$(1 + \frac{i}{n})^{n^2}$的聚点

hbghlyj

\[\lim_{n \to \infty }\left(1 + \frac{i}{n}\right)^{n^2}\color{#f00}\ne\lim_{n \to \infty }\left(\lim_{n \to \infty }\left(1 + \frac{i}{n}\right)^{n}\right)^n=\lim_{n \to \infty }\left(e^i\right)^n=\lim_{n \to \infty }e^{in}\]右边的聚点为$u,∀|u|=1$
但是左边数列的聚点的集合为$\sqrt e\cdot u,∀|u|=1$

取对数

$\log(1+ix)=\frac{1}{2} \log \left(1+x^2\right)+i \arg (1+i x),∀x∈\Bbb R$
代入$x=\frac1n$得$\log(1+\frac in)=n^2 \left(\frac{1}{2} \log \left(1+\frac{1}{n^2}\right)+i \arg \left(1+\frac{i}{n}\right)\right)$
而$n^2 \left(\frac{1}{2} \log \left(1+\frac{1}{n^2}\right)\right)\to\frac12$. 取exp得到$\lim_{n \to \infty }\left(1 + \frac{i}{n}\right)^{n^2}=\sqrt e\cdot u,∀|u|=1$

这里不能把$n^2$写成$n\cdot n$, 然后先求里面的极限:
因为函数$\square^n$不会收敛, 比如$\lim_{n\to\infty}(2^{1/n})^n≠\lim_{n\to\infty}(\lim_{n\to\infty}2^{1/n})^n$.

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连续函数列$f_n(x)$一致收敛到$f(x)$,那么$$\lim_{n\to\infty} f_n(g(n))=f\left(\lim_{n\to\infty}g(n)\right)$$
这个是正确的吗?

hbghlyj

hbghlyj 发表于 2022-11-4 16:24
这里不能把$n^2$写成$n\cdot n$然后先求里面的极限
\[\lim_{n \to \infty }\left(1 + \frac{i}{n}\r ...

math.stackexchange.com/questions/3907414
Consider the sequence of functions defined by
$$ f_n(x)= \left(1+ \frac{x} {n}\right )^{n^2}
$$ for $x \in [-1,0]$.
What can I say about the convergence?

My attempt

For every $n$, $f_n(0)=1$.
Then using the Taylor expression,
for every $x \in [-1,0)$
$$ \lim_{n \to +\infty} f_n (x) = \lim_{n \to +\infty}
e^{\ln(\left(1+ \frac{x} {n}\right)^{n^2}) } = \lim_{n \to +\infty}  e^{n^2 \ln(\left(1+ \frac{x} {n}\right)) } = \lim_{n \to +\infty} e^{n^2 \frac{x} {n}} =
\lim_{n \to +\infty} e^{nx} = 0
$$
We have a limit function which is not continuous, whereas every $f_n$ is continuous.
So we don't have uniform convergence.

("extra" curiosity: I was tempted to write down something like$$ f_n(x)= e^{xn} $$Is such a thing possible or completely wrong? )

hbghlyj

hbghlyj 发表于 2022-11-4 16:37
math.stackexchange.com/questions/3907414
Consider the sequence of functions defined by
$$ ...

根据上面的计算, 当$x>0$时极限是$+∞$, 当$x<0$时极限是0.