Limit of the Recursive Sequence $X_1=1, X_{n+1}= \frac 1{1 + X_n}$

Question

We have that $$ X_1 = 1, \quad X_{n+1} = \frac 1{1 + X_n}. $$ If $X_n$ has a limit find algebraically the value of the limit.

The sequence goes;

$$1,\frac{1}{2},\frac{3}{5},\frac{5}{8}, ...$$ The sequence appears to be converging to 0.6 as $n$ tends to infinity. May you have a say on this one. Thanks

Answer 1

Hint: Your recurrence is of the form $X(n+1) = f(X(n))$, where $f$ is the continuous function $f(x) = \frac 1{1 + x}$. It follows that if the sequence $X(n)$ approaches a limit $L$, then this limit $L$ must satisfy $$ L = f(L). $$ That is, $L$ must be a fixed point of the function.

Answer 2

Hint:

If $X(n) \to L$, where $L$ is the limit in question, then obviously the same is true for $X(n+1)$. Thus in the limit

$$X(n+1) = \frac{1}{1+X(n)} \text{ becomes } L = \frac{1}{1+L}$$

Solve for $L$.

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Warning:

You will find two possible values for $L$. Be mindful of which value makes sense based on the fact that $X(1)$. (Which value of $L$ does the sequence $X(n)$ converge to?)

Answer 3

The sequence converges to $\frac{1}{\varphi},$ where $\varphi$ is the golden ratio. Indeed, notice that $X_n=\frac{F_n}{F_{n+1}},$ with $F_n$ the $n$-th term in the Fibonacci sequence (with $F_{1}=F_2=1.$) Indeed, we can prove this by induction, as $X_{n+1}=\frac{1}{1+X_n}=\frac{1}{(F_{n+1}+F_n)/F_{n+1}}=\frac{F_{n+1}}{F_{n+1}+F_n}=\frac{F_{n+1}}{F_{n+2}}.$

It is a known fact that the ratio of two consecutive Fibonacci terms converges to the golden ratio, and in this case we have the inverse of that ratio.

P.D. your first terms are wrong. You are missing a 2/3 between 1/2 and 3/5.