Convergence rate of $\alpha_n = \ln(1+\frac{(-1)^n}{n})$?

Question

Expanding $\ln(1+\frac{(-1)^n}{n})$ into its Maclaurin series we get:

$\ln(1+\frac{(-1)^n}{n}) = \frac{(-1)^n}{n} - \frac{(-1)^2n}{2!n} + \frac{(-1)^3n}{3!n} + \cdots + R$ for some remainder R.

Letting $\beta_n = \frac{(-1)^n}{n}$ and taking $\lim_{n \to \infty} \frac{|\alpha_n - 0|}{|\beta_n|}$ we get $1.$

But I'm sure the rate of convergence is supposed to be $\frac{1}{n}$ not $\frac{(-1)^n}{n}$. Are they the same? If so, how can I show it?

Thanks!

Answer 1

$ln(1+x)=x-\frac {x^{2}} 2+o(x^{2})$ so $\alpha_n =\frac {(-1)^{n}} n -\frac 1 {2n^2} +o(n^{-2})$. This means $\alpha_n - \frac {(-1)^{n}} n$ tends to $0$ at the rate of $\frac 1 {n^{2}}$.

Answer 2

The definition of rate of convergence uses absolute values... and is a number, not a sequence. A sequence $x_n \to a$ has rate of convergence $\beta$ if $$ \lim \frac{|x_{n+1}-a|}{|x_n-a|^{\beta}} = M $$ for some constant $M>0$. In your case, you must search for a number $\beta$ such that the limit

$$ \lim \frac{\left|\ln\left(1+\frac{(-1)^{n+1}}{n+1}\right)\right|}{\left|\ln\left(1+\frac{(-1)^{n}}{n}\right)\right|^{\beta}} $$

is a positive number. In fact, if you take $\beta=1$, the above limit is $1$, showing that the sequence converges linearly to zero.