Step-by-step explanation:
Pick a function that is the same "family". It needs to be a function that you know diverges or converges. So p-series and geometric series are common choices. Often we make the numerators the same so that it's easy to compare.
For example, if you have an = 1 / (n − 1), you would choose bn = 1 / n. Since n − 1 is less than n, we know an is greater than bn. And since we know bn diverges, that means the larger function an also diverges.
Or, if you have an = 1 / (n + 1), we again choose bn = 1 / n. However, comparison test is inconclusive here (an < bn, bn diverges), so we use limit comparison test instead.
lim(n→∞) an / bn
lim(n→∞) 1 / (n + 1) / (1 / n)
lim(n→∞) n / (n + 1)
1
The limit is greater than 0, and bn diverges, so an also diverges.
Let's try something more complicated. Let's say an = e⁻ⁿ / (n + cos²n). The numerator e⁻ⁿ is always less than 1, and the denominator is always greater than n.
If we again choose p-series bn = 1 / n, we know bn > an, and bn diverges, so comparison test is inconclusive. Limit comparison test is possible, but tricky.
But, if we choose geometric series bn = e⁻ⁿ / 1, we know bn > an, and bn converges, so by comparison test, an converges as well.
We can try one more: an = (n² + 2) / (n⁴ + 5). Let's choose bn = (n² + 2) / n⁴ = 1 / n² + 2 / n⁴.
The numerators are the same, but an has a larger denominator, so bn > an. bn is the sum of two p-series which converge, so bn converges. Therefore, an converges.
