Introduction to Real Analysis: Having trouble proving boundedness/convergence I am stuck on an intro to real analysis question that asks to prove convergence. 
The question follows as "Let $y_1<y_2$ be arbitrary real numbers and 
$y_n =\frac{1}{2}(y_{n-1}+y_{n-2})$ for $n>2$. Prove $y_n$ is convergent."
My first thought was to prove it is increasing so I can use the monotone convergence theorem. However, I keep getting stuck as my proof skills aren't the sharpest yet. This is what I have so far:
We will try to prove $y_n$ is a bounded increasing monotone sequence.                    
We know $y_3= \frac{1}{2}(y_1+y_2$)  which is equivalent to $2y_3=y_1+y_2$
It follows that 
$y_1<y_2<y_1+y_2 =2y_3$
$y_1<y_2<\frac{1}{2}y_1+\frac{1}{2}y_2 = y_3$ 
$y_1<y_2<y_3$
We can continue this pattern infinitely thus concluding $y_n$ is increasing. 
This is where i get stuck because i do not know how to prove $y_n$ is bounded 
Any help would be appreciated.
 A: hint: $y_n - y_{n-1} = -(y_n - y_{n-2})= -((y_n - y_{n-1})+(y_{n-1}-y_{n-2}))$. Thus you have: $z_n = -z_n - z_{n-1}\implies z_n = -\dfrac{z_{n-1}}{2}$ with $z_n = y_n - y_{n-1}$. Can you find a formula for $z_n$ and then for $y_n$ then you can see which value the $y_n$ converges to .
A: The sequence is not monotone; I don't know if there is an easy way to show abstractly that it is convergent. 
If we look at a few terms, it is not hard to deduce that 
$$
y_{n+2}=\frac1{2^n}\left(\frac{2^n-(-1)^n}3\,y_1+\frac{2^{n+1}+(-1)^n}{3} \,y_2\right).
$$
One then confirms this is the right formula by showing that it satisfies the original recursion. 
Finally, taking the limit, 
$$
\lim_{n\to\infty} y_n=\lim_{n\to\infty} y_{n+2}\frac13\,y_1+\frac23\,y_2. 
$$
A: Here is a standard argument (used in the proof of Cantor's Intersection Theorem): use induction to verify that $|y_n-y_{n-1}| \leq |y_1 -y_2|/2^{n-2}$. Once you do this you see that the series $\sum (y_n -y_{n-1})$ is absolutely convergent. Its n-th partial sum is $y_n-y_1$ so $\lim y_n$ exists.
