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Of course, I know that $$ \exp(x)=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n. $$ My question is what happens when instead of $x$, we have a real-valued function depending on $n$, i.e. do we also have

$(1+\frac{f(n)}{n})^n\sim\exp(f(n))$?

For example:

$\exp(\log(r+1))\sim \left(1+\frac{\log(r+1)}{r+1}\right)^{r+1}$?

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    Almost related http://math.stackexchange.com/questions/1460918/is-the-formula-lim-x-to-a-1fxgx-e-lim-x-to-afxgx-a-stand/ – 2017-01-26
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    Depends on how $f(n)$ behaves. If $f(n)$ converges [to $F$] as $n\to\infty$ then the limit as $n\to\infty$ of $(1+f(n)/n)^n$ is $e^{F}$. If for example $f(n) = n^2$ then $(1+f(n)/n)^n \approx n^n = e^{n\log(n)}$ which is not $\sim e^{f(n)} = e^{n^2}$. – 2017-01-26
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    A simple example, which does not seem to appear below: if $f(n)=\alpha n$ with $\alpha>-1$, then $$\left(1+\frac{f(n)}n\right)^n=(1+\alpha)^n$$ while $$e^{f(n)}=\left(e^\alpha\right)^n$$ and $0<1+\alpha$$\left(1+\frac{f(n)}n\right)^n\ll e^{f(n)}$$ – 2017-01-27

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The idea is, for "not so bad" $f$, to compute an asymptotic expansion of $\ln\left(1+\frac{f(n)}{n}\right)$, until we get $o(1)$. Let's consider the following example ...

We would like to obtain a simple equivalent sequence for :

$$u_n=\left(1+\frac{1}{\sqrt{n}}\right)^n$$

which correspond to your question in the special case $f(n)=\sqrt n$.

We have :

$$\ln\left(1+\frac{1}{\sqrt n}\right)=\frac{1}{\sqrt n}-\frac{1}{2n}+o\left(\frac{1}{n}\right)$$

Thus :

$$n\ln\left(1+\frac{1}{\sqrt n}\right)=\sqrt n-\frac{1}{2}+o(1)$$

and hence :

$$\boxed{\left(1+\frac{1}{\sqrt n}\right)^n\sim\exp\left(\sqrt n-\frac{1}{2}\right)}$$

By the way, we see that the equivalence $\left(1+\frac{1}{\sqrt n}\right)^n\sim e^\sqrt n$ is false ...

EDIT :

For your final question, i will rather consider (for convenience) :

$$\left(1+\frac{\ln(r)}{r}\right)^r$$

We have, as $r\to+\infty$ :

$$r\ln\left(1+\frac{\ln(r)}{r}\right)=r\left(\frac{\ln(r)}{r}-\frac{\ln^2(r)}{2r^2}+o\left(\frac{\ln^2(r)}{r^2}\right)\right)$$

and, a fortiori :

$$r\ln\left(1+\frac{\ln(r)}{r}\right)=\ln(r)+o(1)$$

Hence :

$$\left(1+\frac{\ln(r)}{r}\right)^r\sim r$$

In this case, the formula is correct !

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    @Mark Fischler - Almost the same answer, almost in the same time :) – 2017-01-26
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    How do you get $\ln\left(1+\frac{1}{\sqrt{n}}\right)=\frac{1}{\sqrt{n}}-\frac{1}{2n}+o(\frac{1}{n})$? And whats with my example with the logaritzhm? – 2017-01-26
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    I am using the Taylor expansion of $\ln(1+t)$ as $t\to0$ : $$\ln(1+t)=t-\frac{t^2}{2}+o(t^2)$$ – 2017-01-26
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    How do you get from $r\ln\left(1+\frac{\ln(r)}{r}\right)^r=\ln(r)+o(1)$ to $\left(1+\frac{\ln(r)}{r}\right)^r\sim r$? The $\sim$ means - by the other answers, that $\lim_{r\to\infty}\frac{\left(1+\frac{\ln(r)}{r}\right)^r}{r}=L$ for some $L>0$. – 2017-01-26
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    If you suppose that $f(r)=g(r)+o(1)$ as $r\to\infty$, then you have $\frac{\exp(f(r))}{\exp(g(r))}=\exp(f(r)-g(r))=\exp(o(1))\to1$, so that $\exp(f(r))\sim\exp(g(r))$ – 2017-01-26
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    BTW, notation $F(r)\sim G(r)$ (as $r\to\infty$) means $\lim_{t\to\infty}\frac{F(r)}{G(r)}=1$ – 2017-01-26
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    Great thanks! Got it. – 2017-01-26
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For example, if $f(n)=n^2$, your approximation leads to $ (1+n)^n\sim e^{n^2}$. But $e^{n^2}/n^{n}\rightarrow\infty$.

If, for example, $f$ is bounded, then your conjecture is true. Moreover, since $\log(1+x)=x+o(x)$, we have $$\left( 1+ \frac{f(n)}{n} \right)^n= \exp \left( n\cdot \log(1+\frac{f(n)}{n}) \right) = \exp \left( f(n)+f(n)o(n) \right)$$ So, if $f(n)$ is "little" in comparison with $o(x)$, your approximation is true; like when $f$ is bounded or $f=O(x)$.

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When you use $\sim$ in $$ \left(1+\frac{f(n)}{n}\right)^n \sim_{n\to\infty} e^{f(n)} $$ What you are saying is that there exists some finite, non-zero $L$ (which may depend on the specific $f(n)$ you are looking at) such that $$ \lim_{n\to\infty} \frac{\left(1+\frac{f(n)}{n}\right)^n }{e^{f(n)}} = L $$

We can show that this is not true in general. For example, when $f(n) = n^2$, $$ \lim_{n\to\infty} \frac{\left(1+\frac{n^2}{n}\right)^n }{e^{n^2}} = \lim_{n\to\infty} \frac{\left(1+n\right)^n }{e^{n^2}} = 0 $$

On the other hand, the statement is sometimes true. Let $f(n) = \sqrt{n}$, then $$ \lim_{n\to\infty} \frac{\left(1+\frac{\sqrt{n}}{n}\right)^n }{e^{\sqrt{n}}} = \lim_{n\to\infty} \frac{\left(1+\frac{1}{\sqrt{n}}\right)^n }{e^{\sqrt{n}}} = \frac{1}{\sqrt{e}} $$ which is a constant.

It appears that if $f(n)$ grows faster than $O(n)$, the statement is false (the limit is zero.

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    There might be different definitions out there, however I think most commonly the $\sim$ symbol is used to denote only the case $L=1$ like in https://en.wikipedia.org/wiki/Asymptotic_analysis – 2017-01-26
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    $\sim$ definitely means $L=1$. If $\lim f/g = L$ for some nonzero $L$ then we would write $f \sim Lg$. – 2017-01-26