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Solve a rational recurrence by taking reciprocals

Invert first, then look for a pattern

If $a_{n + 1} = a_{n} / (1 + c a_{n})$ and the terms are nonzero, define $b_{n} = 1/ a_{n}$ . Then $b_{n + 1} = b_{n} + c$ , an arithmetic sequence that is easy to solve.

Problem

The sequence satisfies $a_{1} = 1/13$ and $a_{n + 1} = a_{n} / (1 + 3 a_{n})$ . If $a_{2026} = 1/ N$ , find $N$ . Also find the smallest positive integer $n$ for which $a_{n} < 1/1000$ .

Answer

N = 6088

; the first qualifying index is

n = 331

The inequality is strict, so n=330 does not qualify.

Step-by-step solution

1. Take reciprocals

Let $b_{n} = 1/ a_{n}$ . The recurrence becomes $b_{n + 1} = (1 + 3 a_{n}) / a_{n} = b_{n} + 3$ . Since $b_{1} = 13$ , we have $b_{n} = 13 + 3 (n - 1) = 3 n + 10$ .

2. Find the distant term

Thus $a_{n} = 1/ (3 n + 10)$ , and $N = 3 (2026) + 10 = 6088$ .

3. Find the threshold

Because the denominators are positive, $1/ (3 n + 10) < 1/1000$ exactly when $3 n + 10 > 1000$ . This is $n > 330$ , so the first integer index is 331.

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