# Does the series $\sum_{n=1}^{\infty} \frac{n^n}{n!} e^{-n}$ converge?

**Problem**: Does the series $\sum_{n=1}^{\infty} \frac{n^n}{n!} e^{-n}$ converge?

**Note**: In many calculus textbook examples, series involving the factorial term $n!$ are typically analyzed using the ratio test. These exercises often skip a detailed examination of how rapidly $n!$ grows. However, the ratio test can often be inconclusive, requiring an understanding of the growth rate of $n!$.

The given series is a case where the ratio test is inconclusive, and there is no straightforward series available for direct comparison.

I will write my answers below. Different viewpoints or approaches to this problem are welcome.

## 2 answers

If unfamiliar with Stirling's formula, one can still get a rough approximation for $n!$ using elementary calculations.

The basic idea is to consider $\log(n!)$, which converts products into a sum that can then be analyzed using integrals. Observe that: $$ \log(n!) = \sum_{m=1}^n\log m $$ This sum is a Riemann sum approximation to the integral $\displaystyle \int_1^n\log(x)\ dx$: $$ \int_1^n\log(x)\ dx \le \sum_{m=1}^n\log m \le \log(n) + \int_1^n\log(x)\ dx $$ By finding an antiderivative for $\log(x)$ and applying the fundamental theorem of calculus, and then rearranging, one gets: $$ en^ne^{-n} \le n! \le en \cdot n^n e^{-n}\tag{*} $$ This leads to: $$ \frac{n^n}{n!}e^{-n} \ge \frac{1}{en} $$ Given the divergence of the harmonic series and comparison test, we can conclude that the series diverges.

**Note**: the lower bound for $n!$ is unnecessary for this problem. Stirling's formula shows that $n!$ actually lies roughly at the geometric mean of the two bounds in (*).

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The basic approach to determining the convergence of a series involves comparing it to a simpler series using either the direct comparison test or the limit comparison test. The latter indicates when a complicated summand can be substituted with a simpler one for easier analysis.

Stirling's formula, $$ \lim _{n \rightarrow \infty} \frac{n!}{\sqrt{2 \pi n}(n / e)^n}=1, $$ provides crucial insight into the rapid growth of $n!$. It enables the immediate determination of convergence using the limit comparison test:

Consider $a_n = \frac{n^n e^{-n}}{n!}$ and $b_n = \sqrt{2\pi n}$. Applying Stirling’s formula and the limit comparison test shows that $\sum a_n$ diverges since the (multiple of) the p-series $\sum b_n$ does.

**Note:** Stirling's formula might be overkill for this particular problem. All that's really needed is the approximation discovered by De Moivre:
$$
n! \sim C\sqrt{n} n^n e^{-n}
$$
where $C$ is *some* constant.

This answer exemplifies a technique that is impossible or very difficult for beginner calculus students to think of, let alone apply, if they do not already have it in their toolkit. Stirling's formula, which often does not appear in standard calculus textbooks, greatly simplifies the analysis of many series involving the factorial term.

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