Summation of an infinite series that $= e^a$

Question

I need to show that $$\sum_{k=0}^{\infty} {a^k\over k!} = e^a.$$ I am stuck... It seems like I know how to do all the other ones I have seen, but for some reason I can't put it all together and figure this one out. Thanks.

Answer 1

One is to assume that $$ \lim_{n\to\infty}\Bigl(1+\frac1n\Bigr)^n=e=\sum_{k=0}^\infty \frac1{k!} $$ is already known. Then show that for the function $$ E(x)=\sum_{k=0}^\infty \frac{x^k}{k!} $$ the identity $E(x)E(y)=E(x+y)$ holds. By standard techniques that implies that for all rational $x=\frac pq$ you get that $$ E\Bigl(\frac pq\Bigr)=E(1)^{\frac pq}=e^{\frac pq} $$ and by continuity of the power series the claim follows.

Answer 2

One might reasonably say it depends on which definition of $e^a$ is used. But maybe what is to be proved is that it satisfies laws of exponents, such as $e^{a+b} = e^a e^b$, etc. Here's a start on that. Let $$ f(a) = \sum_{k=0}^{\infty} \frac{a^k}{k!}. $$ Recall that $$ (a+b)^k = \sum_{j=0}^k \binom k j a^j b^{k-j} \text{ where } \binom k j = \frac{k!}{j!(k-j)!}. $$ Then we have \begin{align} f(a+b) & = \sum_{k=0}^\infty \frac{(a+b)^k}{k!} = \sum_{k=0}^\infty \left( \frac 1 {k!} \sum_{j=0}^k \frac{k!}{j!(k-j)!} a^j b^{k-j} \right) \\[10pt] & = \sum_{k=0}^\infty \left( \sum_{j=0}^k \frac{a^j}{j!} \cdot \frac{b^{k-j}}{(k-j)!} \right) \\[10pt] & = \sum_{m=0}^\infty \left(\sum_{n=0}^\infty \frac{a^m}{m!} \cdot \frac{b^n}{n!}\right) \tag 1 \\[10pt] & = \sum_{m=0}^\infty \left( \frac{a^m}{m!} \sum_{n=0}^\infty \frac{b^n}{n!} \right) \\[10pt] & = \left(\sum_{m=0}^\infty \frac{a^m}{m!} \right) \left( \sum_{n=0}^\infty \frac{b^n}{n!} \right) \\[10pt] & = f(a) f(b). \end{align} How step $(1)$ is justified is something on which I posted an answer here not long ago. I'll see if I can find it.

PS: Here it is: Prove $e^{x+y}=e^{x}e^{y}$ by using Exponential Series

(It's essentially the same question, but it's explicit in explaining the algebra involved in step $(1)$ above.)

Answer 3

Call the series $f(a)$, i.e.,

$$f(a) = 1+\sum_{k=1}^{\infty} \frac{a^k}{k!} $$

Then

$$f'(a) = \sum_{k=1}^{\infty} \frac{k a^{k-1}}{k!} = \sum_{k=1}^{\infty} \frac{ a^{k-1}}{(k-1)!} = \sum_{k=0}^{\infty} \frac{a^k}{k!} = f(a)$$

$$f(0) = 1$$

The term-by-term differentiation is justified by the absolute convergence of the sum.

The unique solution to this equation is $f(a) = e^a$.

Answer 4

I suppose that you can start from: $$ \lim_{n\to\infty}\Bigl(1+\frac an\Bigr)^n=e^a $$ Now, use the binomial formula: $$ \left(1+\dfrac{x}{n} \right)^n =1+\sum_{k=1}^n\binom{n}{k}\dfrac{a^k}{n^k} $$ where the coefficient of $a^k$ is: $$ \dfrac{n!}{k!(n-k)!\,n^k}=\dfrac {1}{k!} \, \dfrac{1\times 2 \times 3 \cdots \times n}{[1\times 2 \cdots \times (n-k)]\,\times \,\underbrace {n\times n \cdots \times n}_{k \,\mbox{times}}} $$ that can be simplified as: $$ \dfrac{n!}{k!(n-k)!\,n^k}= \dfrac {1}{k!} \,\dfrac{\overbrace{(n-k+1)\times(n-k+ 2) \cdots \times (n-1)}^{(k-1)\,\mbox{factors}}}{\underbrace {n\times n \cdots \times n}_{(k-1) \,\mbox{factors}}}= $$ $$ =\dfrac {1}{k!} \left[ \dfrac{(n-k+1)}{n}\times \dfrac{(n-k+ 2)}{n} \cdots \times \dfrac{(n-1)}{n}\right]= $$ $$ =\dfrac {1}{k!}\,\left(1-\dfrac{k-1}{n}\right)\,\left(1-\dfrac{k-2}{n}\right)\cdots\,\left(1-\dfrac{1}{n}\right) $$

now, for $n \to \infty$ all the factors in the parenthesis $\to 1$ so we have: $$ \lim_{n \to \infty}\dfrac{n!}{k!(n-k)!n^k}=\dfrac{1}{k!} $$ and: $$ e^a=\lim_{n \to \infty}\left(1+\dfrac{a}{n} \right)^n=\lim_{n \to \infty}\left[1+\sum_{k=1}^n\binom{n}{k}\dfrac{a^k}{n^k}\right] = 1+\sum_{k=1}^\infty\dfrac{a^k}{k!}=\sum_{k=0}^\infty\dfrac{a^k}{k!} $$

Note that this proof the series definition of $e^a$ starting simply from the sequence definition, without other assumptions.