Factorial Powers and Terms Involving n to the n

Why These Terms Need Careful Simplification

A term such as \(n^n\) changes with the index in two ways: both the base and the exponent increase. A factorial such as \(n!\) also grows rapidly. Comparing these quantities by intuition alone can be misleading, so the ratio test is a natural tool: it compares consecutive terms and often reduces the changing powers to a familiar exponential limit.

The important step is to form the quotient exactly before taking a limit. In particular, \((n+1)!\) contains an extra factor of \(n+1\), while \((n+1)^{n+1}\) changes both its base and exponent. Keeping those changes visible prevents common index and exponent errors.

Factorial Powers and the Ratio Test

Definition: For a fixed real number \(p\), the factorial power in a term of this tutorial is \((n!)^p\). Since \(n!>0\), this real power is defined for every real \(p\). We consider series whose terms have the form \(a_n=c^n n^n/(n!)^p\), where \(c\) is a fixed nonzero real number and \(n\in\mathbb N\).

For \(c=1\), this gives the basic family \(\sum_{n=1}^{\infty} n^n/(n!)^p\). The constant \(c^n\) allows us to see how an exponential factor changes the result. If \(c\) is negative, the terms may alternate in sign, but the ratio test is applied to their absolute values.

Ratio test: Suppose \(a_n\) is nonzero for all sufficiently large \(n\), and \(\lim_{n\to\infty}|a_{n+1}/a_n|=L\). If \(L<1\), then \(\sum a_n\) converges absolutely. If \(L>1\), including \(L=\infty\), then \(\sum a_n\) diverges. If \(L=1\), the test gives no conclusion.

The test is useful here because the factorial quotient cancels almost completely. What remains is a power of \(n+1\) multiplied by \((1+1/n)^n\).

Deriving the Successive-Term Quotient

Start with \(a_n=c^n n^n/(n!)^p\). The next term is \(a_{n+1}=c^{n+1}(n+1)^{n+1}/((n+1)!)^p\). Using \((n+1)!=(n+1)n!\), we obtain

\begin{aligned} \left|\frac{a_{n+1}}{a_n}\right| &= |c|\frac{(n+1)^{n+1}}{n^n} \left(\frac{n!}{(n+1)!}\right)^p\\ &= |c|\frac{(n+1)^{n+1}}{n^n(n+1)^p}\\ &= |c|(n+1)^{1-p}\left(1+\frac{1}{n}\right)^n. \end{aligned}

The last equality follows because \((n+1)^{n+1}/n^n=(n+1)((n+1)/n)^n\), and \((n+1)/n=1+1/n\). Thus the quotient is exact for every positive integer \(n\); no approximation has been made.

The standard exponential limit is \(\lim_{n\to\infty}(1+1/n)^n=e\). One way to justify it is to take logarithms. For \(x>0\), the integral formula for the logarithm gives \(\frac{x}{1+x}\leq \ln(1+x)\leq x\), because \(1/(1+x)\leq 1/(1+t)\leq1\) for \(0\leq t\leq x\). Setting \(x=1/n\) and multiplying by \(n\) gives

\frac{n}{n+1} \leq n\ln\left(1+\frac{1}{n}\right) \leq 1.

Both bounds tend to \(1\), so the middle expression tends to \(1\). Exponentiating yields \((1+1/n)^n\to e\). In particular, this factor approaches a positive finite number; it does not itself grow without bound or shrink to zero.

How the Factorial Power Determines the Limit

Because \((1+1/n)^n\to e\), the remaining factor \((n+1)^{1-p}\) determines the behavior except for the fixed multiplier \(|c|\). If \(p>1\), its exponent is negative, so it tends to zero. If \(p=1\), it is identically \(1\). If \(p<1\), its exponent is positive, so it tends to infinity. Therefore,

\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| = \begin{cases} 0, & p>1,\\[4pt] |c|e, & p=1,\\[4pt] \infty, & p<1. \end{cases}

For \(p>1\), the ratio test proves absolute convergence for every fixed nonzero \(c\). For \(p<1\), the ratio test proves divergence for every such \(c\). At \(p=1\), the outcome depends on the size of \(|c|e\): a value below \(1\) proves absolute convergence, while a value above \(1\) proves divergence. If \(|c|e=1\), the ratio test is inconclusive.

Key takeaway: For \(\sum n^n/(n!)^p\), the ratio limit is \(0\) when \(p>1\), \(e\) when \(p=1\), and infinity when \(p<1\). Adding \(c^n\) multiplies each of these ratio limits by \(|c|\), with the usual care when the limit is zero or infinity.

Worked Examples

Worked Example: A Squared Factorial in the Denominator

Determine whether \(\sum_{n=1}^{\infty} n^n/(n!)^2\) converges. Set \(a_n=n^n/(n!)^2\), so the terms are positive and nonzero. Using the exact quotient with \(p=2\) and \(c=1\),

\begin{aligned} \frac{a_{n+1}}{a_n} &= \frac{(n+1)^{n+1}}{((n+1)!)^2}\frac{(n!)^2}{n^n}\\ &= \frac{(n+1)^{n+1}}{(n+1)^2n^n}\\ &= \frac{1}{n+1}\left(1+\frac{1}{n}\right)^n. \end{aligned}

The exponential factor tends to \(e\), while \(1/(n+1)\to0\). Hence

\lim_{n\to\infty}\frac{a_{n+1}}{a_n} = \lim_{n\to\infty}\frac{1}{n+1}\left(1+\frac{1}{n}\right)^n =0\cdot e=0<1.

The ratio test gives absolute convergence, which here is simply convergence because all terms are positive. As a check, substituting \(p=2\) into the general quotient gives \((n+1)^{-1}(1+1/n)^n\), exactly the expression calculated directly.

Worked Example: One Factorial Gives a Ratio Limit Above One

Determine whether \(\sum_{n=1}^{\infty} n^n/n!\) converges. Let \(a_n=n^n/n!\). Then

\begin{aligned} \frac{a_{n+1}}{a_n} &= \frac{(n+1)^{n+1}}{(n+1)!}\frac{n!}{n^n}\\ &= \frac{(n+1)^{n+1}}{(n+1)n^n}\\ &= \left(1+\frac{1}{n}\right)^n. \end{aligned}

Therefore

\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=e>1.

The ratio test proves that the series diverges. This is not a case where the test is inconclusive: its limit is \(e\), not \(1\). Checking against the general expression with \(p=1\) and \(c=1\) gives \(|c|e=e\), in agreement with the direct calculation.

Worked Example: An Exponential Factor Changes the Boundary

Consider \(\sum_{n=1}^{\infty}(2/e)^n n^n/n!\). Here \(c=2/e\) and \(p=1\). For \(a_n=(2/e)^n n^n/n!\), the quotient is

\begin{aligned} \left|\frac{a_{n+1}}{a_n}\right| &= \frac{2}{e} \frac{(n+1)^{n+1}}{(n+1)n^n}\\ &= \frac{2}{e}\left(1+\frac{1}{n}\right)^n. \end{aligned}

Taking the limit gives

\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| = \frac{2}{e}\,e=2>1.

The ratio test proves divergence. The answer agrees with the \(p=1\) rule: the limit is \(|c|e=(2/e)e=2\). The calculation also shows why the factor \(c^n\) cannot be ignored; it contributes the constant \(|c|\) to the ratio.

The Boundary Case and a Common Mistake

For \(p=1\) and \(c=1/e\), the ratio limit is \(|c|e=1\). The ratio test then gives no conclusion. It would be incorrect to report convergence or divergence from this test alone. The same caution applies to any series for which the quotient limit is exactly \(1\).

A frequent algebra error is to forget that \((n+1)!\) contributes a factor of \(n+1\), or to treat \((n+1)^{n+1}/n^n\) as though it were only \((1+1/n)^n\). For example, for \(p=1\), the incorrect cancellation

\frac{(n+1)^{n+1}}{(n+1)n^n} \overset{\text{wrong}}{=} \frac{1}{n+1}\left(1+\frac{1}{n}\right)^n

would introduce an extra factor \(1/(n+1)\). The correction is to cancel the factorial factor first:

\frac{(n+1)^{n+1}}{(n+1)n^n} = \frac{(n+1)^n}{n^n} = \left(1+\frac{1}{n}\right)^n \longrightarrow e.

That missing factor changes the limit from \(e\) to \(0\), and could reverse the conclusion. A reliable check is to write \((n+1)!=(n+1)n!\) explicitly before simplifying any powers.

Why This Comparison Matters

These examples illustrate how a factorial power and a variable-base power interact. The exact quotient turns their complicated appearance into two simple components: \((n+1)^{1-p}\), which records the effect of the factorial power, and \((1+1/n)^n\), which approaches \(e\). The ratio test can then classify most of the family without estimating the original terms directly.

The limit still has to be interpreted according to the ratio test’s conditions. A limit below \(1\) proves absolute convergence; a limit above \(1\) proves divergence; and a limit equal to \(1\) leaves the question open. The algebra may be exact and the limit may be easy to compute, but neither fact removes that final distinction.

Check Your Understanding

For each question, write the successive-term quotient before applying the ratio test.

For \(\sum_{n=1}^{\infty} n^n/(n!)^3\), what is the ratio limit, and what does the ratio test conclude?
For \(\sum_{n=1}^{\infty}(1/(2e))^n n^n/n!\), find the ratio limit and state whether the series converges absolutely.
For \(\sum_{n=1}^{\infty}(1/e)^n n^n/n!\), what does the ratio test say? Explain why its conclusion is limited.
In simplifying \((n+1)^{n+1}/((n+1)n^n)\), which factor cancels, and what expression remains?
For fixed nonzero \(c\) and \(p<1\), what is the ratio limit for \(c^n n^n/(n!)^p\), and what does it imply?