Consider this sequence
$$ a_0 = 2,$$
$$ a_{n+1} = \frac 1 a_n $$
Now to prove for convergence I simply put in values ($a_1, a_2, a_3$) and got the the sequence has this behaviour: $2, 1/2, 2, 1/2,\dots$. Now since it isn't approaching any value it shouldn't be converging.
But I would like to prove this, so I looked up how to do it and you're supposed to prove monotony and limitations of the sequence. Now to prove monotony we would have to assume that the sequence is falling or raising. But how can I do that here? Since the sequence is changing its value from <1 to >1 in every iteration what should I assume? Is there a way to solve this without the "obvious" answer?
Thank you!
Best Answer
Let us solve $A_n A_{n+1}=1$, take $A_n=\frac{B_{n+1}}{B_{n}}$, then We get $B_{n+2}=B_{n}$. Next, let $B_n=x^n$, then we get $x^2=1 \implies x=\pm 1$ We can write $$B_n=P +Q(-1)^n \implies A_n=\frac{P+Q(-1)^{n+1}}{P+Q(-1)^{n}}=\frac{R+(-1)^{n+1}}{R+(-1)^{n}}.$$ $$A_0=2 \implies \frac{R-1}{R+1}=2 \implies R=-3$$ Hence $$A_n=\frac{-3-(-1)^n}{-3+(-1)^n}$$
Therefore $A_n$ does not converge but it oscillates:$2,1/2,2,1/2,...$