unicwaan
  • unicwaan
Determine whether the sequence converges or diverges. If it converges, give the limit. 60, -10, 5/3, -5/18, ...
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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SolomonZelman
  • SolomonZelman
ok, your common ratio is>?
unicwaan
  • unicwaan
Okay
SolomonZelman
  • SolomonZelman
your common ratio is not "okay" lol

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SolomonZelman
  • SolomonZelman
do you know what a common ratio is?
unicwaan
  • unicwaan
XD Oh I thought you were telling me that, I'm sorry. I do not know what a common ratio is.
SolomonZelman
  • SolomonZelman
common ratio, is a number by which you multiply to get to the next term each time.
unicwaan
  • unicwaan
Oh okay so would it be (-1/6)?
SolomonZelman
  • SolomonZelman
So, if I had 1, 2, 4, 8, 16 .... then my common ratio is 2 (and common ratio is denoted as r, so in this example r=2) --------------------------------------------- Yes, r=-1/6 is right
unicwaan
  • unicwaan
Okay! I undertsand that!
SolomonZelman
  • SolomonZelman
Note: This is a geometric series (because it is multiplied times some number (referred to as common ratio (r) ).
SolomonZelman
  • SolomonZelman
Do you know what it means for a series to converge and for a series to diverge?
SolomonZelman
  • SolomonZelman
oh, for a sequence to diverge and converge
unicwaan
  • unicwaan
No I do not, that's what I'm coversed on. My lesson never went over it :(
unicwaan
  • unicwaan
confused**
SolomonZelman
  • SolomonZelman
sequence, is just a list of terms. When you say that "sequence converges to 3" (for example 3) you are saying that as you take 1000th 100000th (and roughly speaking infinity-th terms) they all will be approximately 3 (and each time they will get closer and closer to 3)
SolomonZelman
  • SolomonZelman
Say you started from 60, and you were to just divide by 1/6 (not -1/6). \(a_1=60\) \(a_2=10\) \(a_3=5/3\) \(a_4=5/18\) \(a_5=5/108\) and roughly speaking \(a_\infty=5/\infty=0\)
unicwaan
  • unicwaan
Ohhhhhh that makes sense okay
SolomonZelman
  • SolomonZelman
No, you won't actually hit 0. Never! But you closer and closer approach to zero every time when you divide by 1/6. (this is the idea of a "limit" that you approach some value)
unicwaan
  • unicwaan
What about if something diverges?
SolomonZelman
  • SolomonZelman
now, in our case we are multiplying (in the first post, should say multiplying too) by -1/6 so, all we change is that \(a_1=60\) \(a_2=-10\) \(a_3=5/3\) \(a_4=-5/18\) \(a_5=5/108\) and so on.... we still get \(a_\infty=\pm5/\infty=0\) (whether a positive or negative number is divided by infinity, you get 0 in either case)
SolomonZelman
  • SolomonZelman
So the more terms you take, the close and close you approach what value?
unicwaan
  • unicwaan
0, since the number values get smaller and smaller
SolomonZelman
  • SolomonZelman
yes, that is correct
SolomonZelman
  • SolomonZelman
So that means that "sequence converges to zero"
unicwaan
  • unicwaan
^-^ Thanks a lot that makes sense!
SolomonZelman
  • SolomonZelman
(and that is the only case - i.e. - sequence converges to zero - for a series to converge. If you want we can talk about the convergence of a series as well (later on in this thread))
SolomonZelman
  • SolomonZelman
And you asked about divergence, i will answer that now
unicwaan
  • unicwaan
When I do reach series in my lesson I will come to you! and yes
SolomonZelman
  • SolomonZelman
Say you got 1, 2, 4, 8, 16, 32 ....... what is the pattern here, can you tell me?
unicwaan
  • unicwaan
the next number is being multiplied by 2
SolomonZelman
  • SolomonZelman
yes, r=2
SolomonZelman
  • SolomonZelman
And as you take more and more terms you multiply times 2 more and more.... so you are going to go into infinity. (makes sense?)
unicwaan
  • unicwaan
Yes!
SolomonZelman
  • SolomonZelman
that means that sequence diverges.
unicwaan
  • unicwaan
Oh okay so diverges means the increase (or decrease) infinitely while converges ia to get closer and closer to a certain number?
SolomonZelman
  • SolomonZelman
so sequence can diverge in 2 cases. (1) if you have an alternating sequence that is not approaching zero. Such that \(A_n=4(-1)^n\), so your terms are going to be 4, -4, 4, -4, and so on.... and as n approaches infinity (n→∞) you don't know what your term is going to be because it can be either 4 or -4. (2) if your sequence goes into infinity (or negative infinity).
SolomonZelman
  • SolomonZelman
yes, what you said is correct
SolomonZelman
  • SolomonZelman
that is a good definition of convergence of a SEQUENCE.
unicwaan
  • unicwaan
Okay only for a sequence, got it.
SolomonZelman
  • SolomonZelman
now, a series is the sum of terms in a sequence.
SolomonZelman
  • SolomonZelman
think about it. Series is a sum of all terms in the sequence. So if sequence converges to 3 to -8 (or to anything besides 0) then you are going to be adding 3's -4's or something's forever and that will not have a defined sum (will go into ±∞)
SolomonZelman
  • SolomonZelman
is this making some sense?
unicwaan
  • unicwaan
I believe so, yes
SolomonZelman
  • SolomonZelman
yes, so for example in your case, your sequence converges to what?
unicwaan
  • unicwaan
The sequence converges to zero
SolomonZelman
  • SolomonZelman
yes
SolomonZelman
  • SolomonZelman
and the rule is that a geometric SERIES (a series that has all terms follow the pattern of multipling times some common ratio) will always converge if common ratio r is: -1>r>1
unicwaan
  • unicwaan
But in this case, the rule does not apply
SolomonZelman
  • SolomonZelman
this rule always applies provided that ratio is: -1
SolomonZelman
  • SolomonZelman
and so it is in your case, cuz r=-1/6 so it is between -1 and 1
unicwaan
  • unicwaan
Oh okay, I was thinking 0 was r, but r is the common ratio which was -1/6, That makes sense
SolomonZelman
  • SolomonZelman
lets see, but why does the common ratio r has to satisfy -1
SolomonZelman
  • SolomonZelman
for any geometric sequence you multiply times r to get to the next term. So lets say you start from \(a_1=5\) (or from any non-zero first term) \(\large\color{black}{ \displaystyle a_1=5}\) \(\large\color{black}{ \displaystyle a_2=5\cdot r}\) \(\large\color{black}{ \displaystyle a_3=5\cdot r^2}\) \(\large\color{black}{ \displaystyle a_4=5\cdot r^3}\) so on.... \(\large\color{black}{ \displaystyle a_n=5\cdot r^{n-1} }\)
SolomonZelman
  • SolomonZelman
and then you get, roughly speaking \(\large\color{black}{ \displaystyle a_\infty=5\cdot r^{\infty -1}}\)
unicwaan
  • unicwaan
Makes sense
SolomonZelman
  • SolomonZelman
so, if r\(\ge\)1 or if r\(\le\)-1 then the \(\large\color{black}{ \displaystyle r^{\infty -1}}\) part will go to either ∞ or -∞. but if -1
SolomonZelman
  • SolomonZelman
your teacher will go over in detail, but the bottom line is that series is convergent, ONLY when sequence converges to 0. (there are many more tests for various series to converge/diverge that you will learn later in calc, but for geometric series, all you need to know is sequence converges to 0) and sequence convergence is when the \(a_\infty\) approaches one particular value.
SolomonZelman
  • SolomonZelman
And for any geometric series \(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ A_n}\) the sum is given by \(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ A_n~=\frac{a_1}{1-{\rm r}}}\) in your case, the terms 60 is \(a_1\) and r=-1/6 so, \(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ A_n~=\frac{60}{1-{\rm -\dfrac{1}{6}}}=\frac{60}{\dfrac{7}{6}}=360/7 }\)
unicwaan
  • unicwaan
:D Thank you for explaining so much to me! This really helped a lot. and I learn about that and summantion notation in my next lesson so thanks!
SolomonZelman
  • SolomonZelman
Anytime!

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