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?

  • $\begingroup$ 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

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.


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 $$ f(x)=\frac{-x}{x^2+x-1} \\ =\frac{x}{1 -x -x^2} $$

  • $\begingroup$ 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

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