The given problem involves finding the limit of a sum as \( n \) approaches infinity. The expression represents a partial sum of a harmonic series scaled by \( ab \):

\[
S_n = ab \left( \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{n} \right).
\]

This can be written as:

\[
S_n = ab \sum_{k=2}^{n} \frac{1}{k}.
\]

This is the harmonic series starting from \( k = 2 \), which can be approximated as:

\[
\sum_{k=2}^{n} \frac{1}{k} = H_n - 1,
\]
where \( H_n \) is the \( n \)-th harmonic number:

\[
H_n \approx \ln(n) + \gamma,
\]
where \( \gamma \) is the Euler-Mascheroni constant (\( \gamma \approx 0.577 \)).

Thus, the sum can be approximated as:

\[
S_n \approx ab (\ln(n) + \gamma - 1).
\]

Now, let's compute the limit as \( n \to \infty \):

\[
\lim_{n \to \infty} ab (\ln(n) + \gamma - 1) = \lim_{n \to \infty} ab \ln(n) = \infty,
\]

since \( \ln(n) \) grows without bound as \( n \to \infty \).

### Conclusion:
The limit of the given series is \( \infty \).

Would you like more details or further clarification? 

Here are 5 related questions:
1. What is the harmonic series and how does it behave?
2. How do logarithmic functions grow compared to polynomial functions?
3. What is the Euler-Mascheroni constant and where does it come from?
4. Can you explain different types of divergent series?
5. How can we estimate sums of harmonic series for large \( n \)?

**Tip:** The harmonic series grows very slowly, but it does diverge, meaning its sum increases without bound as \( n \to \infty \).

Explore the limit of the harmonic series ab(1/2 + 1/4 + ... + 1/n) as n approaches infinity. Understand key concepts like the growth of logarithms and the Euler-Mascheroni constant. This calculus problem highlights the divergence of the harmonic series, with a detailed explanation of the steps involved. Suitable for university-level students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation