<s>[INST] Find $a_1+\frac{a_2}{2}+\frac{a_3}{2^2}+\cdots\infty$ A sequence $\left\{a_n\right\}$ is defined as $a_n=a_{n-1}+2a_{n-2}-a_{n-3}$ and $a_1=a_2=\frac{a_3}{3}=1$

Find the value of $$a_1+\frac{a_2}{2}+\frac{a_3}{2^2}+\cdots\infty$$

I actually tried this using difference equation method.Let the solution be of the form $a_n=\lambda^n$
$$\lambda^n=\lambda^{n-1}+2\lambda^{n-2}-\lambda^{n-3}$$ which gives the cubic equation $\lambda^3-\lambda^2-2\lambda+1=0$. But i am not able to find the roots manually.
 [/INST] Let $S=a_1+\dfrac{a_2}{2}+\dfrac{a_3}{2^2}+\dfrac{a_4}{2^3}+\ldots $
Then $\dfrac{S}{2} = \dfrac{a_1}{2}+\dfrac{a_2}{2^2}+\dfrac{a_3}{2^3}+ \ldots$
Subtracting we get
$\dfrac{S}{2} = a_1 +\dfrac{a_2-a_1}{2}+\dfrac{a_3-a_2}{2^2}+\dfrac{a_4-a_3}{2^3}+ \ldots$
Now $a_4-a_3 = 2a_2-a_1, a_5-a_4=2a_3-a_2$ etc.
So $\dfrac{S}{2} = 1 +\dfrac{1-1}{2}+\dfrac{3-1}{2^2}+\dfrac{2a_2-a_1}{2^3}+ \dfrac{2a_3-a_2}{2^4}+\ldots $
$=1+\dfrac{1}{2}-\dfrac{1}{8}+3\left(\dfrac{a_2}{2^4}+\dfrac{a_3}{2^5}+\ldots \right)$
$=\dfrac{11}{8}+\dfrac{3}{8} (S-1) =1+\dfrac{3S}{8}$
$ \Rightarrow \dfrac{S}{8} = 1 $ so that $S=8$
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