Find the infimum and supremum of f(x) = (1+x+

Find the infimum and supremum of $\displaystyle f(x) = \left(1+x+\frac{x^{2}}{2!}+\cdots+\frac{x^{n}}{n!}\right)e^{-x} = S_{n}(x)\,e^{-x}$ on $\bigl(0,+\infty\bigr)$, where $S_{n}$ is the $n$-th partial sum of $e^{x}$.

Step 1: Positivity

For $x\gt 0$ every term of $S_{n}(x)$ is positive, and $e^{-x}\gt 0$; hence $f(x)\gt 0$ throughout.

Step 2: The derivative telescopes

Differentiating the partial sum shifts it down by one degree: \[S_{n}'(x) = \sum_{k=1}^{n}\frac{x^{k-1}}{\bigl(k-1\bigr)!} = S_{n-1}(x),\] so that $S_{n}'-S_{n} = S_{n-1}-S_{n} = -\dfrac{x^{n}}{n!}$, and

\[ f'(x) = e^{-x}\bigl(S_{n}'(x)-S_{n}(x)\bigr) = -\frac{x^{n}}{n!}e^{-x}. \tag{1} \]

Everything cancels except the single last term — the tail that $S_{n}$ fails to carry.

Step 3: Monotonicity

By (1), $f'(x)\lt 0$ for every $x\gt 0$: the function is strictly decreasing, with no critical point anywhere in the interval.

Step 4: The two limits

\[ \lim_{x\to0^{+}}f(x) = S_{n}(0)\cdot e^{0} = 1, \qquad \lim_{x\to+\infty}f(x) = \lim_{x\to+\infty}\frac{S_{n}(x)}{e^{x}} = 0, \] the second because $S_{n}$ is a polynomial of degree $n$ and the exponential outgrows every power (Problem 393).

Step 5: Bounds

A strictly decreasing function on $\bigl(0,+\infty\bigr)$ takes values strictly between its two limits: $0\lt f(x)\lt 1$ for every $x\gt 0$, with $\sup f = 1$ and $\inf f = 0$, neither of them attained — the first because $x = 0$ is not in the interval, the second because $f\gt 0$. On the closed half-line $\bigl[0,+\infty\bigr)$ the supremum becomes a genuine maximum $f(0) = 1$, while the infimum is still not attained.

Step 6: Verification

\[n = 3:\quad f(0.1) = 0.9999961; \qquad f(1) = 0.9810118; \qquad f(3) = 0.6472319; \qquad f(10) = 0.0103361 \;\;\checkmark\]

\[f'(x) = -\frac{x^{3}}{6}e^{-x}\lt 0 \text{ for } x\gt 0,\ \text{matching the decrease} \;\;\checkmark\]

\[1-f(x) = e^{-x}\bigl(e^{x}-S_{n}(x)\bigr)\gt 0,\ \text{since } S_{n}(x)\lt e^{x} \text{ for } x\gt 0 \;\;\checkmark\]