We have recursively given sequence:
$$
a_1 = 0\\
a_{2n}=\frac{1}{2}a_{2n-1}\\
a_{2n+1}=a_{2n}+\frac{1}{2}
$$
What I did:
I separated the two analysis of two sequences into even and odd sequence convergence tests, if they are equal, then the whole sequence converges.
$$a_{2n}=\frac{1}{2}a_{2n-1}=\frac{1}{2}(a_{2n-2}+\frac{1}{2})$$
We now prove that the sequence is increasing:
$$a_{2(n+1)}\geq a_{2n}$$
Base:
$$a_2 = 0, a_4 = \frac{1}{4}\implies a_4 \geq a_2$$
Induction step:
$$a_{2n}\geq a_{2n-2}\\
\frac{1}{2}(a_{2n-2}+\frac{1}{2})\geq a_{2n-2}\\
\vdots\\
\frac{-1}{2}a_{2n-2}\geq\frac{-1}{4} \iff a_{2n-2}\leq \frac{1}{2}
$$
So now we know that the even sequence increases if the terms are bellow $\frac{1}{2}$.
Let's see if that is true:
$$a_{2n}\leq \frac{1}{2}\\
a_0=0\leq\frac{1}{2}\\
a_{2n+2}=\frac{1}{2}(a_{2n}+\frac{1}{2})\leq \frac{1}{2}(\frac{1}{2}+\frac{1}{2})=\frac{1}{2}
$$
Now we know that the sequence is increasing and bounded above, so the limit must exist:
$$
\lim_{n\rightarrow \infty}a_{2n+2} = \frac{1}{2}(\lim_{n\rightarrow \infty}a_{2n}+\frac{1}{2})\\
\vdots\\
2a = a + \frac{1}{2}\\
a = \frac{1}{2} = L_1
$$
Proof ended for even sequences. Then I would proof for odd sequences in the same way and see if $L_1$ is equal to $L_2$, which I already know it isn't since I already solved the whole problem.
But my question is, is my process of proving OK? Are there any mistakes ?
(the dots and other trivial calculations that I skipped would be of course sufficiently done)
Best Answer
What you have done is correct. You can even show that $a_{2n} = 1/2 - (1/2)^n$. This implies that $a_{2n-1} = 2a_{2n} = 1 - (1/2)^{n-1}$. Hence $a_{2n} \to 1/2$ and $a_{2n-1} \to 1$.
But is immediately clear, without the above formulae, that $(a_n)$ does not converge. If it would, then $a = \lim_{n \to \infty} a_{2n+1} = \lim_{n \to \infty} (a_{2n} + 1/2) = a +1/2$ which is absurd.