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What is $\sum_{n=1}^{\infty} \frac{a_{n}}{4^{n+1}}$ if $a_{n+2}=a_{n+1}+a_{n}$ and $a_{1}=a_{2}=1$?

scubnoob
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4 Answers4

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$a_{n+2}=a_{n+1}+a_{n}$ with $a_1=a_2 = 1$.

Let $f(x) =\sum_{n=1}^{\infty} a_nx^n $.

$xf(x) =\sum_{n=1}^{\infty} a_nx^{n+1} =\sum_{n=2}^{\infty} a_{n-1}x^{n} $ and $x^2f(x) =\sum_{n=1}^{\infty} a_nx^{n+2} =\sum_{n=3}^{\infty} a_{n-2}x^{n} $ so

$\begin{array}\\ xf(x)+x^2f(x) &=\sum_{n=2}^{\infty} a_{n-1}x^{n}+\sum_{n=3}^{\infty} a_{n-2}x^{n}\\ &=a_1x^2+\sum_{n=3}^{\infty} (a_{n-1}+a_{n-2})x^{n}\\ &=a_1x^2+\sum_{n=3}^{\infty} a_{n}x^{n}\\ &=a_1x^2+f(x)-a_1x-a_2x^2\\ \text{so}\\ f(x)(x^2+x-1) &=(a_1-a_2)x^2-a_1x\\ \end{array} $

Put in the initial $a_1, a_2$ and $x = \frac14$.

Note that this does not need the explicit formula for the $a_n$.

marty cohen
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First, $a_{n}$ has the same definition as the Fibonacci numbers, so that part is easy enough.

$a_{n} = \frac{\phi^n-\psi^n}{\sqrt{5}}$.

https://en.wikipedia.org/wiki/Fibonacci_number

With a bit of re-arranging, you will see that its the difference of two geometric series.

$\sum_{n=1}^{\infty} \frac{a_{n}}{4^{n+1}}=\frac{1}{4}\sum_{n=1}^{\infty} \frac{a_{n}}{4^{n}}=\frac{1}{4\sqrt{5}}\sum_{n=1}^{\infty} \frac{\phi^n-\psi^n}{4^{n}}=\frac{1}{4\sqrt{5}}\sum_{n=1}^{\infty} (\frac{\phi}{4})^n-\frac{1}{4\sqrt{5}}\sum_{n=1}^{\infty} (\frac{\psi}{4})^n$

https://en.wikipedia.org/wiki/Geometric_series

Plug and chug!

Charlie S
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Hint $S=\sum\limits_{n=1}^{\infty}\dfrac{a_n}{4^{n+1}}=\frac 1{4\sqrt{5}}\sum\limits_{n=1}^{\infty}\left(\frac{\phi}4\right)^n-\left(\frac{\bar\phi}4\right)^n=\dfrac 4{11}$

zwim
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I found a nice answer, you guys might be curious to see it as well:

Let S be the sum.

Now $S=\frac{1}{16}+\frac{1}{64}+\frac{2}{256}+\frac{3}{1024}+\frac{5}{4096}$...

Multiply S by 4 to get that

$4S = \frac{1}{4}+\frac{1}{16}+\frac{2}{64}+\frac{3}{256}+\frac{5}{1024}$...

$3S = \frac{1}{4}+\frac{1}{64}+\frac{1}{256}+\frac{2}{1024}+\frac{3}{4096}$...

$3S = \frac{1}{4}+\frac{1}{4}S$

The rest is simplification... $S=\frac{1}{11}$.

Thanks for all the help guys!

scubnoob
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