# two limits: 0<b<a: x_0=1, x_1=a+b, x_{n+2}=(a+b)*x_{n+1}-ab*x_n

#### Vali

##### Junior Member
Let $$\displaystyle 0<b<a$$ and $$\displaystyle (x_{n})_{n\in \mathbb{N}}$$ with$$\displaystyle x_{0}=1, \ x_{1}=a+b$$
$$\displaystyle x_{n+2}=(a+b)\cdot x_{n+1}-ab\cdot x_{n}$$
a) If $$\displaystyle 0<b<a$$ and $$\displaystyle L=\lim_{n\rightarrow \infty }\frac{x_{n+1}}{x_{n}}$$ then $$\displaystyle L= ?$$
solutions to choose from 1. L=a 2. L=b 3. L=a/b 4. L=b/a 5. can't calculate ( right answer L=a)
b) If $$\displaystyle 0<b<a<1$$ and $$\displaystyle L=\lim_{n\rightarrow \infty }\sum_{k=0}^{n}x_{k}$$ then $$\displaystyle L= ?$$
the right answer is $$\displaystyle L=\frac{1}{(1-a)(1-b)}$$
I don't know how to start.I tried to write the first terms x1 x2... but I didn't get too far

#### pka

##### Elite Member
Let $$\displaystyle 0<b<a$$ and $$\displaystyle (x_{n})_{n\in \mathbb{N}}$$ with$$\displaystyle x_{0}=1, \ x_{1}=a+b$$
$$\displaystyle x_{n+2}=(a+b)\cdot x_{n+1}-ab\cdot x_{n}$$
a) If $$\displaystyle 0<b<a$$ and $$\displaystyle L=\lim_{n\rightarrow \infty }\frac{x_{n+1}}{x_{n}}$$ then $$\displaystyle L= ?$$
solutions to choose from 1. L=a 2. L=b 3. L=a/b 4. L=b/a 5. can't calculate ( right answer L=a)
b) If $$\displaystyle 0<b<a<1$$ and $$\displaystyle L=\lim_{n\rightarrow \infty }\sum_{k=0}^{n}x_{k}$$ then $$\displaystyle L= ?$$
the right answer is $$\displaystyle L=\frac{1}{(1-a)(1-b)}$$
I don't know how to start.I tried to write the first terms x1 x2... but I didn't get too far
I cheated due to the limitation of time. Look at this.

Once the recursion has been solved, you do the rest. Post your results.

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#### Jomo

##### Elite Member
Let $$\displaystyle 0<b<a$$ and $$\displaystyle (x_{n})_{n\in \mathbb{N}}$$ with$$\displaystyle x_{0}=1, \ x_{1}=a+b$$
$$\displaystyle x_{n+2}=(a+b)\cdot x_{n+1}-ab\cdot x_{n}$$
a) If $$\displaystyle 0<b<a$$ and $$\displaystyle L=\lim_{n\rightarrow \infty }\frac{x_{n+1}}{x_{n}}$$ then $$\displaystyle L= ?$$
solutions to choose from 1. L=a 2. L=b 3. L=a/b 4. L=b/a 5. can't calculate ( right answer L=a)
b) If $$\displaystyle 0<b<a<1$$ and $$\displaystyle L=\lim_{n\rightarrow \infty }\sum_{k=0}^{n}x_{k}$$ then $$\displaystyle L= ?$$
the right answer is $$\displaystyle L=\frac{1}{(1-a)(1-b)}$$
I don't know how to start.I tried to write the first terms x1 x2... but I didn't get too far
x0 = 1, x1=a+b. x2= x0+2 (so n=0)= (a+b)x1-abx0 = (a+b)2-ab
x3=x1+2 (so n=1) = (a+b)2-ab -ab(a+b) continue (if you like)

What is $$\displaystyle \frac{x_{n+1}}{x_{n}}$$?

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#### pka

##### Elite Member
x0 = 1, x1=a+b. x2= x0+2 (so n=0)= (a+b)x1-abx0 = (a+b)2-ab
x3=x1+2 (so n=1) = (a+b)2-ab -ab(a+b) continue (if you like)
What is $$\displaystyle \frac{x_{n+1}}{x_{n}}$$?
I am just curious. Are you saying that this can be solved without first solving the recursion?

#### Vali

##### Junior Member
I solved both limits!I used the characteristic equation to find $$\displaystyle x_{n}$$
Thanks!