Problem 44 - Calculus Early Transcendentals: Pearson New International Edition

Question

How large must \(N\) be in order for \(S_{N}=\sum_{k=1}^{N}(1 / k)\) just to exceed 4? Note: Computer calculations show that for \(S_{N}\) to exceed \(20, N=272,400,600\), and for \(S_{N}\) to exceed 100 , \(N \approx 1.5 \times 10^{43}\).

Step-by-Step Solution

Answer

The smallest \( N \) for \( S_N \) to exceed 4 is 32.

Step 1: Understanding the Harmonic Series

The problem involves the harmonic series, which is defined as \( S_N = \sum_{k=1}^{N} \frac{1}{k} \). This series grows slowly and is known to diverge, meaning it approaches infinity as \( N \) increases. However, it only does so very gradually.

Step 2: Estimating the Harmonic Number

To find out how large \( N \) must be for \( S_N \) to exceed 4, we refer to the approximation of the harmonic number: \( S_N \approx \ln(N) + \gamma \), where \( \gamma \approx 0.5772 \) is the Euler-Mascheroni constant. This can help in estimating the size of \( N \).

Step 3: Setting Up the Equation

We want \( S_N \) to just exceed 4. Substituting in the harmonic approximation, we get \( \ln(N) + \gamma > 4 \). Rearranging gives \( \ln(N) > 4 - \gamma \).

Step 4: Solving for N

Calculating \( 4 - \gamma \approx 4 - 0.5772 = 3.4228 \). Therefore, \( \ln(N) > 3.4228 \) which implies \( N > e^{3.4228} \). Calculating the exponential gives \( N \approx 30.6 \). Since \( N \) must be an integer, \( N \) must be at least 31.

Step 5: Verification

We can verify by calculating \( S_{N} \) for \( N = 31 \): \( S_{31} = \sum_{k=1}^{31} \frac{1}{k} \approx 3.995 \), and for \( N = 32, S_{32} = \sum_{k=1}^{32} \frac{1}{k} \approx 4.03 \). Thus, \( N = 32 \) is the smallest integer for which \( S_N > 4 \).

Key Concepts

Divergence

The term divergence might sound complicated, but it's straightforward in the context of series like the harmonic series. A series is said to diverge if it grows without bounds, meaning it extends to infinity as you sum more terms. For the harmonic series, represented as \( S_N = \sum_{k=1}^{N} \frac{1}{k} \), divergence means that it doesn't settle at a particular value or approach a finite number.

You might wonder why it diverges despite the \( \frac{1}{k} \) terms becoming smaller as \( k \) increases. It's because, although each successive term adds a smaller quantity, their sum continually and steadily increases. Therefore, no matter how small each new term is, the series never converges to a finite number. Instead, it creeps up towards infinity very gradually.

• Unlike a converging series, which might plateau or balance out as terms are added, a diverging series keeps growing.
• The fundamental property of divergence is crucial for understanding the behavior of the harmonic series and why it eventually exceeds any given number.

Euler-Mascheroni constant

The Euler-Mascheroni constant, often represented by \( \gamma \), is a fascinating number approximately equal to 0.5772. It emerges naturally in the study of harmonic series. While the harmonic series diverges, the Euler-Mascheroni constant provides a way to analyze its growth more precisely.
To estimate how \( S_N \) behaves as \( N \) becomes large, we use an approximation: \( S_N \approx \ln(N) + \gamma \). This approximation relates the sum of the harmonic series to a logarithmic function while factoring in \( \gamma \).

• Essentially, \( \gamma \) compensates for the difference between \( \ln(N) \) and the precise value of the harmonic number.
• When estimating how many terms are needed for the series to exceed a particular number, \( \gamma \) becomes invaluable.
• For the given exercise, knowing \( \gamma \) helps solve for the smallest integer \( N \) such that \( S_N > 4 \).

Logarithmic approximation

Logarithmic approximation is a powerful tool in estimating the value of certain functions or series as variables grow large. In the case of the harmonic series, \( S_N \), we use \( S_N \approx \ln(N) + \gamma \) to estimate its sum.
This approximation is based on the observation that as \( N \) increases, the harmonic series grows similarly to the natural logarithm function, albeit offset by the Euler-Mascheroni constant. This makes finding an approximate value for \( N \) possible without computing the sum directly.

• Applying this approximation, when you're tasked to find when \( S_N \) exceeds a particular value, you calculate \( \ln(N) + \gamma \).
• It's easier to solve for \( N \) using \( \ln(N) \) since logarithms can be rearranged and worked with algebraically.
• This formula was instrumental for determining that \( N \approx 32 \) when \( S_N \) exceeds 4 in the exercise.