The expression you provided is the definition of the
Euler-Mascheroni constant ($\gamma$). Let's carefully analyze and explain the solution step by step.
Problem Statement:
The Euler-Mascheroni constant $\gamma$ is defined as:
$$
\gamma = \lim_{n \to \infty} \left( \sum_{k=1}^n \frac{1}{k} - \ln n \right).
$$
Step-by-Step Explanation:
1.
Harmonic Series:
The term $\sum_{k=1}^n \frac{1}{k}$ represents the $n$-th partial sum of the harmonic series. The harmonic series is given by:
$$
H_n = \sum_{k=1}^n \frac{1}{k}.
$$
As $n \to \infty$, the harmonic series diverges, but its growth rate can be approximated using the natural logarithm.
2.
Asymptotic Behavior of the Harmonic Series:
It is a well-known result in mathematics that the harmonic series grows asymptotically like the natural logarithm:
$$
H_n \sim \ln n + \gamma,
$$
where $\gamma$ is a constant. This means:
$$
H_n = \ln n + \gamma + o(1),
$$
where $o(1)$ represents a term that approaches 0 as $n \to \infty$.
3.
Definition of $\gamma$:
The Euler-Mascheroni constant $\gamma$ is defined as the limiting difference between the harmonic series and the natural logarithm:
$$
\gamma = \lim_{n \to \infty} \left( H_n - \ln n \right).
$$
Substituting $H_n = \sum_{k=1}^n \frac{1}{k}$, we get:
$$
\gamma = \lim_{n \to \infty} \left( \sum_{k=1}^n \frac{1}{k} - \ln n \right).
$$
4.
Interpretation:
The constant $\gamma$ captures the discrepancy between the discrete sum (harmonic series) and the continuous approximation (natural logarithm) as $n$ becomes very large. It is a fundamental constant in mathematics, appearing in various contexts, including number theory, analysis, and probability.
5.
Numerical Value:
The value of $\gamma$ is approximately:
$$
\gamma \approx 0.5772156649.
$$
Final Answer:
The expression you provided is the definition of the Euler-Mascheroni constant. Therefore, the solution is:
$$
\boxed{\gamma}
$$
Parent Tip: Review the logic above to help your child master the concept of hard math.
