just written above?

 

[CAF123]

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 $S_n$.

 

[Jbreezy]

Got it thanks to all.