Prove that the sequence 
be defined by the recursive relationship,
![\[ x_0 = 1, \qquad x_1 = 1, \qquad \frac{1}{x_{n+2}} = \frac{1}{x_{n+1}} + \frac{1}{x_n} \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-55b12c65fffa7adf58717297c622b624_l3.png "Rendered by QuickLaTeX.com")
converges and find the limit of the sequence.
Proof. First, we show that the sequence is monotonically decreasing for all 
. For the base case we have 
and
![\[ \frac{1}{x_2} = \frac{1}{1} + \frac{1}{1} = 2 \quad \implies \quad x_2 = \frac{1}{2}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-10579fbadbe7185a1e6f1753b039f90f_l3.png "Rendered by QuickLaTeX.com")
Hence, 
. Assume then that for all positive integers up to some 
we have 
. 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*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-ff7d5fdbac90135e14a6a9e9c455bff2_l3.png "Rendered by QuickLaTeX.com")
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
To compute the limit of the sequence, assume the sequence converges to a finite limit 
(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*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-2cc1f3ab34079545e6d6c6a1a8c3929c_l3.png "Rendered by QuickLaTeX.com")
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.
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.
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)?
Yes, that’s exactly how I solved this problem
I meant I proved the terms are reciprocals of Fibonacci numbers and used their divergence to get the limit