Question

How do you test if this converges or diverges?

Answer 1

\[\frac{ n(n+1) }{ n^2 }\]

Answer 2

As noted in the previous section most of what we were doing there won’t be done much in this chapter. So, it is now time to start talking about the convergence and divergence of a series as this will be a topic that we’ll be dealing with to one extent or another in almost all of the remaining sections of this chapter.



So, let’s recap just what an infinite series is and what it means for a series to be convergent or divergent. We’ll start with a sequence and again note that we’re starting the sequence at only for the sake of convenience and it can, in fact, be anything.



Next we define the partial sums of the series as,







and these form a new sequence, .



An infinite series, or just series here since almost every series that we'll be looking at will be an infinite series, is then the limit of the partial sums. Or,









If the sequence of partial sums is a convergent sequence (i.e. its limit exists and is finite) then the series is also called convergent and in this case if then, . Likewise, if the sequence of partial sums is a divergent sequence (i.e. its limit doesn’t exist or is plus or minus infinity) then the series is also called divergent.



Let’s take a look at some series and see if we can determine if they are convergent or divergent and see if we can determine the value of any convergent series we find.



Example 1 Determine if the following series is convergent or divergent. If it converges determine its value.



Solution

To determine if the series is convergent we first need to get our hands on a formula for the general term in the sequence of partial sums.





This is a known series and its value can be shown to be,





Don’t worry if you didn’t know this formula (I’d be surprised if anyone knew it…) as you won’t be required to know it in my course.



So, to determine if the series is convergent we will first need to see if the sequence of partial sums,



is convergent or divergent. That’s not terribly difficult in this case. The limit of the sequence terms is,





Therefore, the sequence of partial sums diverges to and so the series also diverges.



So, as we saw in this example we had to know a fairly obscure formula in order to determine the convergence of this series. In general finding a formula for the general term in the sequence of partial sums is a very difficult process. In fact after the next section we’ll not be doing much with the partial sums of series due to the extreme difficulty faced in finding the general formula. This also means that we’ll not be doing much work with the value of series since in order to get the value we’ll also need to know the general formula for the partial sums.



We will continue with a few more examples however, since this is technically how we determine convergence and the value of a series. Also, the remaining examples we’ll be looking at in this section will lead us to a very important fact about the convergence of series.



So, let’s take a look at a couple more examples.



Example 2 Determine if the following series converges or diverges. If it converges determine its sum.

Solution

This is actually one of the few series in which we are able to determine a formula for the general term in the sequence of partial fractions. However, in this section we are more interested in the general idea of convergence and divergence and so we’ll put off discussing the process for finding the formula until the next section.



The general formula for the partial sums is,



and in this case we have,





The sequence of partial sums converges and so the series converges also and its value is,





Example 3 Determine if the following series converges or diverges. If it converges determine its sum.

Solution

In this case we really don’t need a general formula for the partial sums to determine the convergence of this series. Let’s just write down the first few partial sums.







So, it looks like the sequence of partial sums is,



and this sequence diverges since doesn’t exist. Therefore, the series also diverges.



Example 4 Determine if the following series converges or diverges. If it converges determine its sum.

Solution

Here is the general formula for the partial sums for this series.





Again, do not worry about knowing this formula. This is not something that you’ll ever be asked to know in my class.



In this case the limit of the sequence of partial sums is,







The sequence of partial sums is convergent and so the series will also be convergent. The value of the series is,





As we already noted, do not get excited about determining the general formula for the sequence of partial sums. There is only going to be one type of series where you will need to determine this formula and the process in that case isn’t too bad. In fact, you already know how to do most of the work in the process as you’ll see in the next section.



So, we’ve determined the convergence of four series now. Two of the series converged and two diverged. Let’s go back and examine the series terms for each of these. For each of the series let’s take the limit as n goes to infinity of the series terms (not the partial sums!!).











Notice that for the two series that converged the series term itself was zero in the limit. This will always be true for convergent series and leads to the following theorem.



Theorem

If converges then .



Proof

First let’s suppose that the series starts at . If it doesn’t then we can modify things as appropriate below. Then the partial sums are,





Next, we can use these two partial sums to write,





Now because we know that is convergent we also know that the sequence is also convergent and that for some finite value s. However, since as we also have .



We now have,



Pf_Box





Be careful to not misuse this theorem! This theorem gives us a requirement for convergence but not a guarantee of convergence. In other words, the converse is NOT true. If the series may actually diverge! Consider the following two series.









In both cases the series terms are zero in the limit as n goes to infinity, yet only the second series converges. The first series diverges. It will be a couple of sections before we can prove this, so at this point please believe this and know that you’ll be able to prove the convergence of these two series in a couple of sections.



Again, as noted above, all this theorem does is give us a requirement for a series to converge. In order for a series to converge the series terms must go to zero in the limit. If the series terms do not go to zero in the limit then there is no way the series can converge since this would violate the theorem.



This leads us to the first of many tests for the convergence/divergence of a series that we’ll be seeing in this chapter.

Answer 3

Please don't say that D: I almost Eureka this problem....

Answer 4

lol

Answer 5

I am supposed to prove this...


'Suppose that circles of equal diameter are packed tightly in n rows inside an equilateral triangle. If A is the area of the triangle and A_n is the total area occupied by the n rows of the circles, show that \[\lim_{n \rightarrow infinity } \frac{ A_{n} }{ A } = \frac{ \pi }{ 2\sqrt(3) }\]

Answer 6

I'm so closed...

I have

\[A_n = \frac{ n(n+1) }{ 2 } * \frac{ L^2 }{ 4n^2 }\pi \]

Answer 7

let me get a picture

Answer 8

https://goo.gl/photos/MD2nuuHYNbBqQvtN9

Answer 9

I looked at the pattern, I set L to be the length of each side of the triangle.

Answer 10

I got r in terms of L and n # of rows. (n approches infinity

Answer 11

Area of triangle:

\[A= \frac{ 1 }{ 2 }(L)(\frac{ L }{ \sqrt(3)/2) }) = \frac{ L^2\sqrt(3) }{ 4 }\]

Answer 12

@agent0smith You might love this one lol

Answer 13

\[\lim_{n \rightarrow infinity} \frac{ n(n+1) }{ 2n^2 \sqrt(3) } \pi\] You see almost lol If only the sum part converged to one.

Answer 14

by the way. I get that L= 2*n*r. If you look at the image you'll see the pattern.

Answer 15

i am confused perhaps, but the sequence certainly converges
it converges to one
there is no series written there

Answer 16

lol when I simplify and take the limit, for the n part of the expression, I get (1+ 1/n) ..... wait brahhhhhhhhhhhh

Answer 17

I'm sorry, I meant afer simplifying... not taking limit yet lol

Answer 18

after taking the limit, yes it converges omg

Answer 19

brahhhhhhhhhhh why didn't I see this

Answer 20

eureka

Answer 21

Course it converges lol, it's n^2/n^2 + n/n^2 haha

Answer 22

You guys give me some medals :p

Answer 23

well the terms converge at least.

Answer 24

=D