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

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Answer

Since the integral diverges, the series also $diverges$

Steps

Step 1 ： $f (x) = \frac{2}{3 x + 3}$

Step 2 ： $f^{'} (x) = \frac{- 6}{(3 x + 3)^{2}}$

Step 3 ：Since $f^{'} (x) < 0$ for all $x > 1$ , the function is decreasing.

Step 4 ： $\int_{1}^{\infty} \frac{2}{3 x + 3} d x = \int_{1}^{\infty} \frac{2}{3 (x + 1)} d x$

Step 5 ：Substitute $u = x + 1$ , then $d u = d x$ and the integral becomes:

Step 6 ： $\int_{2}^{\infty} \frac{2}{3 u} d u$

Step 7 ： $= \frac{2}{3} \int_{2}^{\infty} \frac{1}{u} d u$

Step 8 ： $= \frac{2}{3} [\ln (u)]_{2}^{\infty}$

Step 9 ：Since the integral diverges, the series also $diverges$