There are many arguments I have seen using $\ln-$ arguments and other properties of the exponential function to show the existence of this limit $\exp(-x) \rightarrow 0 $ for $x \rightarrow \infty$. But does somebody have a good idea how to show the existence of this limit just by having the pure series definition of the exponential function?

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    $\begingroup$ I think that it is not easily possible. The sanest argument I know is that $e^{-x}=1/e^x$, and it is easy to show that $e^x$ becomes large for $x$ large positive. I myself have tried and failed to find "direct/elementary" arguments that the power series for the exponential function gives tiny values for negative arguments... (Differential equation arguments can also apply, but these are not less sophisticated than $e^{-x}=1/e^x$, I think...) $\endgroup$ – paul garrett Apr 16 '16 at 0:23

We have for $x > 0$, $\exp(x) > 1 +x$ and $0 \leqslant \exp(-x) < (1+x)^{-1}$, since $\exp(x)\exp(-x)=1$.


$$\lim_{x \to \infty}\exp(-x) = 0.$$

All of this follows from the power series representation which converges for all $x \in \mathbb{R}$:

$$\exp(x) = \sum_{j=0}^\infty\frac{x^j}{j!}.$$

By Mertens', theorem the Cauchy product of the series for $\exp(x)$ and $\exp(y)$ converges to the product of the sums

$$\exp(x) \exp(y) = \sum_{j=0}^\infty\frac{x^j}{j!}\sum_{k=0}^\infty\frac{y^k}{k!} = \sum_{j=0}^\infty \sum_{k=0}^j \frac{x^{j-k}}{(j-k)!} \frac{y^{k}}{k!} \\ = \sum_{j=0}^\infty \frac{1}{j!}\sum_{k=0}^j \frac{j!}{(j-k)!k!} x^{j-k}y^k .$$

By the binomial theorem

$$\sum_{k=0}^j \frac{j!}{(j-k)!k!} x^{j-k}y^k = (x+y)^j.$$

Hence, $\exp(x)\exp(y) = \exp(x+y).$ With $y = -x$ we have $\exp(x)\exp(-x) = \exp(0) = 1.$

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  • $\begingroup$ nice argument, thanks. $\endgroup$ – user331288 Apr 16 '16 at 0:26
  • $\begingroup$ @Landauer: You're welcome. The multiplicative rule is the key which as I said can be proved by taking the product of the two series. $\endgroup$ – RRL Apr 16 '16 at 0:35
  • $\begingroup$ Yes, indeed, and/but, if one were to quibble, this line does not actually address the power series expansion of $e^{-x}$ itself to show that it becomes small, etc. That is, (which is the only approach I know...) it uses the point that $e^{-x}=1/e^x$, and then the easy natural lower bound(s) for $e^x$ with $x>0$. $\endgroup$ – paul garrett Apr 16 '16 at 0:52
  • $\begingroup$ @paulgarrett You can establish $e^xe^{-x}=1$ using only the power series. Establish absolute convergence so summation order can be played with, and the following does this, where $k$ reindexes as $k=m+n$: $$\sum_{n=0}^\infty\frac{x^n}{n!}\cdot\sum_{m=0}^\infty\frac{(-x)^m}{m!}=\sum_{m=0}^\infty\sum_{n=0}^\infty\frac{(-1)^mx^{m+n}}{m!n!}$$ $$=\sum_{k=0}^\infty\sum_{m=0}^k\frac{(-1)^mx^{k}}{m!(k-m)!}$$ $$=\sum_{k=0}^\infty\frac{x^k}{k!}\sum_{m=0}^k\frac{(-1)^mk!}{m!(k-m)!}$$ $$=\sum_{k=0}^\infty\frac{x^k}{k!}\sum_{m=0}^k(-1)^m\binom{k}{m}$$ $$=\sum_{k=0}^\infty\frac{x^k}{k!}\cdot\delta_{k=0}=1$$ $\endgroup$ – alex.jordan Apr 16 '16 at 22:42
  • $\begingroup$ @alex.jordan, yes, of course. What is not clear is how to show "directly" that $\sum_n (-1)^n x^n/n!$ is rapidly decreasing as $x\to +\infty$ ... by estimating partial sums directly, etc. (And, indeed, the only sane and enduring argument about $e^x\cdot e^{-x}=1$ of course goes via the power series.) $\endgroup$ – paul garrett Apr 17 '16 at 2:42

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