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Does $\lim_{n \to \infty} a_n = \alpha$ imply $\lim_{n \to \infty} \left(1+ \frac {a_n}{n} \right)^n =e^\alpha$?

I'm only considering real value
and also, does the value of $\alpha$ makes a difference?\

I tried considering $$\left|\left(1+ \frac {a_n}{n} \right)^n-e^\alpha\right|\leq\left|\left(1+ \frac {a_n}{n} \right)^n-\left(1+ \frac {\alpha}{n} \right)^n\right|+\left|\left(1+ \frac {\alpha}{n} \right)^n-e^\alpha\right|$$ I know that for any fixed $\alpha$, $\left|\left(1+ \frac {\alpha}{n} \right)^n-e^\alpha\right|$ tends to zero.

But I don't know how to deal with $$\left|\left(1+ \frac {a_n}{n} \right)^n-\left(1+ \frac {\alpha}{n} \right)^n\right|$$, I tried binomial expansion but didn't get anywhere. Is this where the value of $\alpha$ come to matter？ Thanks

Thomas Andrews
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fourofour_un
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    Look at the lemma at the top of this answer. https://math.stackexchange.com/a/1451245/7933 – Thomas Andrews May 25 '21 at 20:28
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    In particular, it is easier to show $$\left(\frac{1+\frac{a_n}{n}}{1+\frac{\alpha}n}\right)^n\to 1$$ than it is to prove the difference goes to zero. – Thomas Andrews May 25 '21 at 20:30
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    Here is another answer in the case $a_n$ is a complex convergence sequence. – Mittens May 25 '21 at 20:39
  • Thank you! That lemma is a really cool approach to me, I wonder where I can learn more about this? I could understand the lemma, but I don’t feel like I’ve grasped its full meaning, and the minus 1 in n*g(n)-1 seem somewhat not so straightforward to me. – fourofour_un May 25 '21 at 20:48

2 Answers2

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HINT: If $a_n\to \alpha > 0$ as $n\to\infty$, then $b_n := na_n\to\infty$ as $n\to\infty$. Therefore, make the substitution $n \mapsto na_n$ in the limit, which yields the following equality:

$$\lim_{n\to\infty} \bigg(1+\frac{a_n}{n}\bigg)^n = \lim_{n\to\infty} \bigg(1+\frac{a_n}{na_n}\bigg)^{na_n} = \lim_{n\to\infty} \bigg(1+\frac{1}{n}\bigg)^{na_n}$$

Does this help?

Franklin Pezzuti Dyer
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It requires breaking up the expansion of $e^a$ into two parts depending on $\epsilon$. Let $N_1$ be defined by $\sum\limits_{k=N_1+1}^\infty \frac{a^k}{k!}\lt \epsilon$. Then the binomial expansion can be used on the sum to $N_1$. Let $N_2$ be defined by $\sum\limits_{k=0}^{N_1} \binom{n}{k}|a_n^k-a^k|\lt \epsilon$ for $n\gt N_2$. Let $N=max(N_1,N_2)$ Expansion differences should be $\lt 2\epsilon$ for $n\gt N$.

To handle the tails for $a_n$ would require a little work showing $|\sum\limits_{k=N_1}^n\binom{n}{k}a_n^k-\sum\limits_{k=N_1+1}^\infty \frac{a^k}{k!}|\lt \sum\limits_{k=N_1+1}^\infty \frac{a^k}{k!}$ for $n\gt N$.

herb steinberg
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