Question

Is the series convergent or divergent? $$ \sum_{n=1}^{\infty} n^{-(n+1 / n)} $$

Short answer

The series is convergent by the Comparison Test.

Step-by-step solution

Understand the Given Series

We are given the series \( \sum_{n=1}^{\infty} n^{-(n+1/n)} \). This is an infinite series where the general term is \( a_n = n^{-(n+1/n)} \). We need to determine if this series is convergent or divergent.

Simplify the General Term

The exponent in the series term is \(-(n + 1/n)\). This can be rewritten as \(-n - 1/n\). So, \(a_n = n^{-n - 1/n} = \frac{1}{n^{n + 1/n}}\).

Apply the n-th Term Test for Divergence

For a series \( \sum a_n \) to be convergent, \( \lim_{{n \to \infty}} a_n = 0 \) is a necessary condition. Let's evaluate \( \lim_{{n \to \infty}} \frac{1}{n^{n + 1/n}} \).

Evaluate the Limit of the General Term

Note that as \( n \to \infty \), \( n^{n + 1/n} \to \infty \). Therefore, \( \frac{1}{n^{n + 1/n}} \to 0 \). This tells us \( \lim_{{n \to \infty}} a_n = 0 \).

Consider the Exponential Growth of the Denominator

Although the limit of \( a_n \) is 0, we observe that each term \( \frac{1}{n^{n + 1/n}} \) grows smaller faster than any geometric progression, indicating rapid decrease due to \( n^n \) in the denominator.

Use the Comparison Test with a Known Convergent Series

Consider the series \( \sum \frac{1}{n^n} \). We compare \( a_n = \frac{1}{n^{n + 1/n}} \) to \( b_n = \frac{1}{n^n} \). Since \( n^{n + 1/n} > n^n \) for all \( n \geq 1 \), it follows that \( 0 < a_n < b_n \). The series \( \sum \frac{1}{n^n} \) is known to converge.

Conclude by Comparison Test

Since \( a_n < b_n \) and \( \sum b_n \) converges, by the Comparison Test, the series \( \sum a_n = \sum \frac{1}{n^{n + 1/n}} \) also converges.

Key concepts

Infinite Series

An infinite series is when you sum up a sequence of numbers indefinitely, without an endpoint. It is expressed in the form \( \sum_{n=1}^{\infty} a_n \), where \( a_n \) represents the terms of the sequence. The challenge with infinite series is determining whether they "settle down" or not, as you keep adding more and more terms.

Convergence and divergence are key concepts to understand here. If the sum of an infinite series approaches a specific number as more terms are added, it is said to converge. If the sum doesn't approach a specific number, the series diverges.

In our exercise, we explore the series \( \sum_{n=1}^{\infty} n^{-(n+1/n)} \) and our task is to determine whether it converges or diverges.

Comparison Test

The Comparison Test is a helpful tool for determining the convergence or divergence of series. The idea is to compare your series with another series whose convergence behavior you already know. If your series is smaller term-by-term than a known convergent series, then your series also converges.

Conversely, if your series is larger term-by-term than a known divergent series, then your series diverges.

In our exercise, we used the known convergent series \( \sum \frac{1}{n^n} \) as a benchmark. We showed that our series \( \sum \frac{1}{n^{n + 1/n}} \) could be compared favorably—since every term in our series is less than those in the known convergent series, we could conclude that our series converges.

n-th Term Test

The n-th Term Test is a straightforward way to begin analyzing a series. It states that if \( \lim_{{n \to \infty}} a_n eq 0 \), then the series \( \sum a_n \) must diverge. However, if \( \lim_{{n \to \infty}} a_n = 0 \), this alone doesn't confirm convergence—it just means the test is inconclusive.

For the series in our example, we found that \( \lim_{{n \to \infty}} \frac{1}{n^{n + 1/n}} = 0 \), satisfying the condition for potential convergence. But this result by itself wasn't sufficient, hence we had to apply further tests like the Comparison Test to definitively determine convergence.

Limit of a Sequence

A sequence is an ordered list of numbers, and studying the limit of a sequence involves understanding where its terms "go" as the sequence progresses towards infinity. Practically, this means looking at what happens to \( a_n \) as \( n \) becomes very large.

If the terms of a sequence get arbitrarily close to a number, we say the sequence converges to that number. If they don't, the sequence diverges. The limit of the sequence is crucial when analyzing series because it is often used to test whether the terms of the series go to zero, a necessary condition for series convergence.

In our exercise, recognizing that \( \lim_{{n \to \infty}} \frac{1}{n^{n + 1/n}} = 0 \) helped guide us in further investigating the behavior of the series.