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The question at hand is this:

Let $\{a_n\}$ be a sequence satisfying $a_0 = \frac{1}{5}$, $a_1 = \frac{4}{25}$ and for $n \geq 2$, $a_n = \frac{4}{5}a_{n-1} - \frac{1}{5}a_{n-2} $.

a)Consider the power series $f(z) = \sum_{n} a_n z^n $ of radius of convergence R. Determine $f$ explicitly.

b)Use (a) to find the explicit expression of $a_n$ and the radius of convergence of $\sum_{n} a_n z^n$.

I dont even know how to start with this one. Any help would be great...

2 Answers 2

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For all $n$, $$a_{n+2}z^{n+2}=\frac{4}{5} za_{n+1}z^{n+1} -\frac{1}{5}z^2a_{n}z^{n}.$$ Provided $R>0$, we can sum and then have for all $|z|

To make this a proof you need to prove $R>0$ in a non-circular way. It is easy to prove by induction that all $|a_n|\leqslant 1$, so $R\geqslant 1$.

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    I see, thank you good sir – 2017-02-19
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    i am getting a different fraction that cannot be separated by partial fractions. Can you please elaborate on the binomial theorem part – 2017-02-19
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    Apologies. Carelessness. Let me re-do it. – 2017-02-19
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    i'm really sorry to ask again, but I dont think I understand how to get the sequence in part b – 2017-02-19
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Write $$f(z) = \sum_{n = 0}^\infty a_n z^n = \frac{1}{5} + \frac{z}{25} + \sum_{n = 2} a_n z^n = \frac{1}{5} + \frac{z}{25} + \frac{4z}{5} \sum_{n = 1} a_n z^n - \frac{z^2}{5} \sum_{n = 0} a_n z^n.$$ This yields $$f(z) = \frac{1}{5} + \frac{z}{25} + \frac{4z}{5} (f(z) - \frac{1}{5}) - \frac{z^2}{5} f(z).$$ Solving this for $f$ yields (if the nominator is non-zero) $$f(z) = \frac{\frac{1}{5} - \frac{19z}{25}}{1 - \frac{4z}{5} + \frac{z^2}{5}} = \frac{5 - 19z}{25 - 20z + 5z^2}.$$ (I hope i didn't make a mistake)