Question

suppose that a sequence is defined by

a1 = 1, an + 1  =  9an/2 + an assuming an is convergent find its limit

Answer 1

why is the limit 7 i found this by subbing in values continuously but i know that's not the way to find the limit

Answer 2

is it \[\large a_{n+1} = \frac{9}{2+a_n} \cdot a_n \]

Answer 3

can't read the LaTeX it's not coming up properly and nor can i use the draw tool for some reason

Answer 4

I have the same issue here, hmm ok, I will try to come up with something, give me a minute.

Answer 5

Let me just type it here as clear as possible. Do you understand that if a sequence converges, the following must be true FOR ALL natural numbers:

lim_( n to infinity) a_n = L where L is the Limit and also the following must be true:
lim_(n to infinity) a_(n+1) = L where L is the limit

Answer 6

if you do understand this concept, then write down your equation and take the limit on both sides as n tends to infinity and substitute the above identities into your equation, solve for L and you will obtain your answer.

Answer 7

im still confused i have no idea what im doing with limits of sequences :(

Answer 8

you plugin L wherever it says a_n and a_{n+1} and solve for L.

Answer 9

use your given equation:
a_{n+1}= 9a_n/(2+a_n)

Answer 10

take the limits on both sides of this equation and substitute L wherever you have a_{n+1} and a_n, we know that we can do that because the question says the sequence is convergent (usually we have to prove that manually first, but it can get tricky).

When you take the limit on both sides as n tends to infinity you're left with a new equation
L=9L/(2+L) and now solve for L.

Answer 11

sequences and series are driving my off the edge
I'm not sure i understand the concepts behind it :(
maybe i need some time to think this over
my mind is clouded

Answer 12

**me

Answer 13

don't worry about that @ngott3, it sometimes sure is a good thing to step back from a topic, let it sink and then start over from scratch again.

Allow me to write a few key concepts anyway. A sequence is nothing like an infinite patter of numbers, a very famous one would be

a_n= 1/n, we write n for a sequence because we want to highlight that we can plugin natural numbers into the expression n. We do not want to plugin pi in there for example.
With that knowledge you can compute already a few terms of the sequence explicitly.

you can say that a_1=1, a_2= 1/2, a_3=1/3 and so on and so forth. Just by inspecting that sequence you see that the terms get smaller and smaller on each step right? For instance a_5 is smaller than a_4, and likewise, a_301 will be smaller than a_300, we call such a sequence monotone decreasing.

You can maybe, by mere inspecting, also tell that the sequence will for very LARGE n, imagine n being like a quadrillion, or whatever big number comes into your head. Then we talk about the limit as n becomes a very huge number and you can see that 1 divided by a very huge number (much bigger than a quadrillion, that was just an example) becomes zero.

In this case we say that the sequence converges and that it's limit L is equal to zero. The other scenario would be a sequence that does not have a limit, it diverges. That would for instance be the case with the sequence b_n=(-1)^n

Answer 14

try to understand these concepts, maybe reread them in a textbook in case my elaborations confuse you. They are the basic concepts of what a sequence is. An infinite pattern of numbers that can a) converge, or b) diverge.

A Series is like the big brother (or sister) of a sequence. The sequence is a pattern of infinite list of numbers, the series is a summation of that pattern.

Answer 15

To make the connection between my elaboration and your question make sure that you understand that your problem given above is just yet another sequence. It is a bit special though isn't it?

It is what we call a recursive sequence. What does that mean?
Remember my example, it was a_n= 1/n. A very easy and demonstrative basic sequence.

If I ask you to tell me what the 500th term of the above sequence is, you would just plugin n=500 into the above equation and get a_500=1/500, now that was easy.

With a recursive sequence that is a bit harder, you can no longer do that because you always need to know the previous term, for instance to compute a_3, you must also know what a_2 is. To know what a_47 is, you must know what a_46 is.

a recursive sequence means in a blunt way of saying it, to know what comes next, you must know what comes before.

Answer 16

hmm I think this concept is slowly sinking in :)
would u mind helping me with this one
an=nsin(10/n)
solution from the textbook says 10
i got up till: lim an approaches infinity = sin(10/n)/(1/n)
usually if it was sin(1/n) then i could have let x=(1/n) let it approach infinity so x approaches zero
then so using the property of sin(x)/x approaching zero would equal one
but that doesn't seem to be the case here

Answer 17

you did very well with that so far, just consider a different substitution, only a slightly different one, choose x=10/n, you can apply the same logic to it, as n tends to infinity, x tends to zero, so you can use the fundamental limes as you have said you wanted to.

Answer 18

thanks for your testimonial @ngott3, if you have any further questions you can forward them to me or ping me with @spacelimbus

Answer 19

aha that seems to be it for now thank you so much for all clearing the confusion up
i'll prob have to bother u some other time :P