What is the result of applying the integral test to the series [math: ∑n=1∞12n−1] \displaystyle\sum\limits_{n=1}^ \infty \frac{1}{2n-1} ?

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We start by restating the problem: applying the integral test to the series sum from n = 1 to infinity of 1/(2n - 1). The integrand corresponding to the terms a_n = 1/(2n - 1) is f(x) = 1/(2x - 1), defined for x ≥ 1. For the integral test, we compare the series to the improper integral ∫ from 1 to ∞ of f(x) dx = ∫_1^∞ 1/(2x - 1) dx. Evaluating this integral, we substitute u = 2x - 1, so du = 2 dx, and the integral becomes (1/2) ∫  from u = 1 to ∞ of (1/u) du, which equals (1/2) [ln u]_{1}^{∞}. The natural logarithm grows without bound as u → ∞, so the integral diverges to infinity. By the integral test, if the improper integral diverges, the corresponding series also diverges. Thus the statement that 'The series diverges' is supported by the integral test. The key idea here is recognizing that the series is essentially a subsequence of the harmonic series (the odd terms), which shares the same divergence behavior, and the integral test confirms this divergence by showing the integral does not converge. If one were to consider alternative viewpoints, noting that the terms 1/(2n - 1) decrease to zero is necessary but not sufficient for convergence; the fact that the integral diverges provides a decisive criterion. In short, since ∫_1^∞ 1/(2x - 1) dx diverges, the series ∑_{n=1}^∞ 1/(2n - 1) also diverges.

What is the result of applying the integral test to the series [math: ∑n=1∞12n−1] \displaystyle\sum\limits_{n=1}^ \infty \frac{1}{2n-1} ?

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