Determine if the sequence converges or diverges
𝑎𝑛 = cos(𝜋𝑛)
it does not converge, but oscillates between ±1
To determine if the sequence 𝑎𝑛 = cos(𝜋𝑛) converges or diverges, we need to examine the behavior of the terms of the sequence as 𝑛 approaches infinity.
Let's first evaluate the first few terms of the sequence:
𝑎₁ = cos(𝜋)
𝑎₂ = cos(2𝜋)
𝑎₃ = cos(3𝜋)
𝑎₄ = cos(4𝜋)
𝑎₅ = cos(5𝜋)
We can notice a pattern here. By observing the argument of the cosine function, we can see that the values of 𝑎𝑛 repeat after every two terms since cos(𝑛𝜋) takes the same values at 𝑛 and 𝑛 + 2 (for integer values of 𝑛).
Now, let's consider the limit as 𝑛 approaches infinity. Since the values of 𝑎𝑛 repeat after every two terms, we can focus on the behavior of the terms with 𝑛 = 2𝑘 and 𝑛 = 2𝑘 + 1, where 𝑘 is an integer.
For 𝑛 = 2𝑘 (even), the sequence becomes:
𝑎₂𝑘 = cos(2𝜋𝑘) = cos(0) = 1
For 𝑛 = 2𝑘 + 1 (odd), the sequence becomes:
𝑎₂𝑘₊₁ = cos(2𝜋𝑘₊₁) = cos(𝜋) = -1
Thus, the terms alternate between 1 and -1 as 𝑛 approaches infinity. As a result, the sequence 𝑎𝑛 = cos(𝜋𝑛) does not converge since it does not approach a single value. Instead, it diverges between 1 and -1.
To determine if the sequence 𝑎𝑛 = cos(𝜋𝑛) converges or diverges, we need to examine its behavior as 𝑛 approaches infinity.
First, let's recall that the cosine function has a period of 2𝜋. This means that for any integer value of 𝑛, cos(𝜋𝑛) will take the same value as cos(𝜋(𝑛+2)).
Now, let's consider the values of cos(𝜋𝑛) as 𝑛 increases:
For 𝑛 = 0, cos(𝜋𝑛) = cos(0) = 1.
For 𝑛 = 1, cos(𝜋𝑛) = cos(𝜋) = -1.
For 𝑛 = 2, cos(𝜋𝑛) = cos(2𝜋) = 1.
For 𝑛 = 3, cos(𝜋𝑛) = cos(3𝜋) = -1.
For 𝑛 = 4, cos(𝜋𝑛) = cos(4𝜋) = 1.
From this, we can observe that the sequence alternates between the values of 1 and -1 as 𝑛 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 𝑎𝑛 = cos(𝜋𝑛) diverges.