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Let $f : \Bbb N \to [0, \infty[$ and $g : \Bbb N \to [0, \infty[$ be two functions, such that $f(n) \to 0$ and $g(n) \to +\infty$ when $n \to \infty$.

What are some (necessary/sufficient) conditions on $f$ and $g$ for the limit $$a_{f,g} := \lim_{n \to \infty} (1+f(n))^{g(n)}$$ to exist and be finite?

For instance, if $f(n) = x/n, g(n)=n$, then the limit is $e^x$. On the other hand, if $f(n)=1/n, g(n)=n^2$, then the limit is $+\infty$.

If $f(n) = a^n, g(n)=n^k$ with $a \in [0,1[$ and $k \geq 1$, then the limit exists since $(1+a^n)^n$ is decreasing for $n$ large enough (take the log and then the derivative wrt $n$), and the sequence is bounded below by $1$. On the other hand, if $f(n) = a^n, g(n)=e^n$ (with $a \in [0,1[$) then the sequence $(1+f(n))^{g(n)}$ is unbounded.

Any help would be appreciated to understand better what conditions I should have on $f$ and $g$.

Alphonse
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    Since $(1+f)^g=[(1+f)^{1/f}]^{fg}$ and $(1+f)^{1/f}\to e$, all is determined by whether $fg$ has either a finite limit $L$ or if it tends to $-\infty$. In the first case the limit will be $e^L$, in the second it will be $0$. If $fg$ is unbounded so will be $(1+f)^g$. If $fg$ accumulates at two values $a$ and $b$ then $(1+f)^{g}$ will accumulate at $e^a$ and $e^b$, which are different since $e^x$ is injective. – conditionalMethod Dec 04 '19 at 18:55
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    Related question, almost a duplicate. – rtybase Dec 04 '19 at 19:09
  • Related: https://math.stackexchange.com/questions/29816 – Watson Mar 28 '22 at 12:51

2 Answers2

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We have that

$$(1+f(n))^{g(n)}=e^{g(n)\log(1+f(n))}$$

and since

$$g(n)\log(1+f(n))=g(n)\cdot f(n)\cdot \frac{\log(1+f(n))}{f(n)}$$

with

$$\frac{\log(1+f(n))}{f(n)}\to 1$$

all boils down in $g(n)f(n)$.

user
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You have, since $\lim_{n\to\infty}f(n)=0$, $$ (1+f(n))^{g(n)} = e^{g(n)\ln(1+f(n))} = \exp\left( {g(n)(f(n)+o(f(n)))} \right) = \exp\left( {g(n)f(n)+o(f(n)g(n))} \right) $$ This limit of $e^{a_n}$ exists if, and only if, the limit of $a_n$ exists in $\mathbb{R}\cup \{-\infty\}$. so a necessary and sufficient condition is that $\lim_{n\to\infty} f(n)g(n)$ exists in $\mathbb{R}\cup \{-\infty\}$.

Clement C.
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