1
$\begingroup$

Let $ f(x) = \sum_{n=1}^{\infty}a_nx^n $ where $a_n$ is the nth Fibonacci number.

Find the values for which the series is convergent, and find $f(x)$ for those values. (i.e - find the limit when it's convergent, in terms of $x$)

I have found out, using the ratio test, and the fact that $ \frac{a_{n+1}}{a_n} \leq 2 \space \forall n$ that the series converges for $|x| \leq \frac{1}{2}$

I also know from another source that when the series does converge that $f(x) = \frac{-1}{x^2+x-1}$, but I can't seem to prove this myself.

Could someone show me how this might be accomplished.

Thanks for the help!

I'm guessing this has something to do with Taylor series, maybe?

$\endgroup$
  • $\begingroup$ For the ratio test, you will need the limit of the ratio. The easiest way to do it is to use the "Binet" formula for the $n$-th Fibonacci number. From the formula you can also get the sum as a sum of two geometric series. $\endgroup$ – André Nicolas Dec 5 '15 at 17:58
2
$\begingroup$

First of all, your convergence interval is mistaken: using the expression $a_n = \dfrac {\varphi^n - (-\varphi)^{-n}} {\sqrt 5} = \dfrac {\varphi^{2n} - (-1)^{n}} {\sqrt 5 \varphi^n}$ where $\varphi = \dfrac {1 + \sqrt 5} 2$, the radius of convergence can be computed as $\dfrac 1 {\lim \limits _n \sqrt[n] {a_n} } = \dfrac 1 \varphi$. Note that when $x = \dfrac 1 \varphi$ the series becomes $\sum \dfrac {\varphi^{2n} - (-1)^{n}} {\sqrt 5 \varphi^{2n}}$ which is obviously divergent (its term tends to $1$) and when $x = -\dfrac 1 \varphi$ the series becomes $\sum (-1)^n \dfrac {\varphi^{2n} - (-1)^{n}} {\sqrt 5 \varphi^{2n}}$ which is again divergent for the same reason. Therefore, the convergence set is $(-\dfrac 1 \varphi, \dfrac 1 \varphi)$.

Note that, at least formally, you can split the series in two subseries: $\sum \limits \dfrac {\varphi^n} {\sqrt 5} x^n - \sum \dfrac {(-\varphi)^{-n}} {\sqrt 5} x^n$; the radius of convergence of the first one is $\dfrac 1 \varphi$ and the radius of convergence of the second one is $\varphi$, so each one converges on $(-\dfrac 1 \varphi, \dfrac 1 \varphi)$. On this interval common to both, they are geometric series which have very simple sums: $\dfrac 1 {\sqrt 5} \left( \dfrac 1 {1 - \varphi x} - \dfrac {1} {1 + \frac x \varphi} \right) = \dfrac x {1 - x - x^2}$.

Note that you may find an alternative derivation of this on Wikipedia.

$\endgroup$
0
$\begingroup$

assuming convergence, define $$ f(x)=\sum_{j=0}^{\infty} a_jx^j $$ so $$ (1+x)f(x) = a_0+\sum_{j=1}^{\infty} (a_{j-1}+a_j)x^j $$ if $a_0=0$ $$ = \sum_{j=1}^{\infty} a_{j+1}x^j \\ =\frac{f(x)-x}x $$ giving $$ f(x)=\frac{-x}{x^2+x-1} \\ =\frac{x}{1 -x -x^2} $$

$\endgroup$
  • $\begingroup$ Somewhere there's something wrong, because the correct answer is known to be $\dfrac x {1 - x - x^2}$. $\endgroup$ – Alex M. Dec 5 '15 at 18:10
  • $\begingroup$ thank you Alex. the mistake was located and corrected! $\endgroup$ – David Holden Dec 5 '15 at 19:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.