Proving sum of recurrence sequence converges to $2^{-55}$

Question

$$24a(n)=26a(n-1)-9a(n-2)+a(n-3)$$ $$a(0)=46, a(1)=8, a(2)=1$$ $$\sum\limits_{k=3}^{\infty}a(k)=2^{-55}$$ How can I prove it?

Answer 1

$$24r^3=26r^2-9r+1$$

solutions $$r=\frac{1}{2},\frac{1}{3},\frac{1}{4}$$

$$a_n=x\left(\frac{1}{2}\right)^n+y\left(\frac{1}{3}\right)^n+z\left(\frac{1}{4}\right)^n$$ Determine $x,y,z$ using base conditions

$$a_0=46=x+y+z$$ $$a_1=8=x/2+y/3+z/4$$ $$a_2=1=x/4+y/9+z/16$$

$$\fbox{ x=4, y=-54,z=96 }$$

$$a_n=4\left(\frac{1}{2}\right)^n-54\left(\frac{1}{3}\right)^n+96\left(\frac{1}{4}\right)^n$$

$$\sum\limits_{k=3}^{\infty}a(k)=\sum\limits_{k=3}^{\infty}\left(4\left(\frac{1}{2}\right)^k-54\left(\frac{1}{3}\right)^k+96\left(\frac{1}{4}\right)^k\right)$$

$$\sum_{k=3}^{\infty}4\left(\frac{1}{2}\right)^k=4\sum_{k=3}^{\infty}\left(\frac{1}{2}\right)^k=4\cdot\frac{\left(\frac{1}{2}\right)^3}{1-\frac{1}{2}}=1$$

$$54\sum_{k=3}^{\infty}\left(\frac{1}{3}\right)^k=54\frac{\left(\frac{1}{3}\right)^3}{1-\frac{1}{3}}=3$$ $$96\sum_{k=3}^{\infty}\left(\frac{1}{4}\right)^k=96\cdot\frac{\left(\frac{1}{4}\right)^3}{1-\frac{1}{4}}=2$$

$$\sum\limits_{k=3}^{\infty}a_k=1-3+2=0$$

Answer 2

Let $S=\sum\limits_{k=3}^{\infty}a(k)$. Then $$ 24S = 26(S+a(2))-9(S+a(1)+a(2))+(S+a(0)+a(1)+a(2)) $$ This gives $S=0$.

Answer 3

Consider the generating function $$f(x)=a_0x^0+a_1x^1+a_2x^2+\cdots+a_nx^n+\cdots.\tag1$$

We have $$f(x)\cdot x^3=a_0x^3+a_1x^4+a_2x^5+\cdots+a_nx^{n+3}+\cdots;\tag2$$ $$f(x)\cdot x^2=a_0x^2+a_1x^3+a_2x^4+\cdots+a_nx^{n+2}+\cdots;\tag3$$ $$f(x)\cdot x=a_0x^1+a_1x^2+a_2x^3+\cdots+a_nx^{n+1}+\cdots.\tag4$$ Thus, by $(2)-9\times(3)+26\times(4)-(1)$, we obtain \begin{align*} f(x)(x^3-9x^2+26x-1)&=-9a_0x^2+26a_0x^1+26a_1x^2-a_0x^0-a_1x^1-a_2x^2\\ &=-207x^2+1188x-46 \end{align*}

Hence $$\boxed{f(x)=\frac{-207x^2+1188x-46}{x^3-9x^2+26x-1}}.$$

Thus $$\sum_{k=0}^\infty a_k=f(1)=55.$$

As a result, $$\sum_{k=3}^\infty a_k=\sum_{k=0}^\infty a_k-\sum_{k=0}^2 a_k=55-55=0.$$