Determine the limit to which the following expression
Determine the limit to which the following expression tends as $n \to \infty$: $$ a_n = \frac{1}{n}\sum_{k=0}^{n} e^{-k}. $$
Explanation of Method
The inner sum $\sum_{k=0}^{n} e^{-k}$ is a finite geometric series with ratio $e^{-1}$, so it has an exact closed form and — more importantly — stays bounded as $n\to\infty$ (converging to $\frac{e}{e-1}$). Since $a_n$ divides this bounded quantity by $n$, which grows without bound, the whole expression is forced to zero regardless of the precise value of the bounded sum.
- Sum the finite geometric series $\sum_{k=0}^{n} e^{-k}$ in closed form.
- Divide by $n$ to get an explicit formula for $a_n$.
- Take the limit as $n\to\infty$, using that the numerator stays bounded.
Step 1 · Recognize and sum the geometric series
$\sum_{k=0}^{n} e^{-k}$ is geometric with first term $1$, ratio $r=e^{-1}$, and $n+1$ terms:
$$ \sum_{k=0}^{n} e^{-k} = \frac{1-e^{-(n+1)}}{1-e^{-1}} = \frac{e\left(1-e^{-(n+1)}\right)}{e-1}. $$
Step 2 · Express $a_n$ explicitly
$$ a_n = \frac1n\sum_{k=0}^{n} e^{-k} = \frac{e\left(1-e^{-(n+1)}\right)}{n(e-1)}. $$
Step 3 · Evaluate the limit
As $n\to\infty$, $e^{-(n+1)} \to 0$, so the numerator $e\left(1-e^{-(n+1)}\right) \to e$ — a fixed finite constant. Dividing a bounded quantity by $n\to\infty$:
$$ \lim_{n\to\infty} a_n = \lim_{n\to\infty} \frac{e\left(1-e^{-(n+1)}\right)}{n(e-1)} = \frac{e}{e-1}\cdot\lim_{n\to\infty}\frac1n = \frac{e}{e-1}\cdot 0 = 0. $$
Summary Table
| Quantity | Value / behavior |
|---|---|
| $\sum_{k=0}^{n} e^{-k}$ | bounded, $\to \dfrac{e}{e-1}\approx1.582$ |
| $a_n = \dfrac1n\sum_{k=0}^{n} e^{-k}$ | bounded quantity $\div\, n$ |
| $\lim_{n\to\infty} a_n$ | $0$ |
Additional Notes
- The result follows from a general principle: if $b_n$ is any bounded sequence, then $\frac{b_n}{n}\to0$ automatically, regardless of how $b_n$ behaves internally — here $b_n=\sum_{k=0}^n e^{-k}$ happens to converge, but even a bounded oscillating $b_n$ would give the same conclusion.
- This contrasts sharply with Problem 19, where the analogous averaged sum $\frac1n\sum\ln k$ diverges to infinity — the difference is that $\sum e^{-k}$ stays bounded (its terms shrink geometrically) while $\sum \ln k$ grows without bound (its terms grow, however slowly).
- Numerically, $a_n$ decays roughly like $\frac{e/(e-1)}{n} \approx \frac{1.582}{n}$, so reaching $a_n<0.01$ requires $n$ on the order of $158$ or more.
Result
The limit of the expression is: