Sure, let's solve the given limit step by step:
The problem is to find the limit:
[tex]\[ \gamma = \lim_{n \rightarrow \infty}\left(\sum_{k=1}^n \frac{1}{k} - \ln n\right) \][/tex]
1. Understanding the Harmonic Series:
The sum [tex]\(\sum_{k=1}^n \frac{1}{k}\)[/tex] is known as the [tex]\(n\)[/tex]-th Harmonic number, denoted as [tex]\(H_n\)[/tex]. As [tex]\(n\)[/tex] grows large, the harmonic series [tex]\(H_n\)[/tex] diverges. However, it diverges logarithmically, and is asymptotically close to [tex]\(\ln(n)\)[/tex].
2. Integral Approximation:
We will leverage the approximate relationship between the harmonic series and the natural logarithm. For large [tex]\(n\)[/tex], we have:
[tex]\[ H_n = \sum_{k=1}^n \frac{1}{k} \approx \ln(n) + \gamma \][/tex]
where [tex]\(\gamma\)[/tex] is the Euler-Mascheroni constant.
3. Subtracting [tex]\(\ln(n)\)[/tex]:
To find the specific limit given in the problem, we subtract [tex]\(\ln(n)\)[/tex] from both sides of the approximation:
[tex]\[ \sum_{k=1}^n \frac{1}{k} - \ln(n) \approx \ln(n) + \gamma - \ln(n) = \gamma \][/tex]
4. Taking the Limit:
Since the approximation becomes more accurate as [tex]\(n \rightarrow \infty\)[/tex], the limit is simply the constant term [tex]\(\gamma\)[/tex]:
[tex]\[ \gamma = \lim_{n \to \infty} \left( \sum_{k=1}^n \frac{1}{k} - \ln n \right) \][/tex]
Therefore, the limit is:
[tex]\[ \gamma \][/tex]
This [tex]\(\gamma\)[/tex] is a well-known mathematical constant called the Euler-Mascheroni constant, approximately equal to 0.57721.
