skylsn

2022-06-26

The sequence is defined by recurrence: ${x}_{1}=1$, ${x}_{2}=2$

${x}_{n+1}=\frac{1}{2}({x}_{n}+{x}_{n-1})$

${x}_{n+1}=\frac{1}{2}({x}_{n}+{x}_{n-1})$

Harold Cantrell

Beginner2022-06-27Added 21 answers

${x}_{n+2}$ is the middle of $[{x}_{n},{x}_{n+1}]$, 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 $({x}_{2n})$ and $({x}_{2n+1})$ are of opposite monotony). Therefore

$\frac{{x}_{n+1}-{x}_{n}}{2}={(-\frac{1}{2})}^{n-1}\frac{{x}_{2}-{x}_{1}}{2}=\frac{(-1{)}^{n-1}}{{2}^{n}}$

Thus ${x}_{n+1}-{x}_{n}=\frac{(-1{)}^{n-1}}{{2}^{n-1}}$, summing this gives that

${x}_{n}={x}_{1}+\sum _{k=1}^{n-1}({x}_{k+1}-{x}_{k})=1-\sum _{k=1}^{n-1}{(-\frac{1}{2})}^{k-1}=1+\frac{1-(-1/2{)}^{n-1}}{1+1/2}=1+2\frac{1-(1/2{)}^{n-1}}{3}$

Therefore $\underset{n\to +\mathrm{\infty}}{lim}{x}_{n}=\frac{5}{3}$ (you could also solve the characterisic equation of $({x}_{n})$ and do the standard calculations.)

$\frac{{x}_{n+2}-{x}_{n+1}}{2}=\frac{{x}_{n+1}+{x}_{n}-2{x}_{n+1}}{4}=-\frac{1}{2}\frac{{x}_{n+1}-{x}_{n}}{2}$

(there is a - because $({x}_{2n})$ and $({x}_{2n+1})$ are of opposite monotony). Therefore

$\frac{{x}_{n+1}-{x}_{n}}{2}={(-\frac{1}{2})}^{n-1}\frac{{x}_{2}-{x}_{1}}{2}=\frac{(-1{)}^{n-1}}{{2}^{n}}$

Thus ${x}_{n+1}-{x}_{n}=\frac{(-1{)}^{n-1}}{{2}^{n-1}}$, summing this gives that

${x}_{n}={x}_{1}+\sum _{k=1}^{n-1}({x}_{k+1}-{x}_{k})=1-\sum _{k=1}^{n-1}{(-\frac{1}{2})}^{k-1}=1+\frac{1-(-1/2{)}^{n-1}}{1+1/2}=1+2\frac{1-(1/2{)}^{n-1}}{3}$

Therefore $\underset{n\to +\mathrm{\infty}}{lim}{x}_{n}=\frac{5}{3}$ (you could also solve the characterisic equation of $({x}_{n})$ and do the standard calculations.)

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