asymptotic behavior of the two sequences defining exponential function

Question

There are two definitions of exponential function: $$e^x=\lim_{n\to\infty} S_n=\lim_{n\to \infty} a_n \text{ ,}$$ where $$S_n=1+x+\frac{x^2}{2!}+\dots+\frac{x^n}{n!}$$ and $$a_n=(1+\frac{x}{n})^n \text{ .}$$

Since the two sequence have the same limit, I guess they are somehow related and reflect different aspects of $e^x$. So my first question is: are there any relationships between $S_n$ and $a_n$?

My other questions come from the following observations:

When x is positive, obviously $S_n$ is increasing. Is $a_n$ also increasing?

When x is negative, $S_n$ goes up and down since it keeps adding numbers of alternating signs as as $n$ grows. Eventually $S_n$ "squeezes" to its limit. What about the behavior of $a_n$ in this case? When $n$ is smaller than $|x|$, I can see that $a_n$ changes signs very often. When $n$ is large, $a_n$ is always positive, and is $a_n$ increasing when $n$ is large?

Answer 1

Define $$ E(x)=\sum_{k=0}^\infty\frac{x^k}{k!} $$ We know that this power series has infinite radius of convergence and is thus a continuous (and differentiable, etc.) function. By careful but elementary application of the Cauchy product an binomial theorem one finds $$ E(x)·E(y)=E(x+y) $$ which as functional equation has the solution $$ E(x)=E(1)^x=e^x $$

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By multiplying out $$ \left(1+\frac xn\right)\left(1+\frac yn\right)\left(1-\frac {x+y}n\right)=1-\frac {x^2+xy+y^2}{n^2}-\frac {xy(x+y)}{n^3} $$ one finds that for $e(x)=\lim_{n\to\infty}\left(1+\frac xn\right)^n$ it holds that $$ e(x)e(y)e(-x-y)=\lim_{n\to\infty}\left(1+\frac xn\right)^n\left(1+\frac yn\right)^n\left(1-\frac {x+y}n\right)^n=1 $$ which again implies the functional equation $$ e(x)e(y)=e(x+y) $$ For the Cauchy functional equation one would need to establish for instance continuity at $x=0$ to then again arrive at $$ e(x)=e(1)^x=e^x $$

Answer 2

Use the binomial formula: $$ \left(1+\dfrac{x}{n} \right)^n =1+\sum_{k=1}^n\binom{n}{k}\dfrac{x^k}{n^k} $$ where the coefficient of $x^k$ 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: $$ \lim_{n \to \infty}\left(1+\dfrac{a}{n} \right)^n=\lim_{n \to \infty}\left[1+\sum_{k=1}^n\binom{n}{k}\dfrac{a^k}{n^k}\right] $$

Answer 3

Let $x > 0$ then it can be easily proven using binomial theorem (for positive integer index) that \begin{align} \left(1 + \frac{x}{n}\right)^{n} &= 1 + x + \dfrac{1 - \dfrac{1}{n}}{2!}\cdot x^{2} + \dfrac{\left(1 - \dfrac{1}{n}\right)\left(1 - \dfrac{2}{n}\right)}{3!}\cdot x^{3} + \cdots\notag\\ &\leq 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \cdots + \frac{x^{n}}{n!}\notag \end{align} and therefore $a_{n} \leq S_{n}$. Moreover from the expansion of $a_{n}$ (via binomial theorem above) you can see that as $n$ increases each term in the expansion increases as well as the number of terms also increases. Thus $a_{n}$ is an increasing sequence if $x > 0$.

The story of $a_{n}$ and $S_{n}$ is not complete without the introduction of another sequence $b_{n}$ defined by $$b_{n} = \left(1 - \frac{x}{n}\right)^{-n}$$ For $0 < x < n$ we can use the general binomial theorem (for any index) to get $$b_{n} = \left(1 - \frac{x}{n}\right)^{-n} = 1 + x + \dfrac{1 + \dfrac{1}{n}}{2!}\cdot x^{2} + \dfrac{\left(1 + \dfrac{1}{n}\right)\left(1 + \dfrac{2}{n}\right)}{3!}\cdot x^{3} + \cdots$$ and this shows clearly that for $0 < x < n$ we have $$a_{n} \leq S_{n} \leq b_{n}$$ It is easy to prove that $S_{n}$ represents a convergent series and since $a_{n}$ is increasing it follows that $a_{n}$ also is convergent for $x > 0$. Let $S_{n} \to S(x)$ and $a_{n} \to a(x)$. Note also that $b_{n}$ is decreasing sequence for $0 < x < n$ and is bounded below by $S(x)$ and hence $b_{n}$ also is convergent and let $b_{n} \to b(x)$. We then have $a(x) \leq S_{x} \leq b(x)$. The magic happens when we see that $a_{n}/b_{n} = (1 - x^{2}/n^{2})^{n} \to 1$ as $n \to \infty$ so that $a(x) = b(x)$ and therefore all the sequences $a_{n}, b_{n}, S_{n}$ tend to the same limit which is traditionally denoted by $e^{x}$.

To handle negative values of $x$ note that the limit $S(x)$ of $S_{n}$ is an infinite convergent series which has the interesting property that $S(x + y) = S(x)S(y)$ (this is proved via multiplication of series). It thus means that $S(-x) = 1/S(x)$. Note that the sequence $a_{n}, b_{n}$ are such that $b_{n}(x) = 1/a_{n}(-x)$ (added $x$ to show the dependence of $x$) and hence it follows that $b(x) = 1/a(-x)$ or $a(-x) = 1/a(x)$ and this establishes the relation $a(x) = b(x) = S(x)$ for negative values of $x$ also.

Answer 4

Following the outline given by Henry in the comments to the question:

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Proposition: If $\sum_{k=0}^\infty a_k(t)$ is a series with parameter $t$ so that $\sum_{k=0}^\infty a_k(0)$ converges absolutely, $|a_k(t)|\le|a_k(0)|$ and $\lim_{t\to 0}a_k(t)=a_k(0)$, then $$ \lim_{t\to 0}\sum_{k=0}^\infty a_k(t)=\sum_{k=0}^\infty a_k(0) $$

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Proof: Let $ε>0$ be arbitrary. Determine $N$ such that $\sum_{k=N}^\infty |a_k(0)|<ε/3$. Then select $\bar t$ such that $\sup_{k=0,…,N-1}|a_k(t)-a_k(0)|<ε/(3N)$ for all $t\in(0,\bar t)$. Then $$ \left|\lim_{t\to 0}\sum_{k=0}^\infty a_k(t)-\sum_{k=0}^\infty a_k(0)\right|\le\sum_{k=0}^{N-1}|a_k(t)-a_k(0)|+ \sum_{k=N}^\infty |a_k(t)|+\sum_{k=N}^\infty |a_k(0)|<ε. $$

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Application: Write the binomial theorem as $$ \left(1+\frac xn\right)^n =\sum_{k=0}^\infty\binom{n}{k}\left(\frac xn\right)^k =\sum_{k=0}^\infty\left(1-\frac1n\right)\left(1-\frac2n\right)…\left(1-\frac{k-1}n\right)\frac{x^k}{k!} $$ thus using $a_k(t)=(1-t)(1-2t)…(1-(k-1)t)\frac{x^k}{k!}$ the assumption of above proposition are satisfied, which establishes pointwise convergence of $\left(1+\frac xn\right)^n$ towards the value of the exponential series.