Calculus – Sequences and series – The Lucas sequence

The Lucas sequence \(1,3,4,7,11,18,29,47,76,123,\ldots\) is defined by \(L_{n+2}=L_n+L_{n+1}\) for \(n=1,2,3,\ldots\) with \(L_1=1\) and \(L_2=3\).

The Lucas numbers satisfy the same recurrence relation as the Fibonacci numbers with different initial values.

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

In the same way as for the Fibonacci sequence we can solve the difference equation

\[L_{n+2}=L_n+L_{n+1}\quad\Longleftrightarrow\quad L_{n+2}-L_{n+1}-L_n=0.\]

Suppose that \(L_n=r^n\) for some \(r\in\mathbb{R}\), then we have:

\[r^{n+2}-r^{n+1}-r^n=0\quad\Longleftrightarrow\quad r^n(r^2-r-1)=0.\]

If we assume that \(r\neq0\), then we have: \(r^2-r-1=0\) with two (different) real solutions:

\[r^2-r-1=0\quad\Longrightarrow\quad r=\frac{1\pm\sqrt{5}}{2}.\]

Due to the linearity (the principle of superposition) it follows that

\[L_n=c_1\left(\frac{1+\sqrt{5}}{2}\right)^n+c_2\left(\frac{1-\sqrt{5}}{2}\right)^n,\quad n=0,1,2,\ldots.\]

The initial values \(L_0=2\) and \(L_1=1\) imply that

\[c_1+c_2=2\quad\text{and}\quad c_1\left(\frac{1+\sqrt{5}}{2}\right)+c_2\left(\frac{1-\sqrt{5}}{2}\right)=1.\]

Hence:

\[\left\{\begin{array}{l}c_1+c_2=2\\[2.5mm]c_1(1+\sqrt{5})+c_2(1-\sqrt{5})=2\end{array}\right.\quad\Longleftrightarrow\quad \left\{\begin{array}{l}c_1+c_2=2\\[2.5mm]c_1\sqrt{5}-c_2\sqrt{5}=0\end{array}\right.\]

and this implies that \(c_1=c_2=1\).

Hence we have

\[L_n=\left(\frac{1+\sqrt{5}}{2}\right)^n+\left(\frac{1-\sqrt{5}}{2}\right)^n,\quad n=0,1,2,\ldots.\]

The number \(\varphi=\displaystyle\frac{1+\sqrt{5}}{2}\) is also known as the golden ratio. Now we have

\[L_n=\varphi^n+(1-\varphi)^n,\quad n=0,1,2,\ldots.\]

In the same way as for the Fibonacci sequence we may conclude that \(\displaystyle\lim\limits_{n\to\infty}\frac{L_{n+1}}{L_n}=\varphi\).

More results on Lucas numbers

Since \(L_{n+2}=L_n+L_{n+1}\) for \(n=1,2,3,\ldots\) with \(L_1=1\) and \(L_2=3\) we use the telescoping property to find that

\[\sum_{k=1}^nL_k=\sum_{k=1}^n\left(L_{k+2}-L_{k+1}\right)=L_{n+2}-L_{n+1}+L_{n+1}-L_n+\cdots+L_3-L_2=L_{n+2}-3.\]

Furthermore, applying partial fractions we obtain

\[\frac{1}{L_nL_{n+2}}=\frac{A}{L_n}+\frac{B}{L_{n+2}}=\frac{AL_{n+2}+BL_n}{L_nL_{n+2}} =\frac{A\left(L_n+L_{n+1}\right)+BL_n}{L_nL_{n+2}}=\frac{(A+B)L_n+AL_{n+1}}{L_nL_{n+2}}.\]

Note that this is satisfied by \(A=\displaystyle\frac{1}{L_{n+1}}\) and \(B=-\displaystyle\frac{1}{L_{n+1}}\), which implies that

\[\frac{1}{L_nL_{n+2}}=\frac{1}{L_nL_{n+1}}-\frac{1}{L_{n+1}L_{n+2}}.\]

Now we use the telescoping property to find that

\[\sum_{n=1}^{\infty}\frac{1}{L_nL_{n+2}}=\lim\limits_{N\to\infty}\sum_{n=1}^N\left(\frac{1}{L_nL_{n+1}}-\frac{1}{L_{n+1}L_{n+2}}\right) =\lim\limits_{N\to\infty}\left(\frac{1}{L_1L_2}-\frac{1}{L_{N+1}L_{N+2}}\right)=\frac{1}{3}\]

and

\[\sum_{n=1}^{\infty}\frac{L_{n+1}}{L_nL_{n+2}}=\lim\limits_{N\to\infty}\sum_{n=1}^{\infty}\left(\frac{1}{L_n}-\frac{1}{L_{n+2}}\right) =\lim\limits_{N\to\infty}\left(\frac{1}{L_1}+\frac{1}{L_2}-\frac{1}{L_{N+1}}-\frac{1}{L_{N+2}}\right)=1+\frac{1}{3}=\frac{4}{3}.\]

Similarly, we have

\begin{align*} \frac{1}{L_nL_{n+4}}&=\frac{A}{L_n}+\frac{B}{L_{n+4}}=\frac{AL_{n+4}+BL_n}{L_nL_{n+4}}=\frac{A\left(L_{n+2}+L_{n+3}\right)+BL_n}{L_nL_{n+4}}\\[2.5mm] &=\frac{A\left(L_{n+2}+L_{n+1}+L_{n+2}\right)+BL_n}{L_nL_{n+4}}=\frac{A\left(2L_{n+2}+L_{n+2}-L_n\right)+BL_n}{L_nL_{n+4}}\\[2.5mm] &=\frac{A\left(3L_{n+2}-L_n\right)+BL_n}{L_nL_{n+4}}=\frac{3AL_{n+2}+(B-A)L_n}{L_nL_{n+4}}\quad\Longrightarrow\quad A=B=\frac{1}{3L_{n+2}}. \end{align*}

This implies that

\begin{align*} \sum_{n=1}^{\infty}\frac{1}{L_nL_{n+4}}&=\sum_{n=1}^{\infty}\left(\frac{1}{3L_nL_{n+2}}+\frac{1}{3L_{n+2}L_{n+4}}\right)\\[2.5mm] &=\frac{1}{3}\left(\sum_{n=1}^{\infty}\frac{1}{L_nL_{n+2}}+\sum_{n=1}^{\infty}\frac{1}{L_nL_{n+2}}-\frac{1}{L_1L_3}-\frac{1}{L_2L_4}\right) =\frac{1}{3}\left(\frac{2}{3}-\frac{1}{4}-\frac{1}{21}\right)=\frac{31}{252}. \end{align*}

Using \(\displaystyle\frac{1}{1-\varphi}+1-\varphi=1-2\varphi\) and \(\displaystyle\frac{1}{\varphi}+\varphi=2\varphi-1=\sqrt{5}\) we find that

\begin{align*} F_{n-1}+F_{n+1}&=\frac{\varphi^{n-1}-(1-\varphi)^{n-1}}{\sqrt{5}}+\frac{\varphi^{n+1}-(1-\varphi)^{n+1}}{\sqrt{5}} =\frac{\varphi^n\left(\frac{1}{\varphi}+\varphi\right)-(1-\varphi)^n\left(\frac{1}{1-\varphi}+1-\varphi\right)}{\sqrt{5}}\\[2.5mm] &=\frac{2\varphi-1}{\sqrt{5}}\left(\varphi^n+(1-\varphi)^n\right)=\varphi^n+(1-\varphi)^n=L_n,\quad n=1,2,3,\ldots. \end{align*}

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Last modified on April 7, 2024