Home » Blog » Prove that the recursive sequence 1 / xn+2 = 1 / xn+1 + 1 / xn converges

Prove that the recursive sequence 1 / xn+2 = 1 / xn+1 + 1 / xn converges

Prove that the sequence \{ x_n \} be defined by the recursive relationship,

    \[ x_0 = 1, \qquad x_1 = 1, \qquad \frac{1}{x_{n+2}} = \frac{1}{x_{n+1}} + \frac{1}{x_n} \]

converges and find the limit of the sequence.


Proof. First, we show that the sequence \{ x_n \} is monotonically decreasing for all n \geq 1. For the base case we have x_1 = 1 and

    \[ \frac{1}{x_2} = \frac{1}{1} + \frac{1}{1} = 2 \quad \implies \quad x_2 = \frac{1}{2}. \]

Hence, x_2 < x_1. Assume then that for all positive integers up to some k we have x_{k+1} < x_k. Then,

    \begin{align*}  && \frac{1}{x_{k+2}} &= \frac{1}{x_{k+1}} + \frac{1}{x_k} \\[9pt]  \implies && \frac{1}{x_{k+2}} &> \frac{1}{x_{k+1}} + \frac{1}{x_{k+1}} &(x_{k+1} < x_k \implies \frac{1}{x_{k+1}} > \frac{1}{x_k}) \\[9pt]  \implies && \frac{1}{x_{k+2}} &> \frac{2}{x_{k+1}} \\[9pt]  \implies && x_{k+2} &< \frac{x_{k+1}}{2} \\[9pt]  \implies && x_{k+2} &< x_{k+1}. \end{align*}

Thus, the sequences is monotonically decreasing. The sequence is certainly bounded below since all of the terms are greater than 0. Therefore, the sequence converges. \qquad \blacksquare

To compute the limit of the sequence, assume the sequence converges to a finite limit L (justified since we just proved that it does indeed converge). Therefore,

    \begin{align*}  && \lim_{n \to \infty} x_n &= L \\[9pt]  \implies && \lim_{n \to \infty} \frac{1}{x_n} &= \frac{1}{L} \\[9pt]  \implies && \lim_{n \to \infty} \left( \frac{1}{x_{n-1}} + \frac{1}{x_{n-2}} \right) &= \frac{1}{L} \\[9pt]  \implies && \frac{1}{L} + \frac{1}{L} &= \frac{1}{L} \\[9pt]  \implies && 2L &= L \\[9pt]  \implies && L &= 0. \end{align*}

2 comments

  1. tom says:

    The second to last line of the IH fails for k=0; 1/2<1/2 is NOT true. These numbers appear to be reciprocals of the fibonacci numbers btw.

    • tom says:

      Also, since the given reciprocals are increasing, can’t we say the sequence is decreasing and bounded by zero (obviously the terms don’t go negative)?

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