Determine if the sequence converges or diverges
šš = cos(šš)
it does not converge, but oscillates between Ā±1
To determine if the sequence šš = cos(šš) converges or diverges, we need to examine the behavior of the terms of the sequence as š approaches infinity.
Let's first evaluate the first few terms of the sequence:
šā = cos(š)
šā = cos(2š)
šā = cos(3š)
šā = cos(4š)
šā
= cos(5š)
We can notice a pattern here. By observing the argument of the cosine function, we can see that the values of šš repeat after every two terms since cos(šš) takes the same values at š and š + 2 (for integer values of š).
Now, let's consider the limit as š approaches infinity. Since the values of šš repeat after every two terms, we can focus on the behavior of the terms with š = 2š and š = 2š + 1, where š is an integer.
For š = 2š (even), the sequence becomes:
šāš = cos(2šš) = cos(0) = 1
For š = 2š + 1 (odd), the sequence becomes:
šāšāā = cos(2ššāā) = cos(š) = -1
Thus, the terms alternate between 1 and -1 as š approaches infinity. As a result, the sequence šš = cos(šš) does not converge since it does not approach a single value. Instead, it diverges between 1 and -1.
To determine if the sequence šš = cos(šš) converges or diverges, we need to examine its behavior as š approaches infinity.
First, let's recall that the cosine function has a period of 2š. This means that for any integer value of š, cos(šš) will take the same value as cos(š(š+2)).
Now, let's consider the values of cos(šš) as š increases:
For š = 0, cos(šš) = cos(0) = 1.
For š = 1, cos(šš) = cos(š) = -1.
For š = 2, cos(šš) = cos(2š) = 1.
For š = 3, cos(šš) = cos(3š) = -1.
For š = 4, cos(šš) = cos(4š) = 1.
From this, we can observe that the sequence alternates between the values of 1 and -1 as š increases. Therefore, the sequence does not approach a specific value and oscillates indefinitely between 1 and -1.
Since the sequence does not converge to a limit, we conclude that the sequence šš = cos(šš) diverges.