Question

Determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \sin \frac{(2 n-1) \pi}{2} $$

Short answer

The given series diverges.

Step-by-step solution

Understand the Series or Sequence

First, look at the series. Here, it's an infinite series based on a sine function, \(\sin \frac{(2 n-1) \pi}{2}\). Note that \(n\) starts from \(1\) and goes up to infinity. Because \((2n - 1)\) gives you an odd number for any integer \(n\), the inside of the sine function is \(\frac{odd \cdot \pi}{2}\). The sine function exhibits a pattern for \(0\), \(\frac{\pi}{2}\), \(\pi\), \(\frac{3\pi}{2}\), and \(2\pi\).

Determine Sine Values

Next, determine the sine values for the pattern observed in step 1. The sine values for \(\frac{odd \cdot \pi}{2}\) are 1, -1, 1, -1, ... constantly alternating between 1 and -1.

Convergence Test

We want to determine if this series converges or diverges. For the series to converge, the terms in the series must approach zero as \(n\) approaches infinity. However, in this case, the terms in this series are constantly alternating between 1 and -1, and therefore do not approach zero.

Conclude Series Divergence

Since the terms do not approach zero, by the nth-term test for divergence (also known as the Test for Divergence), the given series diverges.