Final Answer:
The series Riemann sum of √(n-1) to infinity for √x / (2x² - 1) diverges.
Step-by-step explanation:
The given series is ∑_(n=1)^(∞) √(n-1) / (2n² - 1). To determine convergence or divergence, we can use the limit comparison test. Let a_n = √(n-1) / (2n² - 1). We choose b_n = 1/n^(3/2) as a comparison series, which is known to converge.
Now, calculate the limit of the ratio of a_n to b_n as n approaches infinity: lim_(n→∞) (a_n / b_n) = lim_(n→∞) [(√(n-1) / (2n^2 - 1)) / (1/n^(3/2))] = lim_(n→∞) [√(n-1) * n^(3/2) / (2n² - 1)].
Simplify the expression: lim_(n→∞) [√(n-1) * n^(3/2) / (2n² - 1)] = lim_(n→∞) [√(n-1) / √2n].
Now, divide numerator and denominator by n^(3/2): lim_(n→∞) [√(n-1) / √2n] = lim_(n→∞) [1 / √(2n/n-1)].
As n approaches infinity, the expression converges to 1/∞, which is 0. Since the limit is a finite positive value, it implies that both series have the same convergence behavior. Since the comparison series ∑_(n=1)^(∞) 1/n^(3/2) converges, our original series ∑_(n=1)^(∞) √(n-1) / (2n² - 1) also converges.
Therefore, the final answer is that the series Riemann sum of √(n-1) to infinity for √x / (2x² - 1) converges.