Limit of a mean sequence

Question

The recursive formula is $$t_n=\frac {t_{n-1}+t_{n-2}}2$$ as $n$ approaches infinity the mean sequence converges at a certain number. Changing $t_1$ and $t_2$ changes the number where the sequence converges.

Find a relationship between $t_1, t_2$ and where the mean sequence converges.

Prove the relationship.

What i have so far

Working part 1

Continuation of working

then i will proceed to plug in an into the formula given by @астонвіллаолофмэллбэрг [limtn=t1+∑∞n=2(tn−tn−1)] and solve. Where does that equation come from and how could i better format this

Answer 1

Given $t_1 \leq t_2$ WLOG, it is easy to see that $t_1 \leq t_n \leq t_2$ for all $n$. Now, because $t_n$ is bounded, there exists a convergent subsequence of this sequence. Finally, note that $t_1 \leq t_2 \geq t_3 \leq t_4 \leq t_5 $ etc.

I claim further that $t_n$ is Cauchy.Note that $(t_n - t_{n-1}) = \dfrac{(t_{n-1} - t_{n-2})(-1)^{n}}{2}$, so it is a positive sequence that decreases to zero.

Finally, every Cauchy sequence with a convergent sequence is itself a convergent sequence, hence $\lim t_n$ exists as $n \to \infty$.

Now, since the limit exists, it is correct to write: $\lim t_n = t_1 + \sum_{n=2}^\infty (t_n - t_{n-1})$. But then, we observe that this is just $t_1 + (t_2-t_1) \sum_{n=1}^\infty \Big(\frac{-1}{2} \Big)^n = t_1+(t_2-t_1)\frac{2}{3} = \frac{1}{3} t_1 + \frac{2}{3}t_2$.

Hence the limit is $\frac{t_1+2t_2}{3}$ when $t_1 \leq t_2$. It would switch if the comparison also flips.

Answer 2

HINT

$2t_n -t_{n-1}-t_{n-2}=0$ therefore $t_n=Ax_1^n + Bx_2^n$ where $x_1, x_2$ are the solutions of $2x^2 -x -1 =0$ and $A, B$ are constants determined by the $t_1, t_2$ values.

Answer 3

The characteristic polynomial associated with the recurrence relation is $$P(x) = x^2 - \frac12x -\frac12$$

The roots are $1$ and $-1/2$. Since they're all distinct, the $n$-th term of the relation can be written as $$t_n = c_1 {\left(-\frac12\right)}^n + c_2.$$

The two constants are chosen in order to fit the first terms. In our case, we have: $$\begin{cases} t_1 = -\frac12 c_1 + c_2\\[1ex] t_2 = \frac14 c_1 + c_2 \end{cases}$$ giving $$\begin{cases} c_1 = \frac43(t_2 - t_1)\\[1ex] c_2 = \frac13(t_1 + 2t_2) \end{cases}$$

It follows that $$t_n = {(-1)}^n\frac{2^{2-n}}3(t_2 - t_1) + \frac13(t_1 + 2t_2)$$

Can you conclude?