Series $(1+b^n/n)^n$ at infinity

Question

Is there a series expansion for $$\left ( 1 + \frac{b^n}{n} \right )^n$$

at $n\rightarrow\infty$, assuming $0<b<1$?

Similar to $$ \left ( 1 + \frac{b}{n} \right )^n \rightarrow e^b $$

I want to want to write this as $$\left ( 1 + \frac{b^n}{n} \right )^n \rightarrow e^{b^n} $$

But I am not sure if this is correct. Thanks!

@DacianBonta How can this look OK? When you take a limit as $n\to\infty$, whatever result you get cannot still feature a $n$ anywhere. At the very least, mention to the OP this is not the proper use of the symbol $\to$. — Clement C., Apr 24 '17 at 00:21
And unless $x=0$, you never have $e^x=1+x$, no matter how small $x$ is. You have $e^x = 1+x+o(x)$, but you cannot write that equality for $x\neq 0$ without specifying the $o(x)$ (or $O(x^2)$). — Clement C., Apr 24 '17 at 00:24
More generally if a sequence $x_{n} \to x$ then $(1+(x_{n}/n))^{n}\to e^{x} $. See this answer https://math.stackexchange.com/a/1787254/72031 for proof. Here $b^{n} \to 0$ and hence the desired limit is $e^{0}=1$. — Paramanand Singh, Apr 28 '17 at 18:10

RRL · Accepted Answer · 2017-04-24T16:24:27.703

For $0 < b < 1$, we have $b^n \to 0$ as $n \to \infty$ and it is true that

$$\lim_{n \to \infty} \left(1 + \frac{b^n}{n} \right)^n = e^0 = 1.$$

This follows because

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

converging uniformly for $x$ in a compact set.

Justification for all of this requires showing, first, given sequences $f_n(x)$ and $a_n$, if $f_n(x) \to f(x)$ uniformly and $a_n \to a$, then $f_n(a_n) \to f(a)$. This is a straightforward application of the properties of uniform convergence.

Second, we have to verify the uniform convergence of the limit (*) for $x \in [0,1)$.

If $0 \leqslant x < n$, we have

$$\log(1+x/n) = \frac{x}{n} - \frac{x^2}{2n^2} + \frac{x^3}{3n^3} - \frac{x^4}{4n^4} + \ldots,$$

and, using the triangle inequality

$$\left|n\log(1+x/n) - x\right| \leqslant \frac{x^2}{2n} + \frac{x^3}{3n^2} + \frac{x^4}{4n^3} + \ldots.$$

Since, $0 \leqslant x < 1$, for all $n > 2 \geqslant 2x$ we have $x/n \leqslant 1/2$ and

$$\left|n\log(1+x/n) - x\right| \leqslant \frac{x^2}{n}\left[ \frac{1}{2} + \frac{x}{3n} + \frac{x^2}{4n^2} + \ldots\right] \\ \leqslant \frac{x^2}{n}\left[ \frac{1}{2} + \frac{1}{2 \cdot 3} + \frac{1}{4 \cdot 2^2} + \ldots\right] \\ \leqslant \frac{x^2}{n}\left[ \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \ldots\right] \\ = \frac{x^2}{n} \\ \leqslant \frac{1}{n}.$$

Thus,

$$\lim_{n \to \infty} n\log(1+x/n) = \lim_{n \to \infty} \log(1+x/n)^n = x,$$

uniformly for $0 \leqslant x < 1$.

The function $x \mapsto \exp(x)$ is uniformly continuous on compact sets and it follows that

$$\lim_{n \to \infty} (1+x/n)^n = e^x, $$

uniformly.

Thanks for your answer. A different but related question: in this case we could approximate $(1+x/n)^n \approx 1+ x$ at small $x$; would it also be correct for $x=b^n$? — smörkex, Apr 24 '17 at 02:32
Yes. There are many ways to approach this, all leading to $f_n(b^n) \to f(0)$ where $f_n(x)= (1 +x/n)^n$ and $f(x) = e^x$ since $b^n$ converges to $0$ and stays inside the compact interval $[0,1]$. I tried to give you a general justification based on uniform convergence which I showed by expanding $\ln(1+x/n)$. You could also use the binomial expansion for $(1+x/n)^n$ which reduces to what you are saying since the higher order terms can be neglected as $n$ gets large -- and again because we can bound $|f_n(b^n) - f(0)| = O(1/n)$. — RRL, Apr 24 '17 at 02:43

score 1 · Answer 2 · answered Apr 24 '17 at 02:09

The usual way we show that $$\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n=e^x$$ is by setting a variable to that limit and using algebra like so: $$y=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n\implies\ln(y)=\lim_{n\to\infty}n\ln\left(1+\frac{x}{n}\right)=\lim_{n\to\infty}\dfrac{\ln\left(1+\frac{x}{n}\right)}{\frac{1}{n}}\to\dfrac{0}{0}$$ L'Hopital's: $$\lim_{n\to\infty}\dfrac{\ln\left(1+\frac{x}{n}\right)}{\frac{1}{n}}=\lim_{n\to\infty}\dfrac{\dfrac{-x}{n^2\left(1+\frac{x}{n}\right)}}{-\frac{1}{n^2}}=\lim_{n\to\infty}\dfrac{x}{1+\frac{x}{n}}=x\implies y=e^x$$ Now, we can try practically the same idea with the $b^n$ (remember that $0\le b<1\implies \lim\limits_{n\to\infty}b^n=0$ and hence $\lim\limits_{n\to\infty}\frac{b^n}{n}=0$):

$$y=\lim_{n\to\infty}\left(1+\frac{b^n}{n}\right)^n\implies\ln(y)=\lim_{n\to\infty}n\ln\left(1+\frac{b^n}{n}\right)=\lim_{n\to\infty}\dfrac{\ln\left(1+\frac{b^n}{n}\right)}{\frac{1}{n}}\to\dfrac{0}{0}$$ L'Hopital's: $$\lim_{n\to\infty}\dfrac{\ln\left(1+\frac{b^n}{n}\right)}{\frac{1}{n}}=\lim_{n\to\infty}\dfrac{\dfrac{-\frac{b^n}{n^2}+\frac{\ln(b)b^n}{n}}{\left(1+\frac{b^n}{n}\right)}}{-\frac{1}{n^2}}=\lim_{n\to\infty}\dfrac{b^n-n\ln(b)b^n}{1+\frac{b^n}{n}}\to\frac{0}{1}=0\implies \ln(y)\to0\implies y\to\boxed{1}$$

Series $(1+b^n/n)^n$ at infinity

2 Answers2