QuestionCalculate the value of the following series: $$\sum_{k=2}^{\infty} \frac{1}{(3 k+1)(3 k+4)}$$

Studdy Solution
Observe the telescoping nature of the series: The terms will cancel in a telescoping manner, leaving only a few terms from the beginning and the end.
Write out the first few terms to see the pattern: $$\frac{1}{3} \left( \left( \frac{1}{7} - \frac{1}{10} \right) + \left( \frac{1}{10} - \frac{1}{13} \right) + \left( \frac{1}{13} - \frac{1}{16} \right) + \cdots \right)$$
Notice that all intermediate terms cancel, leaving: $$\frac{1}{3} \left( \frac{1}{7} - \lim_{n \to \infty} \frac{1}{3n+4} \right)$$
As $n \to \infty$, $\frac{1}{3n+4} \to 0$.
Therefore, the sum is: $$\frac{1}{3} \times \frac{1}{7} = \frac{1}{21}$$
The value of the series is:
$$\boxed{\frac{1}{21}}$$