Determine whether the sequence converges or diverges. If it converges, find the limit. an =
cos^2(n)/3^n
clearly it converges, since |cos(x)| < 1 so the terms approach zero as x gets huge, since 3^x gets huge. That is,
an < 1/3^n
To determine whether the sequence converges or diverges, we can take the limit of the terms of the sequence as n approaches infinity.
Let's find the limit of an as n approaches infinity.
lim(n→∞) [cos^2(n)/3^n]
Since the cosine function is bounded between -1 and 1, cos^2(n) is also bounded between 0 and 1 for all values of n.
lim(n→∞) (cos^2(n)/3^n) ≤ lim(n→∞) (1/3^n)
To evaluate lim(n→∞) (1/3^n), we can use the limit of geometric series.
The series 1/3^n is a geometric series with first term a = 1 and common ratio r = 1/3.
The sum of a convergent geometric series is given by S = a/(1 - r), as long as |r| < 1.
In this case, |1/3| < 1, so the series converges.
Therefore,
lim(n→∞) (1/3^n) = 1/(1 - 1/3) = 1/(2/3) = 3/2
Since cos^2(n)/3^n is always less than or equal to 1/3^n, we have
lim(n→∞) (cos^2(n)/3^n) ≤ lim(n→∞) (1/3^n) = 3/2
Therefore, the given sequence converges and its limit is 3/2.
To determine whether the sequence converges or diverges, we need to find the limit of the sequence as n approaches infinity.
First, let's express the nth term of the sequence, an, in terms of n.
an = (cos^2(n)) / (3^n)
To find the limit of the sequence, we will take the limit as n approaches infinity:
lim(n→∞) (cos^2(n)) / (3^n)
To evaluate this limit, we need to analyze the behavior of both the numerator (cos^2(n)) and the denominator (3^n) as n becomes infinitely large.
1. Analyzing the numerator:
The cosine function, cos(n), oscillates between -1 and 1 for any value of n. Squaring this function, cos^2(n), will always yield a positive value between 0 and 1. So, as n becomes infinitely large, cos^2(n) will still remain between 0 and 1.
2. Analyzing the denominator:
The term 3^n grows exponentially as n becomes larger. For example, if we evaluate 3^1, 3^2, 3^3, and so on, we get 3, 9, 27, and so on. As n approaches infinity, 3^n will grow without bound.
Considering both the numerator and denominator, we can see that the numerator remains limited between 0 and 1, while the denominator grows without bound. Dividing a small, bounded number (numerator) by a large, unbounded number (denominator) will yield a value of 0.
Hence, the limit of the sequence as n approaches infinity is 0.
Since the limit exists and is finite, the sequence converges. Therefore, the sequence given by an = (cos^2(n)) / (3^n) converges to 0.