skylsn

2022-06-26

The sequence is defined by recurrence: ${x}_{1}=1$, ${x}_{2}=2$
${x}_{n+1}=\frac{1}{2}\left({x}_{n}+{x}_{n-1}\right)$

Harold Cantrell

${x}_{n+2}$ is the middle of $\left[{x}_{n},{x}_{n+1}\right]$, therefore the distance between ${x}_{n}$ and ${x}_{n+1}$ is divided by 2 at each iteration, namely
$\frac{{x}_{n+2}-{x}_{n+1}}{2}=\frac{{x}_{n+1}+{x}_{n}-2{x}_{n+1}}{4}=-\frac{1}{2}\frac{{x}_{n+1}-{x}_{n}}{2}$
(there is a - because $\left({x}_{2n}\right)$ and $\left({x}_{2n+1}\right)$ are of opposite monotony). Therefore
$\frac{{x}_{n+1}-{x}_{n}}{2}={\left(-\frac{1}{2}\right)}^{n-1}\frac{{x}_{2}-{x}_{1}}{2}=\frac{\left(-1{\right)}^{n-1}}{{2}^{n}}$
Thus ${x}_{n+1}-{x}_{n}=\frac{\left(-1{\right)}^{n-1}}{{2}^{n-1}}$, summing this gives that
${x}_{n}={x}_{1}+\sum _{k=1}^{n-1}\left({x}_{k+1}-{x}_{k}\right)=1-\sum _{k=1}^{n-1}{\left(-\frac{1}{2}\right)}^{k-1}=1+\frac{1-\left(-1/2{\right)}^{n-1}}{1+1/2}=1+2\frac{1-\left(1/2{\right)}^{n-1}}{3}$
Therefore $\underset{n\to +\mathrm{\infty }}{lim}{x}_{n}=\frac{5}{3}$ (you could also solve the characterisic equation of $\left({x}_{n}\right)$ and do the standard calculations.)

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