Confirm that the Integral Test can be applied to the series.
\[
\sum_{n=1}^{\infty} \frac{2}{3 n+3}
\]
Evaluate the following.
\[
\int_{1}^{\infty} \frac{2}{3 x+3} d x=
\]
Use the Integral Test to determine the convergence or divergence of the series.
converges
diverges
Answer
Since the integral diverges, the series also \(\boxed{\text{diverges}}\)
Step 1 ：\(f(x) = \frac{2}{3x+3}\)
Step 2 ：\(f'(x) = \frac{-6}{(3x+3)^2}\)
Step 3 ：Since \(f'(x) < 0\) for all \(x > 1\), the function is decreasing.
Step 4 ：\(\int_{1}^{\infty} \frac{2}{3x+3} dx = \int_{1}^{\infty} \frac{2}{3(x+1)} dx\)
Step 5 ：Substitute \(u = x + 1\), then \(du = dx\) and the integral becomes:
Step 6 ：\(\int_{2}^{\infty} \frac{2}{3u} du\)
Step 7 ：\(= \frac{2}{3} \int_{2}^{\infty} \frac{1}{u} du\)
Step 8 ：\(= \frac{2}{3} [\ln(u)]_{2}^{\infty}\)
Step 9 ：Since the integral diverges, the series also \(\boxed{\text{diverges}}\)
