Series $(1+b^n/n)^n$ at infinity 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!
 A: 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.
A: 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: $$\lim_{n\to\infty}\dfrac{\ln\left(1+\frac{b^n}{n}\right)}{\frac{1}{n}}=\lim_{n\to\infty}\dfrac{\dfrac{-\frac{b^n}{n^2}+\frac{\ln(b)b^n}{n}}{\left(1+\frac{b^n}{n}\right)}}{-\frac{1}{n^2}}=\lim_{n\to\infty}\dfrac{b^n-n\ln(b)b^n}{1+\frac{b^n}{n}}\to\frac{0}{1}=0\implies \ln(y)\to0\implies y\to\boxed{1}$$
