Do $e^{c/n}$ and $1+\frac{c}{n}$ have the same rate of convergence as n grows to infinity?

Question

I am wondering whether the functions $e^{c/n}$ and $1+\frac{c}{n}$ converge to one at the same rate as $n$ grows to infinity? I was trying to establish this based on Wikipedia's definitions of convergence rates but was unable to do so.

I don't think that convergence is linear since the quotient in the first definition here must not be exactly equal to one.

Any help would be much appreciated!

Answer 1

There are surely easier and quicker ways, but sticking to the main road here...

Yes. As someone else mentioned in a hint, consider the Maclaurin series expansion of $e^x$. It's $$ e^x = \sum_{k=0}^{+\infty} \frac{x^k}{k!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots $$ Therefore, $$ e^{c/n} = 1 + \frac{c}{n} + \frac{c^2}{2! \ n^2} + \frac{c^3}{3! \ n^3} + \cdots$$

The $c/n$ in the expansion above is the term that converges the slowest as $n \to +\infty$. To put it another and incredibly informal way, the $\frac{c^2}{2! \ n^2} + \frac{c^3}{3! \ n^3} + \frac{c^4}{4! \ n^4} + \cdots $ terms will all be "done" converging before $c/n$ is "done" converging. (I put "done" in quotes because these expressions all converge to $0$ but technically they're never really done, because $n$ is never actually equal to $+\infty$.)

Therefore $e^{c/n}$ and $1 + \frac{c}{n}$ do indeed converge at the same rate.

And this rate is logarithmic. We can use the definition to show this.

First, it should be clear that the convergence is sublinear for both. Let $x_n = 1 + \frac{c}{n}$. Then we have $$ \lim_{n\to+\infty} \frac{|x_{n+1} - 1|}{|x_n - 1|} = \lim_{n \to +\infty} \frac{|c|/(n+1)}{(|c|/n)} = \lim_{n\to+\infty} \frac{n}{n+1} = 1.$$

Now let $y_n = e^{c/n}$. Then we have $$ \lim_{n \to+\infty} \frac{|y_{n+1}-1|}{|y_n-1|} = \lim_{n\to+\infty} \frac{e^{c/(n+1)}}{e^{c/n}} = \frac{e^0}{e^0} = 1.$$

From the link you provided:

If the sequence converges sublinearly and additionally $$ \lim_{k\to+\infty}\frac{|x_{k+2}-x_{k+1}|}{|x_{k+1}-x_k|} = 1,$$ then it is said the sequence $\{x_k\}$ converges logarithmically to $L$.

For our $x_n = 1 + \frac{c}{n}$, we have $$\frac{|x_{n+2} - x_{n+1}|}{|x_{n+1} - x_n|} = \cdots = \frac{|(n+1)-(n+2)|}{|n - (n+1)|} = 1. $$ The details have been left to the reader.

It can be shown that our $y_n = e^{c/n}$ satisfies the definition of logarithmic convergence. This has also been left to the reader.

Answer 2

If we set $u=\frac{c}{n}$, then $u$ approaches $0$ and we have

$$\lim_{u\rightarrow 0} \frac{e^u}{1+u}=\lim_{u\rightarrow 0} \frac{1+u+u^2/2+...}{1+u}=1$$

Because of the existence of the limit, we can conclude that the functions converge at the same rate towards $1$.