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The problem is the following: suppose $s_1>s_2>0$ and let $s_{n+1}=\frac{1}{2} (s_n+s_{n-1})$ for $n\geq 2$. Show that ($s_n$) converges.

Now, here is what I figured out:

  • $s_2
  • Assume $s_{2n-2}
  • Induction step: $s_{2n}
  • $s_1>s_3$: Base Case for induction that $s_{2n-1}$ is a decreasing sequence.
  • Assume $s_{2n-1}
  • Induction step: $s_{2n+1}

I have proved those two. However arguing in favor of convergence has me going around in circles. Since $s_1>s_2$ and (as discovered during the formulation of Base Cases) $s_3>s_4$, I figured it might be a good idea ot assume that if every odd member of the original sequence ($s_n$) is greater than the following even member, then the limit would be somewhere in between, the two (odd and even) sequences won't cross. Hence the upper and lower bounds would be $s_1$ and $s_2$ respectively. Here is how I approach this:

  • Assume $s_{2n-1}>s_{2n}$
  • Show that $s_{2n+1}>s_{2n+2}$

The proof as I mentioned has me running in circles. Any assistance?

3 Answers 3

1

Consider the characteristic equation of your recurrence:

$X^2-\frac1{2}X-\frac1{2}=0$

Which has solutions $1$ and $-\frac1{2}$.

Therefore the general expression for $s_n$ is:

$s_n=a\cdot (1)^n+b\cdot(-\frac1{2})^n=a+b\cdot(-\frac1{2})^n$

You can determine the value of the constants $a$ and $b$ using $s_1$ and $s_2$.

Therefore

$\lim_{n\to\infty} s_n=a$.

1

From the definition of $s_n$ one has $$ s_{n+1}-s_n=-{1\over2}(s_n-s_{n-1}). $$ It follows that if $s_{2n}-s_{2n-1}<0$ then $$ s_{2n+2}-s_{2n+1}={1\over4}(s_{2n}-s_{2n-1})<0. $$

0

$(s_{2n})$ is increasing and $(s_{2n-1})$ is decreasing and we have $s_2\leqslant s_{2n}