Obtain the convergence order of a succession of real numbers

Question

How can the convergence order of the next succession be calculated?

$$\left(1+\frac{1}{n}\right)^{\frac{1}{2}}$$

Answer 1

Note that

$$x^*=\lim_{n\to\infty} \left(1+\frac{1}{n}\right)^{\frac{1}{2}}= 1$$

We are looking for $p$ such that

$$\lim_{n\to\infty} \frac{|x_{n+1}-x^*|}{|x_n-x^*|^p}= \lim_{n\to\infty} \frac{\big|\left(1+\frac{1}{n+1}\right)^{\frac{1}{2}}-1\big|}{\big|\left(1+\frac{1}{n}\right)^{\frac{1}{2}}-1\big|^p}=C>0$$

since by binomial series

$$\left(1+\frac{1}{n}\right)^{\frac{1}{2}}=1+\frac1{2n}+o\left(\frac1n\right)$$

we have that

$$\left(1+\frac{1}{n}\right)^{\frac{1}{2}}-1=\frac1{2n}+o\left(\frac1n\right)$$

$$\left(1+\frac{1}{n+1}\right)^{\frac{1}{2}}-1=\frac1{2(n+1)}+o\left(\frac1n\right)$$

and thus

$$\lim_{n\to\infty} \frac{\big|\left(1+\frac{1}{n+1}\right)^{\frac{1}{2}}-1\big|}{\big|\left(1+\frac{1}{n}\right)^{\frac{1}{2}}-1\big|^p}= \lim_{n\to\infty} \frac{\frac1{2(n+1)}+o\left(\frac1n\right)}{\left(\frac1{2n}+o\left(\frac1n\right)\right)^p}=\begin{cases}1 \quad p=1\\0 \quad p<1 \\\infty \quad p>1\end{cases}$$

thus the order is $p=1$.