I make one examlpe from the book "Introduction of real analysis" by G. Bartle, R. Sherbert.
We have a sequence $x_1:=1, x_2:=2$ and $(x_n):=\frac{1}{2}(x_{n-2}+x_{n-1})$. So it's not difficult to see, that
$|x_n-x_{n+1}|=\frac{1}{2^{n-1}}$ for $n \ \in \ \mathbb N$. To establish the limit we have to consider the subsequnce of $(x_n)$ with odd indices $(x_{2n+1})=1+\frac{1}{2}+\frac{1}{2^3}+....+\frac{1}{2^{2n-1}}$.
Can somebody explain me, how did we get this sum on the right-hand side? I have tried to play with $(x_n)$, but I didn't get this sum. Thank you in advance!
You can establish both $|x_n-x_{n+1}|=\frac{1}{2^{n-1}}$ $(*)$ and $x_{2n+1}= 1+\frac12 +\dots+\frac{1}{2^{2n-1}}$ inductively. As you are asking about the latter, I will assume the former.
For $n=1$, $x_n=1$ is an assumption and thus the base case is trivial. Now, assume $x_{2n+1}= 1+\frac12 +\dots+\frac{1}{2^{2n-1}}$. Let us show $x_{2(n+1)+1}=x_{2n+3}= 1+\frac12 +\dots+\frac{1}{2^{2n+1}}$. Use the definition, then $(*)$ to write $$x_{2n+3}=\frac12(x_{2n+2}+x_{2n+1}) = \frac12(2x_{2n+1}+\frac{1}{2^{2n}}) = x_{2n+1}+\frac{1}{2^{2n+1}},$$ and the induction hypothesis completes the proof.
I expect it is this way you are expected to see the formula. Depending on your level, it could have been nice if they have added "it is easy by induction to see that..." or something similar, making it easier to accept.