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Claim: $$\sum_{k=0}^{n} \frac{x^k}{k!} < e^x, \qquad \forall x > 0$$

Proof:

Let $f(x)=e^x$. Then the Taylor series gives

$$f(x) = \sum_{k=0}^{\infty} \frac{x^k}{k!}$$

$$= \sum_{k=0}^{n} \frac{x^k}{k!} + \sum_{k=n+1}^{\infty} \frac{x^k}{k!}$$

Now, because $x>0$, $$\sum_{n+1}^{\infty} \frac{x^k}{k!} > 0$$

$$\implies f(x) > f(x) - \sum_{n+1}^{\infty} \frac{x^k}{k!}$$

$$\implies \sum_{k=0}^{n} \frac{x^k}{k!} = f(x) - \sum_{n+1}^{\infty} \frac{x^k}{k!} < f(x)$$

QED.

I would appreciate constructive criticism on this proof.

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  • $\begingroup$ all fine. You showed that this is just a partial (and not the full) sum. $\endgroup$ – Andreas Nov 3 '17 at 15:09
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    $\begingroup$ The inequality should be reversed: $$f(x) < f(x) - \sum_{n+1}^{\infty} \frac{x^k}{k!}$$. $\endgroup$ – Cm7F7Bb Nov 3 '17 at 15:10
  • $\begingroup$ It can help: math.stackexchange.com/questions/2228083/… $\endgroup$ – Anıl B.C.T. Nov 3 '17 at 15:12
  • $\begingroup$ @Cm7F7Bb, thank you. Typo! $\endgroup$ – AndyDufresne Nov 3 '17 at 15:15
  • $\begingroup$ What is $e$ and $e^x$ for the purposes of this task? Can it be that $e=\lim(1+\frac1n)^n$ , $e^x$ is simply the exponential function for that basis, and the equality to the exponential power series is not previously known? $\endgroup$ – Lutz Lehmann Nov 3 '17 at 16:20
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You are assuming $e^x = \sum_{k=0}^{\infty}\frac{x^k}{k!}$. In this case you can observe that the sequence $$f_n(x) := \sum_{k=0}^n \frac{x^k}{k!}, \quad x > 0$$ is strictly increasing, as $f_{n+1}(x)$ is $f_n(x)$ plus a positive number (as $x$ is positive). This implies that $(f_n(x))_{n=0}^{\infty}$ admits a limit, which is $e^x$. Using the definition of limit and that $(f_n(x))_{n=0}^{\infty}$ is an increasing sequence, if you fix $\varepsilon > 0$ then you can find $N \in \mathbb{N}$ such that $$0 < e^x-f_n(x) < \varepsilon \text{ if } n>N$$ and this implies in particular that $$f_n(x) < e^x.$$

Suppose you don't know $e^x = \sum_k \frac{x^k}{k!}.$ Consider $$g_n(x):=e^x - \sum_{k=0}^n \frac{x^k}{k!}, \quad x \geq 0.$$ Observe that $g_n(0) = 0$ for any $n \in \mathbb{N}$. Assume by contradiction that there is an $N \in \mathbb{N}$ such that $$e^x \leq \sum_{k=0}^N \frac{x^k}{k!}$$ and further, that this is the least natural number for which this happens, that is $$e^x > \sum_{k=0}^{N-1} \frac{x^k}{k!}.$$ This is equivalent to saying that \begin{equation} g_N(x) \leq 0 \end{equation} and $$g_{N-1}(x) > 0.$$ But one can compute directly $$g_N'(x) = g_{N-1}(x) > 0.$$ This fact and $g_n(0) = 0$ for every $n \in \mathbb{N}$ yield $g_N(x) > 0$ if $x > 0$, contradiction with $g_N(x) \leq 0$.

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You argument is fine, but there also is a shortcut. By exploiting Taylor's theorem with integral remainder we have:

$$ e^x = \sum_{k=0}^{n}\frac{x^k}{k!}+\frac{1}{n!}\int_{0}^{x} e^t (x-t)^n\,dt\tag{A} $$ and it is pretty obvious that the last integral is positive if $x> 0$, since it is the integral of a continuous and positive function.

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