Question

Determine the convergence or divergence of the series. $$ \frac{1}{201}+\frac{1}{204}+\frac{1}{209}+\frac{1}{216}+\cdots \cdot $$

Short answer

The given series converges.

Step-by-step solution

Identify the Series

Firstly, recognise the pattern of the input series, where the denominator for each term seems to follow the rule \( n^3 + 2n \). Thus, the series can be re-written as \( \sum _{n=3} ^{\infty} \frac{1}{n^3+2n} \)

Select Appropriate Test

To assess the convergence or divergence of the sum of the series, it is suggested to use the comparison test. The comparison test would be favorable since we could easily compare it against a known convergent series - \( \frac{1}{n^3} \) which is a p-series with \( p > 1 \), hence convergent.

Apply the Comparison Test

To apply the comparison test, we compare our given series with the known convergent series. Since \( \frac{1}{n^3+2n} \leq \frac{1}{n^3} \) for all \( n \geq 3 \), and we know that the series \( \sum \frac{1}{n^3} \) is a convergent p-series, we can conclude via the comparison test that our original series \( \sum _{n=3} ^{\infty} \frac{1}{n^3+2n} \) converges.