aanpalendmw

Answered

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:

${S}_{2}=\frac{b\cdot {S}_{2-1}}{b+{S}_{2-1}}+a=\frac{b\cdot {S}_{1}}{b+{S}_{1}}+a=\frac{b\cdot (a+b)}{b+(a+b)}+a$

And for ${S}_{3}$:

${S}_{3}=\frac{b\cdot {S}_{3-1}}{b+{S}_{3-1}}+a=\frac{b\cdot {S}_{2}}{b+{S}_{2}}+a=\frac{b\cdot (\frac{b\cdot (a+b)}{b+(a+b)}+a)}{b+(\frac{b\cdot (a+b)}{b+(a+b)}+a)}+a$

And so on. Now my question is: what happens when $n\to \mathrm{\infty}$?

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:

${S}_{2}=\frac{b\cdot {S}_{2-1}}{b+{S}_{2-1}}+a=\frac{b\cdot {S}_{1}}{b+{S}_{1}}+a=\frac{b\cdot (a+b)}{b+(a+b)}+a$

And for ${S}_{3}$:

${S}_{3}=\frac{b\cdot {S}_{3-1}}{b+{S}_{3-1}}+a=\frac{b\cdot {S}_{2}}{b+{S}_{2}}+a=\frac{b\cdot (\frac{b\cdot (a+b)}{b+(a+b)}+a)}{b+(\frac{b\cdot (a+b)}{b+(a+b)}+a)}+a$

And so on. Now my question is: what happens when $n\to \mathrm{\infty}$?

Answer & Explanation

Mira Spears

Expert

2022-07-18Added 14 answers

Assume that ${S}_{n}=\frac{{p}_{n}}{{q}_{n}}$ is associated with $({p}_{n},{q}_{n})\in {\mathbb{R}}^{2}$. The recurrence

$\begin{array}{}\text{(1)}& {S}_{n}=\frac{(a+b){S}_{n-1}+ab}{{S}_{n-1}+b}\end{array}$

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 $\{{p}_{n}{\}}_{n\ge 0}$ and the $\{{q}_{n}{\}}_{n\ge 0}$ 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 ${\zeta}_{\pm}=\frac{(a+2b)\pm \sqrt{{a}^{2}+4ab}}{2}$.

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 $(A,B,C,D)$ 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 $L=\frac{1}{2}(a\pm \sqrt{{a}^{2}+4ab})$

$\begin{array}{}\text{(1)}& {S}_{n}=\frac{(a+b){S}_{n-1}+ab}{{S}_{n-1}+b}\end{array}$

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 $\{{p}_{n}{\}}_{n\ge 0}$ and the $\{{q}_{n}{\}}_{n\ge 0}$ 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 ${\zeta}_{\pm}=\frac{(a+2b)\pm \sqrt{{a}^{2}+4ab}}{2}$.

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 $(A,B,C,D)$ 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 $L=\frac{1}{2}(a\pm \sqrt{{a}^{2}+4ab})$

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