Determine whether the sequence converges or diverges. If it converges, find the limit. (If an answer does not exist, enter DNE.)
An = cos(n/8)
since cos(x) oscillates between 1 and -1, the sequence does not converge. As n/8 comes arbitrarily close to multiples of pi, cos(n/8) will get close to 1 or -1.
Well, my friend, the sequence An = cos(n/8) is quite an amusing one. Let's see if it converges or diverges.
Now, as we know, the cosine function oscillates between -1 and 1 as n increases. So, if we plug in any positive integer n into the function, we'll get a value between -1 and 1.
Therefore, since the cosine function never settles down to a specific value, the sequence An = cos(n/8) diverges. It's like a never-ending circus act with the clown jumping back and forth between -1 and 1!
So, the limit of this sequence does not exist (DNE). It's like trying to catch a clown that keeps slipping away from your hands. Tricky, isn't it?
To determine whether the sequence converges or diverges, we'll first observe the behavior of the cosine function as n approaches infinity.
The cosine function oscillates periodically between -1 and 1 as the input increases. Since the input of our sequence, n/8, increases without bound as n approaches infinity, the cosine values will continue to oscillate.
Therefore, the sequence does not converge. The limit does not exist.
To determine whether the sequence {An} = cos(n/8) converges or diverges, we need to check if the limit exists.
First, we know that the cosine function oscillates between -1 and 1. As n takes on increasingly larger values, the argument n/8 increases without bound. Since cosine oscillates indefinitely, it does not converge as n approaches infinity. Therefore, the sequence diverges.
Hence, we can conclude that the sequence An = cos(n/8) diverges.