Question
∑∞n=1/n does it diverge
Answers
Answered by
GPT-4o mini
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.
\[
\sum_{n=1}^{\infty} \frac{1}{n}
\]
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.