Recurrence relation - Show that a sum of a sequence is zero

Question

We are given the following sequence: $f(n)=4f(n-1)-5f(n-2)$, $f(0)=f(1)=a$ where $a$ is some value in $\mathbb C$.

We are asked to show that $$\sum_{n=0}^{\infty}\frac{f(n)}{3^n}=0$$

First thing I did was find $f(n)$, its homogenous so it's relatively simple, we get that

$$f(n)=\frac{a[(i-1)(2+i)^n+(i+1)(2-i)^n]}{2i}$$ So what we need to show is that

$$\sum_{n=0}^{\infty} \frac{a[(i-1)(2+i)^n+(i+1)(2-i)^n]}{2i*3^n}=0$$

How can we do this? and let's assume that $a \neq 0$ since that would make the question trivial, so really we just need to show that

$$\sum_{n=0}^{\infty} \frac{(i-1)(2+i)^n+(i+1)(2-i)^n}{2i*3^n}=0$$

Answer 1

If you know that $\displaystyle \sum_{n=0}^\infty \frac{f(n)}{3^n}$ is convergent use that $$\sum_{n=0}^\infty \frac{f(n)}{3^n} = a+\frac{a}{3}+\frac{4}{3}\sum_{n=1}^\infty \frac{f(n)}{3^n}-\frac{5}{9}\sum_{n=0}^\infty \frac{f(n)}{3^n} = \frac{7}{9}\sum_{n=0}^\infty \frac{f(n)}{3^n}$$ This shows $\displaystyle \sum_{n=0}^\infty \frac{f(n)}{3^n}=0$ immediately.

Answer 2

Use the formula for the infinite sum of geometric series :

$$\sum_{k=0}^{\infty} x^k = \frac{1}{1-x}$$

for $|x|<1$

and apply it for $\frac{2+i}{3}$ and $\frac{2-i}{3}$. The terms should cancel out.