How do you determine whether the infinite sequence #a_n=n*cos(n*pi)# converges or diverges?

Question

Paisley Johnson · Accepted Answer

The sequence #a_n=ncos(n pi)# diverges.

Let us look at some details.

Let us take the limit of its subsequence of only even terms.

#\lim_{n to infty}a_{2n}=lim_{n to infty}(2n)cos(2npi)#

since #cos(2npi)=1# for all integer #n#,

#=lim_{n to infty}(2n)=2(infty)=infty#

Since the subsequence diverges, the original sequence also diverges.

Scarlett Allison · Answer

undefined To determine whether the infinite sequence $ a_n = n \cos(n\pi) $ converges or diverges, we can analyze its behavior.

1. Recall that the cosine function oscillates between -1 and 1.
2. For integer values of $ n $, $ \cos(n\pi) $ will be either 1, -1, or 0.
3. For $ n $ even, $ \cos(n\pi) $ is 1.
4. For $ n $ odd, $ \cos(n\pi) $ is -1.

Now, let's look at the sequence $ a_n = n \cos(n\pi) $:
- When $ n $ is even, $ \cos(n\pi) = 1 $, so $ a_n = n \cdot 1 = n $.
- When $ n $ is odd, $ \cos(n\pi) = -1 $, so $ a_n = n \cdot (-1) = -n $.

This means that the sequence alternates between positive and negative terms, with the absolute values increasing without bound as $ n $ increases.

Since the terms do not approach a fixed value as $ n $ approaches infinity, the sequence $ a_n = n \cos(n\pi) $ diverges.