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*}


  1. Artem says:

    I do not thing that substituting limits is actually allowed, since the limit is 0, thus, 1/L = infinity? This is undefined, by Apostol theorem, where it is allowed to apply limit quotient rule only if the denominator is not zero.

    The way to solve this is to use the upper bound of 1/n. The adventurer needs to show that x_n <= 1/n. This can be done by induction. After this use the squeeze theorem 0 <= x_n <= 1/n, which shows that the limit is zero.

  2. 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.

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):