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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.

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  • $\begingroup$ For what it's worth, playing around in Excel with this, the summation seems to converge to $8$, though I'm not sure offhand how one would approach this. $\endgroup$ – Eevee Trainer Jan 26 at 5:03
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    $\begingroup$ Do some people write $\infty$ after the dots? Is that a thing? $\endgroup$ – runway44 Jan 26 at 5:14
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    $\begingroup$ Pops up every now and then on MSE, so I guess it's a notation some people use. Personally not a fan of it, but it is what it is. $\endgroup$ – Eevee Trainer Jan 26 at 5:14
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    $\begingroup$ Alternatively define the sequence $b_k=\frac{a_k}{2^k}$ then we get $$b_k-b_{k-1}=\frac{b_{k-2}-b_{k-1}}{2}-\frac{b_{k-3}}{8}$$ now telescope ... $\endgroup$ – Albus Dumbledore Jan 26 at 5:25
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Let $$S=\sum_{n=1}^{\infty}\frac{a_n}{2^{n-1}}$$ We have, $$S=\frac{1}{1}+\frac{1}{2}+\frac{3}{4}+P$$ Where $$P=\sum_{n=4}^{\infty}\frac{a_n}{2^{n-1}}$$ Using the given recurrence we have, $$P=\sum_{n=4}^{\infty}\frac{a_{n-1}+2a_{n-2}-a_{n-3}}{2^{n-1}}$$ So we get, $$P=\sum_{n=4}^{\infty}\frac{a_{n-1}}{2^{n-1}}+2\sum_{n=4}^{\infty}\frac{a_{n-2}}{2^{n-1}}-\sum_{n=4}^{\infty}\frac{a_{n-3}}{2^{n-1}}$$ By Change of variable for each summation we get $$P=\sum_{k=3}^{\infty}\frac{a_k}{2^{k}}+2\sum_{k=2}^{\infty}\frac{a_k}{2^{k+1}}-\sum_{k=1}^{\infty}\frac{a_k}{2^{k+2}}$$ Which implies $$P=\frac{1}{2}\sum_{k=3}^{\infty}\frac{a_k}{2^{k-1}}+\frac{2}{4}\sum_{k=2}^{\infty}\frac{a_k}{2^{k-1}}-\frac{1}{8}\sum_{k=1}^{\infty}\frac{a_k}{2^{k-1}}$$ $\implies$ $$P=\frac{1}{2}\left(\frac{a_3}{4}+P\right)+\frac{1}{2}\left(\frac{a_2}{2}+\frac{a_3}{4}+P\right)-\frac{S}{8}$$ Using $a_3=3,a_2=1$ $$P=\frac{3}{8}+\frac{P}{2}+\frac{1}{4}+\frac{3}{8}+\frac{P}{2}-\frac{S}{8}$$ Thus we get $$\frac{S}{8}=\frac{3}{8}+\frac{1}{4}+\frac{3}{8}$$ $\implies$ $$S=8$$

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    $\begingroup$ very impressive! $\endgroup$ – A rural reader Jan 26 at 5:19
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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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  • $\begingroup$ Love this awesome way of solving $\endgroup$ – Ekaveera Gouribhatla Mar 26 at 17:07

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