Showing exp(x) definitions are the same.

[DeadOriginal]

Homework Statement

e^{x}=\sum\limits_{k=0}^{\infty}\frac{x^{k}}{k!} and e^{x}=\lim\limits_{n\rightarrow\infty}\left(1+\frac{x}{n}\right)^{n}.

I want to show that \sum\limits_{k=0}^{\infty}\frac{x^{k}}{k!}=\lim\limits_{n\rightarrow\infty}\left(1+\frac{x}{n}\right)^{n}.

2. The attempt at a solution
Let $$t_{n}=\left(1+\frac{x}{n}\right)^{n}.$$ Then by the binomial theorem we have
$$ t_{n}=1+x+\frac{x^{2}}{2!}\left(\frac{n-1}{n}\right)+\frac{x^{3}}{3!}\left(\frac{n-1}{n}\right)\left(\frac{n-2}{n}\right)+\cdots+\frac{x^{n}}{n!}\left(\frac{n-1}{n}\right)\left(\frac{n-2}{n}\right)\cdots\left(\frac{1}{n}\right)=\sum\limits_{k=0}^{n}\frac{x^{k}}{k!}\frac{n!}{n^{k}(n-k)!}.$$
Then
$$
\begin{align*}
\lim\limits_{n\rightarrow\infty}t_{n}=\lim\limits_{n\rightarrow\infty} \sum_{k=0}^{n}\frac{x^{k}}{k!}\frac{n!}{n^{k}(n-k)!}
&=\sum_{k=0}^{\infty}\lim\limits_{n\rightarrow\infty}\frac{x^{k}}{k!}
\frac{n!}{n^{k}(n-k)!}\\
&=1+x+\frac{x^{2}}{2!}\lim\limits_{n\rightarrow\infty}\frac{n-1}{n}+\frac{x^{3}}{3!}\lim\limits_{n\rightarrow\infty}\frac{n-1}{n}\frac{n-2}{n}+\cdots \\
&=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\cdots \\
&=\sum_{k=0}^{\infty}\frac{x^{k}}{k!}.
\end{align*}
$$

Hence
$$\lim\limits_{n\rightarrow\infty}t_{n}=\lim\limits_{n\rightarrow\infty}\left(1+\frac{x}{n}\right)^{n}=\sum\limits_{k=0}^{\infty}\frac{x^{k}}{k!}.$$

Can someone check if my reasoning works? I am particularly worried about the part where I say
$$
\lim\limits_{n\rightarrow\infty} \sum_{k=0}^{n}\frac{x^{k}}{k!}\frac{n!}{n^{k}(n-k)!}=\sum_{k=0}^{\infty}\lim\limits_{n\rightarrow\infty}\frac{x^{k}}{k!}
\frac{n!}{n^{k}(n-k)!}.
$$

 

[mfb]

Can someone check if my reasoning works? I am particularly worried about the part where I say
$$
\lim\limits_{n\rightarrow\infty} \sum_{k=0}^{n}\frac{x^{k}}{k!}\frac{n!}{n^{k}(n-k)!}=\sum_{k=0}^{\infty}\lim\limits_{n\rightarrow\infty}\frac{x^{k}}{k!}
\frac{n!}{n^{k}(n-k)!}.
$$

That is certainly not true for arbitrary expressions.

$$\lim\limits_{n\rightarrow\infty} \sum_{k=0}^{n} \frac{1}{n+1} = \lim\limits_{n\rightarrow\infty} 1 = 1$$
$$\sum_{k=0}^{\infty}\lim\limits_{n\rightarrow\infty} \frac{1}{n+1} = \sum_{k=0}^{\infty} 0 = 0$$

 

[DeadOriginal]

Hmm. Thanks. Could you recommend a way to tackle this problem?

 

[Boorglar]

yeah putting the limit inside the summation requires some proving, because limit sum laws don't necessarily work for infinite sums. I think you could try something like that, for n ≥ 2:

\left(1+\frac{x}{n}\right)^n = \sum_{k=0}^{n}\frac{n!}{k!(n-k)!}\frac{x^k}{n^k} = 1 + x + \sum_{k=2}^{n}\frac{n(n-1)...(n-k+1)}{n^k}\frac{x^k}{k!}
= 1 + x + \sum_{k=2}^{n}\left(1-\frac{1}{n}\right)...\left(1-\frac{k-1}{n}\right)\frac{x^k}{k!}.

Then I am not quite sure how, but you'll need to prove that taking the limit inside the sum is ok. I tried some ideas but they failed miserably, so I'll think about it. It's an interesting problem, though!

 

[verty]

I would try to expand the second definition for n = 1,2... and show how it relates to the first definition. Hopefully it correlates perfectly when n = k.

Oh wait, it won't.

 

[Office_Shredder]

It's actually easier if you don't try to take both limits at the same time. The infinite summation
\sum_{k=0}^{\infty} \frac{x^k}{k!}
is an infinite series that converges for all x. For each fixed n,
\left(1+\frac{x}{n}\right)^n
is a power series (with only finitely many terms) that converges for all x. It's easier to show that as n goes to infinity, the latter converges to the former by taking their difference and noting that for each n, you have a power series that converges for each x - all that is left is to show that for a fixed x, as n goes to infinity the series goes to zero for any fixed x that you pick

 

[DeadOriginal]

It's actually easier if you don't try to take both limits at the same time. The infinite summation
\sum_{k=0}^{\infty} \frac{x^k}{k!}
is an infinite series that converges for all x. For each fixed n,
\left(1+\frac{x}{n}\right)^n
is a power series (with only finitely many terms) that converges for all x. It's easier to show that as n goes to infinity, the latter converges to the former by taking their difference and noting that for each n, you have a power series that converges for each x - all that is left is to show that for a fixed x, as n goes to infinity the series goes to zero for any fixed x that you pick

Interesting. It didn't cross my mind to think of it like that. Thank you.