# Squaring Infinite Series Expansion Of e^x

$Fact$:$$\lim\limits_{n \to \infty}\frac{x^0}{0!}+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots+\frac{x^n}{n!}=e^x$$ so $$\lim\limits_{n \to \infty}\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots+\frac{1}{n!}=e$$ also $$\lim\limits_{n \to \infty}e^2=\frac{2^0}{0!}+\frac{2}{1!}+\frac{4}{2!}+\frac{8}{3!}+\dots+\frac{2^n}{n!}$$ can I safely say... $$L.H.S=(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots)^2=\frac{2^0}{0!}+\frac{2}{2!}+\frac{4}{2!}+\frac{8}{3!}+\dots=R.H.S$$ if yes, how to prove L.H.S=R.H.S (without using the above fact)? It is not a problem from textbook I was just thinking about it.

• so e diverges.... Commented Aug 20, 2016 at 17:42

This boils down to proving that:

$$\sum_{\substack{k_1,k_1\\k_1+k_2 = n}}\frac{1}{k_1!k_2!} = \frac{2^n}{n!}$$

This follows from the binomial theorem, or you could just say that $$\frac{n!}{k_1! k_2!}$$

for $k_1+k_2 = n$ is the number of ways you can distribute $n$ items in two sets of $k_1$ and $k_2$ elements and if you sum over all the set sizes that sum to $n$, then you get the total number of ways you can choose some of the $n$ items. For each of the $n$ items you can choose to either take it or not, so there are two choices for each item, therefore there are $2^n$ choices in total.

Just use the algebraic fact that $(e^x)^2 = e^{2x}$, and substitute that into your first expansion, for a slick way to do it.

The next best option is to just distribute everything out, and try to collect the like terms. This is pretty tedious, in my opinion, but after the first few terms you might be able to see how it works.

$$(1 + 1 + \frac{1}{2} + \frac{1}{6}+....)(1 + 1 + \frac{1}{2} + \frac{1}{6}+....)=$$ $$1(1 + 1 + \frac{1}{2} + \frac{1}{6}+....)+1(1 + 1 + \frac{1}{2} + \frac{1}{6}+....) + \frac{1}{2}(1 + 1 + \frac{1}{2} + \frac{1}{6}+....)+...$$

Now you just stare at this until you see how to group the terms, and follow the pattern that emerges. Clearly the first two terms can be spotted - we have enough 1's. Then with the extra 1, and the two halves, we can get 2, which we will write as $\frac{4}{2}$, for the next term. Then we need $\frac{8}{6}$, which might seem a little harder, if only because we didn't write out the next term of the series. But, really it is not so bad, as we have $\frac{2}{6}$ from the next term of the sum, and then each of the $\frac{1}{2}$ terms in the original sum can be used to acquire three more, so we have the 8 we need.

Write down all the terms like this, as far out as you need to go to convince yourself, and then just keep scratching them out as you use them and group them appropriately.

Ultimately, I think this is not as nice as just using the properties of infinite series, but since you wanted an algebraic way, this is the best I can come up with.

As a last note, you really shouldn't write infinity in those sums. What's happening is that the terms are actually getting smaller and smaller and smaller, which is why they "stop growing" at $e$. They shrink so fast, so that when you add all these guys up, they come right up to a number they can never pass, and we call that number $e$ because it is very special in the rest of mathematics.

Write out a table with the columns labeled $1, 1, 1/2!,\ldots$ and the rows labeled the same. Then in each entry, put the product of the row heading with the column heading. Now look at each upward diagonal. The sum of the first is 1, the next is 2, etc. These are the terms of the series for $e^2.$ The $n$th diagonal needs to be expressed in terms of $1/n!$. When you do this, you'll see that the numerators are the $n$th row of Pascal's Triangle. Then note that the sum of the $n$th row of Pascal's triangle is $2^n$.

Let $$f(z) = \sum_{n = 0}^{\infty}\frac{z^{n}}{n!}\tag{1}$$ then it is easy to see that the series on right is absolutely convergent for all values of $z \in \mathbb{C}$ and hence the function $f(z)$ is well defined for all complex values of $z$. Using Cauchy's product rule for multiplication of infinite series we see that $$f(z)f(w) = \sum_{n = 0}^{\infty}\frac{z^{n}}{n!}\cdot\sum_{n = 0}^{\infty}\frac{w^{n}}{n!} = \sum_{n = 0}^{\infty}c_{n}\tag{2}$$ where $$c_{n} = \sum_{k = 0}^{n}\frac{z^{n - k}}{(n - k)!}\cdot\frac{w^{k}}{k!} = \frac{1}{n!}\sum_{k = 0}^{n}\binom{n}{k}z^{n - k}w^{k} = \frac{(z + w)^{n}}{n!}\tag{3}$$ and hence from $(2), (3)$ we get $$f(z)f(w) = \sum_{n = 0}^{\infty}\frac{(z + w)^{n}}{n!} = f(z + w)\tag{4}$$ for all complex $z, w$. Putting $z = w = 1$ we get $$f(1)\cdot f(1) = f(2)$$ which is the result asked in question.

Using $(4)$ and little bit of algebra it is possible to prove that $\{f(z)\}^{n} = f(nz)$ for all complex $z$ and rational $n$. Putting $z = 1$ and replace $n$ by $x$ we see that $f(x) = \{f(1)\}^{x}$ where $x$ is rational or in grand fashion $$\left(1 + \frac{1}{1!} + \frac{1}{2!} + \cdots\right)^{x} = 1 + x + \frac{x^{2}}{2!} + \cdots$$ where $x$ is rational (this is the fact you mention in the beginning of your question).

Since $e^ne^m = e^{n+m}$ you can simply let $x = 2t$, giving you

$$(e^{t})^2 = e^{2t} = 1 + 2t + \frac{4t^2}{2!}+\frac{8t^3}{3!} + \cdots$$

Now, let $t = 1$ and you have

$$e^2 = 1 + 2 + \frac{4}{2!} + \frac{8}{3!} + \cdots = (1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots )^2 = (e^1)^2.$$

• He said don't use the taylor expansion of the exponential function. Commented Aug 20, 2016 at 17:47