aanpalendmw

2022-07-17

Limit of a sequence of fractions
I have that:
${S}_{n}=\frac{b\cdot {S}_{n-1}}{b+{S}_{n-1}}+a$
where $n\in {\mathbb{N}}^{+}$ and ${S}_{1}=a+b$
So, for ${S}_{2}$ we will get:
${S}_{2}=\frac{b\cdot {S}_{2-1}}{b+{S}_{2-1}}+a=\frac{b\cdot {S}_{1}}{b+{S}_{1}}+a=\frac{b\cdot \left(a+b\right)}{b+\left(a+b\right)}+a$
And for ${S}_{3}$:
${S}_{3}=\frac{b\cdot {S}_{3-1}}{b+{S}_{3-1}}+a=\frac{b\cdot {S}_{2}}{b+{S}_{2}}+a=\frac{b\cdot \left(\frac{b\cdot \left(a+b\right)}{b+\left(a+b\right)}+a\right)}{b+\left(\frac{b\cdot \left(a+b\right)}{b+\left(a+b\right)}+a\right)}+a$
And so on. Now my question is: what happens when $n\to \mathrm{\infty }$?

Mira Spears

Expert

Assume that ${S}_{n}=\frac{{p}_{n}}{{q}_{n}}$ is associated with $\left({p}_{n},{q}_{n}\right)\in {\mathbb{R}}^{2}$. The recurrence
$\begin{array}{}\text{(1)}& {S}_{n}=\frac{\left(a+b\right){S}_{n-1}+ab}{{S}_{n-1}+b}\end{array}$
can be written in the following form
$\begin{array}{}\text{(2)}& \left(\begin{array}{c}{p}_{n}\\ {q}_{n}\end{array}\right)=\left(\begin{array}{cc}a+b& ab\\ 1& b\end{array}\right)\left(\begin{array}{c}{p}_{n-1}\\ {q}_{n-1}\end{array}\right)\end{array}$
hence the closed form of both the $\left\{{p}_{n}{\right\}}_{n\ge 0}$ and the $\left\{{q}_{n}{\right\}}_{n\ge 0}$ sequences just depends on the diagonalization of $M=\left(\begin{array}{cc}a+b& ab\\ 1& b\end{array}\right)$, whose eigenvalues are given by ${\zeta }_{±}=\frac{\left(a+2b\right)±\sqrt{{a}^{2}+4ab}}{2}$.
In particular
$\begin{array}{}\text{(3)}& {S}_{n}=\frac{A{\zeta }_{+}^{n}+B{\zeta }_{-}^{n}}{C{\zeta }_{+}^{n}+D{\zeta }_{-}^{n}}\end{array}$
for some constants $\left(A,B,C,D\right)$ depending on the initial values. By taking the limit as $n\to +\mathrm{\infty }$, we simply get $\frac{A}{C}$. On the other hand, by assuming ${S}_{n}\to L$ as $n\to \mathrm{\infty }$, such limit has to fulfill
$\begin{array}{}\text{(4)}& L=a+\frac{bL}{b+L}\end{array}$
hence the only chances are $L=\frac{1}{2}\left(a±\sqrt{{a}^{2}+4ab}\right)$

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