Question

Use the Direct Comparison Test to determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \frac{1}{n^{2}+1} $$

Short answer

The given series \( \sum_{n=1}^{\infty} \frac{1}{n^{2}+1} \) converges.

Step-by-step solution

Identify a benchmark series

Choose \( \frac{1}{n^2} \) (a p-series with p=2) as the benchmark series. This is because we know that a p-series with p>1 will always converge.

Use the Direct Comparison test

To apply the Direct Comparison Test, try to establish inequality between given series ( \( \frac{1}{n^{2}+1} \) ) and the benchmark series ( \( \frac{1}{n^2} \) ). It is clear that \(0 < \frac{1}{n^{2}+1} < \frac{1}{n^2} \) for all \(n \geq 1\). The benchmark series \( \frac{1}{n^2} \) is convergent, and since the given series is positive and always less than the convergent benchmark series.

Conclude the result

According to the Direct Comparison Test, if 0 ≤ a_n ≤ b_n for all n sufficiently large and the series \( \sum b_n \) converges, then the series \( \sum a_n \) also converges. Therefore, the given series \( \sum \frac{1}{n^{2}+1} \) also converges.