How do you determine what bn should be in a limit comparison test and a comparison test? When do you know that the series should use a limit comparison test or a comparison test to find if it is convergent/divergent? Examples would be greatly appreciated

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.

a=24; b=12

step-by-step explanation:

a/b=2: 1

a=2b

36/(1+2)=12

b=12

a=24

wfirfkfr/jjfi; jn; p/ojkfvlbvmkmgkgkkkgkkg

step-by-step explanation: