The Recurrence x(n+1)=x(n)/(1+x(n))

Introduction

A recurrence can define a sequence whose terms become smaller at every step, but that observation alone does not say what value the sequence approaches. The recurrence \(x_{n+1}=\frac{x_n}{1+x_n}\) is a useful example: for every positive initial value, its terms stay positive and decrease, yet zero is never reached in a finite number of steps.

We will prove that the sequence is bounded below and converges to zero. We will also derive an exact expression for every term. The expression gives more information than convergence alone: it lets us calculate how many steps are needed to get below a chosen positive threshold.

The Recurrence and Its Basic Properties

Let \(a>0\), set \(x_1=a\), and define subsequent terms by \(x_{n+1}=\frac{x_n}{1+x_n}\) for \(n\in\mathbb N\). The recurrence is defined whenever \(1+x_n\ne 0\). We will show that every term is positive, so its denominator is always greater than one.

If \(x_n>0\), then \(1+x_n>1\). Therefore, \(\frac{x_n}{1+x_n}>0\), so \(x_{n+1}>0\). Since \(x_1=a>0\), induction proves that \(x_n>0\) for every \(n\).

For each positive term, comparison with the preceding term gives \(\frac{x_n}{1+x_n}<x_n\): the numerator is positive and the denominator is greater than one. Thus \(x_{n+1}<x_n\) for every \(n\). Also, positivity shows that zero is a lower bound. The sequence is therefore decreasing and bounded below.

Worked Example: Starting from Two

Why the Sequence Converges to Zero

The monotone convergence theorem gives a limit \(L\ge 0\). To identify it, take limits in the recurrence. The function \(f(x)=\frac{x}{1+x}\) is continuous for \(x\ne-1\), and \(L\ge0\), so it is continuous at \(L\). Since both \(x_n\) and its shifted sequence \(x_{n+1}\) converge to \(L\), the recurrence implies \(L=\frac{L}{1+L}\).

This proves convergence, not just that zero is a plausible destination: monotonicity and the lower bound first guarantee that a limit exists, and the recurrence then determines its value.

An Exact Formula from Reciprocals

The recurrence has an especially simple form after taking reciprocals. This is valid because every \(x_n\) is positive. Dividing numerator and denominator by \(x_n\), or directly taking the reciprocal of both sides, gives \(\frac{1}{x_{n+1}}=\frac{1+x_n}{x_n}=\frac{1}{x_n}+1\). Thus each new reciprocal is exactly one greater than the previous reciprocal.

To justify the formula, start with \(\frac1{x_1}=\frac1a\). If \(\frac1{x_n}=\frac1a+n-1\), then the reciprocal recurrence gives \(\frac1{x_{n+1}}=\frac1{x_n}+1=\frac1a+n\). This is the claimed expression with \(n+1\) in place of \(n\), so induction proves it for all \(n\in\mathbb N\). Taking reciprocals yields the displayed formula for \(x_n\).

The formula also verifies the recurrence directly. Write \(D_n=1+(n-1)a\), which is positive because \(a>0\). Then \(x_n=\frac{a}{D_n}\) and \(D_n+a=1+na\), so

Finally, the denominator \(1+(n-1)a\) increases without bound as \(n\) increases, while the numerator \(a\) remains fixed. Therefore the exact formula also gives \(x_n\to0\), in agreement with the monotone-limit proof.

Worked Example: Starting from One Third

How Many Steps to Get Below a Threshold?

The exact expression gives a direct convergence estimate. Given \(\varepsilon>0\), the condition \(x_n<\varepsilon\) becomes \(\frac{a}{1+(n-1)a}<\varepsilon\). All quantities being compared are positive, so multiplying by the denominator preserves the inequality. Rearrangement gives \(n>1+\frac1\varepsilon-\frac1a\). Any natural number \(N\) satisfying this strict inequality ensures \(x_n<\varepsilon\) for \(n\ge N\), since the sequence is decreasing.

A Common Algebraic Pitfall

When showing that the terms decrease, it is easy to lose a factor while subtracting fractions. A tempting but incorrect calculation is

The denominator has not disappeared: both terms must first be written over the same denominator. The correct calculation is

Because \(x_n>0\), the numerator \(x_n^2\) and denominator \(1+x_n\) are both positive, so the final expression is strictly negative. For instance, when \(x_n=2\), the true difference is \(\frac23-2=-\frac43\), whereas the incorrect expression \(-x_n^2\) would give \(-4\). The exact difference confirms strict decrease without confusing the amount of decrease with its sign.

Why This Recurrence Is Useful

This example illustrates a reliable sequence of reasoning for a recursive sequence. First establish that the recurrence is defined at every step and preserves the relevant sign. Then compare consecutive terms and find a lower bound. Those properties yield convergence; the recurrence and continuity identify the limit. Here, the reciprocal substitution adds a further benefit by converting the nonlinear recurrence into the simple relation \(\frac1{x_{n+1}}=\frac1{x_n}+1\).

The exact formula also distinguishes a limit from an attained value. Since \(x_n>0\) for every finite \(n\), the sequence never equals zero, even though its limit is zero. This is consistent with strict decrease and the lower bound: the terms approach the bound without reaching it.