Easy sequence convergence problem

In summary, it was shown that the sequence defined by y_1=1 and y_n+1 = 4-(1/y_n) converges by using induction to prove it is monotonically increasing and bounded above by 4. The next step was to find the limit by taking the limit of each side of the recursive equation, which led to the quadratic equation L^2-4L+1=0. Using the quadratic formula, the two possible values for L were found to be 2+sqrt(3) and 2-sqrt(3). Since the sequence is monotonically increasing, the limit was determined to be 2+sqrt(3).
  • #1
jimmypoopins
65
0

Homework Statement


Show that the sequence defined by y_1=1 and y_n+1 = 4-(1/y_n) converges and find the limit.


Homework Equations


thm: a sequence that is monotone and bounded is convergent.


The Attempt at a Solution


i used induction to show that the sequence is increasing and bounded above by 4, so i showed that it is convergent.

i'm having issues showing mathematically that the limit is 4. maybe I'm forgetting something from calc 2?

the book suggests to "take the limit of each side of the recursive equation to explicitly compute lim y_n," but i don't understand how to do that.

any help in the right direction would be gratefully appreciated :)
 
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  • #2
well you first have to show that this sequence is monotonic and bounded, after that show that the lim exists and find it.

y_1=1, y_2=3, y_3=11/3 now use induction to show that this sequence is monotono increasing.

After that try to show that the sequence is upper bounded by 4. Again you have to use induction to prove this, maybe there is any other what but this is what crosess my mind right now.

After that let lim(n-->infinity)(y_n)=L and see what happenes.

now you want to go like this

lim(n->inf)(y_n+1)=4-1/lim(n->inf)(y_n) , then replace it by L, you will get a quadratic eq, solve it for L, and see which result to take?

do u get it?
 
  • #3
yeah, i understand. i did the first part and proved that it is monotone and bounded.

i'm confused on the part "lim(n->inf)(y_n+1)=4-1/lim(n->inf)(y_n) , then replace it by L, you will get a quadratic eq, solve it for L, and see which result to take?"

if lim(n->inf)(y_n)=L, then what does lim(n->inf)(y_n+1) equal?

if i substitute in L, i get

lim(n->inf)(y_n+1)=4-1/L. how does that help me?
 
  • #4
lim(y_n)(n->inf)=L. this means that:
for every epsylon e>0, there exists some N(e)>0 such that for any n>N Iy_n-LI<e
now n+1 is obviously greater than n, right?
what do u think lim(n->inf)(y_n+1) equals now?
 
  • #5
ah, since it's an infinite series, then |y_n+1-L|<e also. so i get

L=4-(1/L) and i solve for L.

thanks :)

edit: no, I'm clearly wrong because that implies

L+1/L=4
(L^2+1)/(L+1)=4 isn't satisfied if L=4...
 
Last edited:
  • #6
yep, and you will end up with two values for L, but since the sequence is monotono increasing you have to chose only one of them, whichever is closest to 4.
 
  • #7
L+1/L=4
(L^2+1)/(L+1)=4

is only satisfied if L=0 or L=1, so you're saying the limit is 1? i thought it was 4.

edit: no I'm stupid. sorry. I'm bad at algebra. i got it now, thanks :)
 
Last edited:
  • #8
one last question: i get that the equation is satisfied when L=3.7320508... (via calculator)

how can i algebraically solve L=4-(1/L) for L to get an exact answer? i don't think we are allowed to use calculators on homework.
 
  • #9
jimmypoopins said:
one last question: i get that the equation is satisfied when L=3.7320508... (via calculator)

how can i algebraically solve L=4-(1/L) for L to get an exact answer? i don't think we are allowed to use calculators on homework.

well you multiply through by L, and you will get

L^2-4L+1=0, then by applying viettis formulas we get

L_1,2=[-b^2+-sqrt(b^2-4ac)]/2a, where a=1, b=-4, c=1
and your final answer will be 2+sqrt(3)
 
  • #10
This is just some hight school algebra problem, i mean this last part!
 
  • #11
jimmypoopins said:
edit: no, I'm clearly wrong because that implies

L+1/L=4
(L^2+1)/(L+1)=4 isn't satisfied if L=4...

who said that the limit will be 4 at first place?
 

Related to Easy sequence convergence problem

1. What is an easy sequence convergence problem?

An easy sequence convergence problem is a mathematical problem that involves determining whether a sequence of numbers converges or diverges. Convergence refers to the behavior of a sequence as the number of terms increases, while divergence means that the sequence does not have a limit as the number of terms increases.

2. How do you determine if a sequence converges or diverges?

To determine if a sequence converges or diverges, you can use different methods such as the limit comparison test, ratio test, or root test. These methods involve evaluating the limit of the sequence to see if it approaches a specific value or if it diverges to infinity.

3. What is the importance of studying sequence convergence?

Understanding sequence convergence is crucial in many areas of science and mathematics. It helps in analyzing the behavior of functions, solving equations, and approximating values. It is also essential in understanding the convergence of numerical methods used in scientific computing.

4. Can an easy sequence convergence problem have multiple solutions?

Yes, an easy sequence convergence problem can have multiple solutions. This can happen when the sequence has different behaviors depending on the starting point or when the limit is undefined. In such cases, there may be multiple solutions or no solution at all.

5. What are some real-life examples of sequence convergence?

Sequence convergence can be seen in various real-life situations, such as population growth, compound interest, and the decay of radioactive materials. In these cases, the sequence represents the growth or decay of a quantity over time, and understanding its convergence helps in predicting future values.

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