Problem 14 - Calculus: Early Transcendentals

Question

Determine whether the series is convergent or divergent. \( 1 + \frac {1}{2 \sqrt 2} + \frac {1}{ 3 \sqrt 3} + \frac {1}{4 \sqrt 4} + \frac {1}{5 \sqrt 5} + \cdot \cdot \cdot \)

Step-by-Step Solution

Answer

The series is convergent.

Step 1: Identify the series form

The given series is \( \sum_{n=1}^{\infty} \frac{1}{n\sqrt{n}} \). Each term of the series is given by \( a_n = \frac{1}{n\sqrt{n}} \).

Step 2: Simplify the terms

Notice that each term can be expressed as \( a_n = n^{-3/2} \). This helps us compare it to a known series for convergence.

Step 3: Use the p-series test

The p-series test states that a series of the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \) converges if \( p > 1 \) and diverges if \( p \leq 1 \). Here, we have \( p = \frac{3}{2} > 1 \).

Step 4: Conclusion about the series

Since \( p = \frac{3}{2} > 1 \), the series \( \sum_{n=1}^{\infty} \frac{1}{n\sqrt{n}} \) converges by the p-series test.

Key Concepts

p-series test

The p-series test is a powerful tool for determining the convergence or divergence of a series. It applies to series in the form of \[ \sum_{n=1}^{\infty} \frac{1}{n^p} \] where \( n \) is a natural number and \( p \) is a positive constant. This test becomes particularly useful for series where every term behaves like \( \frac{1}{n^p} \) for large \( n \).

• If \( p > 1 \), the series converges. This is considered a convergent series.
• If \( p \leq 1 \), the series diverges, meaning it does not meet a finite sum as \( n \rightarrow \infty \). This is a divergent series.

In the given exercise, the series can be rewritten so that each term is expressed as \( \frac{1}{n^\frac{3}{2}} \). Here, \( p = \frac{3}{2} \), which is greater than \( 1 \), thus confirming that the series is convergent using the p-series test.

convergent series

A convergent series is one where the sum of its infinite terms approaches a finite limit as more terms are added. It is a series that contrasts with a divergent series, which spirals towards infinity or oscillates without settling on a sum. Convergence in mathematical terms means that the sequence of partial sums \( s_n = a_1 + a_2 + a_3 + ... + a_n \) approaches a certain value \( L \) as \( n \) becomes large.

Some signs of convergent series include:

• The absolute value of terms getting smaller and closer to zero.
• Using tests like the p-series test or the ratio test to show that the series converges.

In practical applications, convergent series can model situations where effects accumulate to a limiting value, such as calculating probabilities or physical properties like total resistance in certain electrical circuits. In our case, since \( p = \frac{3}{2} > 1 \), the series given is confirmed to be convergent.

divergent series

Divergent series are series whose terms do not settle towards a finite limit. Instead, they either grow without bound or fail to approach any particular value. Divergence can occur in various forms, such as the series having increasingly large partial sums or the series oscillating.

Key characteristics of divergent series include:

• The partial sums \( s_n = a_1 + a_2 + a_3 + ... + a_n \) moving indefinitely away from any finite value.
• The absolute value of terms not approaching zero.

Tests like the p-series test help to identify divergent series by checking the condition \( p \leq 1 \). Series such as \( \frac{1}{n} \) (the harmonic series) are classic examples of divergence (where \( p = 1 \)). They grow without bound as more terms are added. In contrast, our exercise reveals that since \( p = \frac{3}{2} \), the original series does not diverge but rather converges.