- #1
Lambda96
- 189
- 65
- Homework Statement
- Show that the following applies with tasks b and c ##\lim_{n\to\infty} \Bigl( 1 + \frac{z}{n} \Bigr)^n = e^z##
- Relevant Equations
- Tasks b and c
Hi,
I have problems proving task d
I then started with task c and rewrote it as follows ##\lim_{n\to\infty}\sum\limits_{k=0}^{N}\Bigl( \frac{z^k}{k!} - \binom{n}{k} \frac{z^k}{n^k} \Bigr)=0 \quad \rightarrow \quad \lim_{n\to\infty}\sum\limits_{k=0}^{N} \frac{z^k}{k!} = \lim_{n\to\infty}\sum\limits_{k=0}^{N} \binom{n}{k} \frac{z^k}{n^k}##
I can use the right-hand side of the equation, i.e. ##\lim_{n\to\infty}\sum\limits_{k=0}^{N} \binom{n}{k} \frac{z^k}{n^k}## with the help of the binomial theorem rewrite as follows ##\lim_{n\to\infty}\sum\limits_{k=0}^{N} \binom{n}{k} \frac{z^k}{n^k}= \lim_{n\to\infty} \Big( 1+ \frac{z}{n} \Bigr)^n ##
Now I just have to show that the left-hand side is ##\lim_{n\to\infty}\sum\limits_{k=0}^{N} \frac{z^k}{k!}=e^z##. Unfortunately, I can't get any further here, I assume that I can use task 1b for this, but unfortunately I don't know how.
I have problems proving task d
I then started with task c and rewrote it as follows ##\lim_{n\to\infty}\sum\limits_{k=0}^{N}\Bigl( \frac{z^k}{k!} - \binom{n}{k} \frac{z^k}{n^k} \Bigr)=0 \quad \rightarrow \quad \lim_{n\to\infty}\sum\limits_{k=0}^{N} \frac{z^k}{k!} = \lim_{n\to\infty}\sum\limits_{k=0}^{N} \binom{n}{k} \frac{z^k}{n^k}##
I can use the right-hand side of the equation, i.e. ##\lim_{n\to\infty}\sum\limits_{k=0}^{N} \binom{n}{k} \frac{z^k}{n^k}## with the help of the binomial theorem rewrite as follows ##\lim_{n\to\infty}\sum\limits_{k=0}^{N} \binom{n}{k} \frac{z^k}{n^k}= \lim_{n\to\infty} \Big( 1+ \frac{z}{n} \Bigr)^n ##
Now I just have to show that the left-hand side is ##\lim_{n\to\infty}\sum\limits_{k=0}^{N} \frac{z^k}{k!}=e^z##. Unfortunately, I can't get any further here, I assume that I can use task 1b for this, but unfortunately I don't know how.