Compute the first 5 terms of the sequence: {b(n)}n=1-> infinity = n�sqrt(6n) + (5/6)^n.
((that is:(Bsub(n))from n=1 to infinity = nth root of 6n + (5/6)raised to n.))
If the sequence converges, find the limit.
To compute the first 5 terms of the sequence, we substitute the values of n from 1 to 5 into the given formula.
Let's start with n = 1:
b(1) = 1√(6 * 1) + (5/6)^1 = 1√6 + 5/6
Next, for n = 2:
b(2) = 2√(6 * 2) + (5/6)^2 = 2√12 + 25/36
For n = 3:
b(3) = 3√(6 * 3) + (5/6)^3 = 3√18 + 125/216
For n = 4:
b(4) = 4√(6 * 4) + (5/6)^4 = 4√24 + 625/1296
Finally, for n = 5:
b(5) = 5√(6 * 5) + (5/6)^5 = 5√30 + 3125/7776
To determine whether the sequence converges, we analyze the behavior of the terms as n approaches infinity. One way to do this is to find the limit as n goes to infinity:
lim (n→∞) b(n) = lim (n→∞) n√(6n) + (5/6)^n
To find the limit, we can check the behavior of each term separately.
As n approaches infinity, n√(6n) grows at a faster rate than (5/6)^n. Therefore, the dominant term is n√(6n).
We can simplify the expression by dividing each term by n:
lim (n→∞) n√(6n) + (5/6)^n = lim (n→∞) √(6n) + (5/6)^n/n
Now let's analyze the two terms separately:
The first term, √(6n), as n approaches infinity, becomes √(∞), which is infinity.
The second term, (5/6)^n/n, as n approaches infinity, converges to zero because the exponential term (5/6)^n decreases much faster than n increases.
Therefore, the limit of the sequence as n approaches infinity is infinity.