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 $n \to \infty$.