Calculus – Sequences and series – A generating function for the Lucas numbers

Earlier we have seen that the sequence of Lucas numbers \(1,3,4,7,11,18,29,47,76,123,\ldots\) is defined by \(L_{n+2}=L_n+L_{n+1}\) for \(n=1,2,3,\ldots\) with \(L_1=1\) and \(L_2=3\).

An alternative definition is: \(L_{n+2}=L_n+L_{n+1}\) for \(n=0,1,2,\ldots\) with \(L_0=2\) and \(L_1=1\).

Now we consider the generating function \(F(x)=\displaystyle\sum_{n=0}^{\infty}L_nx^n\). Then we have

\begin{align*} x^2F(x)&=\sum_{n=0}^{\infty}L_nx^{n+2}=\sum_{n=0}^{\infty}L_{n+2}x^{n+2}-x\sum_{n=0}^{\infty}L_{n+1}x^{n+1}\\[2.5mm] &=\sum_{n=0}^{\infty}L_nx^n-L_0-L_1x-x\left(\sum_{n=0}^{\infty}L_nx^n-L_0\right)=F(x)-2-x-x\left(F(x)-2\right). \end{align*}

This implies that

\[(1-x-x^2)F(x)=2+x-2x=2-x\quad\Longrightarrow\quad F(x)=\frac{2-x}{1-x-x^2}.\]

In order to find the radius of convergence, we apply the ratio test: for \(x\neq0\) let \(a_n=L_nx^n\), then we have:

\[\lim\limits_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim\limits_{n\to\infty}\left|\frac{L_{n+1}x^{n+1}}{L_nx^n}\right| =\lim\limits_{n\to\infty}\frac{L_{n+1}}{L_n}|x|=\varphi|x|,\]

where \(\displaystyle\lim\limits_{n\to\infty}\frac{L_{n+1}}{L_n}=\varphi=\frac{1+\sqrt{5}}{2}\approx1.618\) denotes the golden ratio.

Now the ratio test implies that the generating series is absolutely convergent for \(|x| < \displaystyle\frac{1}{\varphi}=\varphi-1\approx0.618\).

For instance, this implies that \(\displaystyle\sum_{n=0}^{\infty}\frac{L_n}{2^n}=F(\tfrac{1}{2})=\frac{\frac{3}{2}}{1-\frac{1}{2}-\frac{1}{4}}=6\). This implies that \(\displaystyle\sum_{n=1}^{\infty}\frac{L_n}{2^n}=4\).

Differentiation once leads to

\[\sum_{n=1}^{\infty}nL_nx^{n-1}=F'(x)=\frac{-1+x+x^2+(1+2x)(2-x)}{(1-x-x^2)^2}=\frac{1+4x-x^2}{(1-x-x^2)^2}.\]

For instance, this implies that \(\displaystyle\sum_{n=1}^{\infty}\frac{nL_n}{2^n}=\tfrac{1}{2}F'(\tfrac{1}{2})=\frac{1+2-\frac{1}{4}}{2\left(1-\frac{1}{2}-\frac{1}{4}\right)^2}=22\).

Differentiating once more we obtain

\begin{align*} \sum_{n=2}^{\infty}n(n-1)L_nx^{n-2}=F''(x)&=\frac{(4-2x)(1-x-x^2)^2-2(1-x-x^2)(-1-2x)(1+4x-x^2)}{(1-x-x^2)^4}\\[2.5mm] &=\frac{2(2-x+1+4x-x^2+2x+8x^2-2x^3)}{(1-x-x^2)^3}=\frac{2(3+3x++6x^2-x^3)}{(1-x-x^2)^3}. \end{align*}

For instance, this implies that \(\displaystyle\sum_{n=2}^{\infty}\frac{n(n-1)L_n}{2^n}=\tfrac{1}{4}F''(\tfrac{1}{2})=\frac{2\left(3+\frac{3}{2}+\frac{3}{2}-\frac{1}{8}\right)}{4\left(1-\frac{1}{2}-\frac{1}{4}\right)^3}=188\).

For \(x=-\frac{1}{2}\) we have: \(\displaystyle\sum_{n=0}^{\infty}\left(-\tfrac{1}{2}\right)^nL_n=F(-\tfrac{1}{2})=2\). This implies that \(\displaystyle\sum_{n=1}^{\infty}\left(-\tfrac{1}{2}\right)^nL_n=0\).

Finally we have for \(x=\frac{1}{3}\): \(\displaystyle\sum_{n=0}^{\infty}\frac{L_n}{3^n}=F(\tfrac{1}{3})=3\). This implies that \(\displaystyle\sum_{n=1}^{\infty}\frac{L_n}{3^n}=1\).

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Last modified on April 7, 2024