The series you are referring to is the harmonic series, which is given by:
\[ \sum_{n=1}^{\infty} \frac{1}{n} \]
This series is known to diverge. One common way to demonstrate this is by using the comparison test or the integral test.
Integral Test:
Consider the function \( f(x) = \frac{1}{x} \). We can check the integral:
\[ \int_1^{\infty} \frac{1}{x} , dx \]
Calculating this integral, we have:
\[ \int_1^{\infty} \frac{1}{x} , dx = \lim_{t \to \infty} \left[ \ln(x) \right]1^{t} = \lim{t \to \infty} (\ln(t) - \ln(1)) = \lim_{t \to \infty} \ln(t) \]
Since \( \ln(t) \) approaches infinity as \( t \) approaches infinity, the integral diverges. By the Integral Test, since this improper integral diverges, the harmonic series also diverges.
Conclusion:
Therefore, the series
\[ \sum_{n=1}^{\infty} \frac{1}{n} \]
diverges.
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