The preamble of my analysis book defines $\exp$ in terms of its Taylor series around $0$.

$$ \exp(z) \stackrel{def}{=} \sum_{n=0}^\infty \frac{z^n}{n!} \tag{1} $$

I'm trying to show that this is equal to $e^z$, using the more familiar limit definition of $e$ .

$$ e \stackrel{def}{=} \lim_{n \rightarrow \infty} \left( 1+\frac{1}{n}\right) ^n \tag{2} $$

For one of steps in the proof I want to invoke the binomial theorem and translate $(3)$ to $(4)$

$$ \lim_{n \rightarrow \infty} \left(1 + \frac{x}{n} \right)^{n} \tag{3} $$


$$ \lim_{n \in \mathbb{N} \rightarrow \infty} \sum_{i=0}^n {n \choose i} \frac{x^i}{n^i} \tag{4} $$

I've changed $n$ from ranging over real numbers $\mathbb{R}$ to ranging over natural numbers $\mathbb{N}$.

In this particular case, I think the step "works" because $(1+ \frac{x}{n})^n$ is fairly well behaved with respect to $n$.

In general, however, I could have a pathological function that's defined differently on $\mathbb{N} + 0.5$, for example. In which case the original expression would not converge, but the rewritten one with the "sparse" limit would converge.

Is there a name in general for such "sparse limits" and criteria for when you're allowed to use them?

  • $\begingroup$ Yes, it is called continuity. Your $f(n)=(1+x/n)^n$ as a function of real variable $n$ (here $x$ is a constant) is continuous hence if the limit $\lim_{n\to m}f(n)$ exists then for any $a_n\to m$ the limit $\lim f(a_n)$ exists and is equal to $\lim_{n\to m}f(n)$. Note that your pathological function might be continuous but the limit $\lim_{n\to \infty}f(n)$ surely doesn't exist. $\endgroup$ – freakish Oct 12 '18 at 5:43
  • $\begingroup$ @freakish if I know that $\lim_{n\rightarrow m} f(n)$ exists, do I need continuity? $\endgroup$ – Gregory Nisbet Oct 12 '18 at 5:49
  • 1
    $\begingroup$ If the limit $\lim_{n\to m}f(n)$ exists then either $f$ is continuous at $m$ or you can refine it to be continuous by putting $f(m):=\lim_{n\to m} f(n)$ (i.e. the discontnuity is removable). Either way these two concepts are extremely close to each other. Note that even though in your case $m=\infty$ is not in domain it actually can be by extending the real line with $\infty$. In that situation $f$ becomes continuous at $\infty$. So you don't need $f$ to be globally continuous, you only need it to be continuous at $m$ in order to freely juggle with sequences and limits. $\endgroup$ – freakish Oct 12 '18 at 6:19

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