Finding the limit of the recursive sequence defined as $5a_{n+2}=3a_{n+1}+2a_n$

Question

Find the limit of the recursive sequence defined as follows: $$5a_{n+2}=3a_{n+1}+2a_n\;\;\text{for $n\geq 0$, $a_1=1/2$ and $a_2=2/3$}.$$

Could someone give me a hint as to how to start this problem? I think it is increasing.

Answer 1

This is a second-order homogeneous linear recurrence with constant coefficients whose characteristic polynomial is $$5z^2-3z-2=5(z+2/5)(z-1).$$ Hence the general solution is $$a_n=A(-2/5)^n+B$$ where the constants $A$ and $B$ depend on the initial terms $a_1$ and $a_2$.

Are you able to find the limit of $a_n$ as $n$ goes to infinity given $a_1=1/2$ and $a_2=2/3$?

Answer 2

$$5a_{n+2}-5a_{n+1}=2a_{n}-2a_{n+1}$$ or $$a_{n+2}-a_{n+1}=-\frac{2}{5}(a_{n+1}-a_n),$$ which says that $b_n=a_{n+1}-a_n$ is a geometric progression.

Thus, $$a_{n+1}-a_n=(a_2-a_1)\left(-\frac{2}{5}\right)^{n-1}=\frac{1}{6}\left(-\frac{2}{5}\right)^{n-1}.$$ Thus, $$a_n=a_1+\sum_{k=1}^{n-1}(a_{k+1}-a_k)=\frac{1}{2}+\frac{1}{6}\sum_{k=1}^{n-1}\left(-\frac{2}{5}\right)^{k-1}=$$ $$=\frac{1}{2}+\frac{1}{6}\cdot\frac{\left(-\frac{2}{5}\right)^{n-1}-1}{-\frac{2}{5}-1}\rightarrow\frac{1}{2}+\frac{1}{6}\cdot\frac{-1}{-\frac{2}{5}-1}=\frac{13}{21}.$$