Question

Using the first 10 terms graph the sequence, and use the graph to discuss the convergence or divergence of the sequence an=cos(npi/2)?

Answer 1

Diverges due to oscillation of the cosine.

Explanation:

Determine the first `10` terms.

`a_1=cos(pi/2)=0`

`a_2=cospi=-1`

`a_3=cos((3pi)/2)=0`

`a_4=cos(2pi)=1`

It's worth noting here that we've completed one period of the cosine function. From here on, it'll be obvious that the values repeat.

`a_5=cos((5pi)/2)=0`

`a_6=cos(3pi)=-1`

`a_7=cos((7pi)/2)=0`

`a_8=cos(4pi)=1`

`a_9=cos((9pi)/2)=0`

`a_10=cos(5pi)=-1`

We must then plot the following points:

`(1, 0), (2, -1), (3, 0), (4, 1), (5, 0), (6, -1), (7, 0), (8, 1), (9,0), (10, -1):`

Don't connect them, although this graphing program might. Sequences aren't continuous, they only exist at integers. This program has done connecting and shading for visualization purposes only.

A sequence `a_n` converges if, for increasing values of `n, a_n` nears some finite value, IE, `lim_(n->oo)a_n=L ne +-oo`

In the above plot, we do not approach positive or negative infinity, but we do not quite approach a finite value either. The sequence is perpetually oscillating between `-1, 1,` not approaching any finite value, so it diverges.