Question

Find all values of \(\boldsymbol{p}\) and \(q\) such that \(\sum_{n=1}^{\infty} \frac{n^{p}}{(n !)^{q}}\) converges.

Short answer

The series converges for \( q > p \).

Step-by-step solution

Consider the Series

We need to determine the conditions on \(p\) and \(q\) for which the series \(\sum_{n=1}^{\infty}\frac{n^p}{(n!)^q}\) converges. The terms are given by \( \frac{n^p}{(n!)^q} \), where \( n! \) is the factorial of \( n \).

Analyze Growth of Factorial

The factorial function \( n! \) grows very quickly, much faster than exponential or polynomial functions. This suggests that for large \( n \), \( (n!)^q \) will become very large, which helps in convergence of the series.

Consider Ratio Test

We apply the ratio test to determine convergence: \[\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{(n+1)^p}{(n+1)!^q} \cdot \frac{n!^q}{n^p} = \lim_{n \to \infty} \frac{(n+1)^p n!^q}{(n+1)!^q n^p}.\]

Simplify Ratio

Simplifying the expression:\[\frac{(n+1)^p}{n^p} \cdot \frac{n!^q}{(n+1)!^q} = \frac{(n+1)^p}{n^p} \cdot \frac{1}{(n+1)^q} = \frac{(n+1)^p}{n^p (n+1)^q} = \frac{1}{(n+1)^{q-p}} \cdot \left(\frac{n+1}{n}\right)^p.\]

Simplify Further

The term \( \left(1 + \frac{1}{n}\right)^p \) approaches 1 as \( n \to \infty \). Therefore, the limit becomes:\[\lim_{n \to \infty} \frac{1}{(n+1)^{q-p}} \cdot 1 = \lim_{n \to \infty} \frac{1}{(n+1)^{q-p}} = 0. \]This limit is 0 when \( q > p \).

Convergence Condition

Since the limit is 0 for \( q > p \), by the ratio test, the series converges for \( q > p \). If \( q \leq p \), the terms don’t diminish fast enough and the series diverges.

Key concepts

Factorial Growth

Factorial growth is one of the key concepts in understanding why certain series converge. The factorial of a number, denoted by \( n! \), represents the product of all positive integers up to \( n \). This number grows very rapidly as \( n \) becomes large. For example, \( 5! = 120 \) and \( 10! = 3,628,800 \). As you can see, factorial growth is faster than both polynomial and exponential growth.

When evaluating series that feature terms with factorials, such as \( (n!)^q \), we observe that larger \( q \) significantly boosts how fast the denominator of the sequence grows. Consequently, this rapid escalation often leads to the terms diminishing quickly enough to allow for series convergence. In essence, the powerful growth rate of factorials helps control the series' sum by making the individual terms shrink swiftly.

Ratio Test

The ratio test is a fundamental tool in determining the convergence of series. To apply the ratio test, we evaluate the ratio of successive terms in the series and take the limit as \( n \) approaches infinity. Specifically, given a series \( \sum a_n \), we look at:

• \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \)

If this limit is less than 1, the series converges absolutely. If it's greater than 1, the series diverges. If exactly 1, the test is inconclusive.

In our exercise, applying the ratio test involves the expression \( \lim_{n \to \infty} \frac{(n+1)^p}{n^p (n+1)^q} \). Simplifying this is crucial in establishing when the series converges. As derived, the ratio test here simplifies to \( \frac{1}{(n+1)^{q-p}} \), which tends to zero as \( n \to \infty \) when \( q > p \). Thus, confirming convergence when \( q > p \).

Convergence Conditions

Convergence conditions are pivotal in determining whether a series will sum to a finite value. These conditions depend heavily on the behavior of the terms within the series.

For the series \( \sum \frac{n^p}{(n!)^q} \), derived from the factorial growth and use of the ratio test, we discovered that the condition \( q > p \) is necessary for convergence. It means the growth rate of \( (n!)^q \) must exceed that of \( n^p \) enough to make the terms shrink as \( n \) grows.

To intuitively understand why, consider that beyond controlling growth, \( q \) reflects the rapidity in reducing the sequence terms' size. A larger \( q \) means exponentially stronger diminishment, assisting in ensuring the whole series remains manageable and converges to a finite value. Conversely, if \( q \leq p \), the shrinkage isn't sufficient, causing the series to diverge.