# `a_n=cos(pin)/n^2` Determine the convergence or divergence of the sequence with the given n'th term. If the sequence converges, find its limit.

`cos(pi n)={(1 if n=2k-1),(-1 if n=2k):},` `k in ZZ` i.e. it is equal to `1` for odd `n` and `-1` for even `n.` Therefore, we can break this into two cases.

`n=2k-1` (n is odd)

`lim_(n to infty)a_n=lim_(n to infty)1/n^2=1/infty^2=1/infty=0`

`n=2k` (n is even)

`lim_(n to infty)a_n=lim_(n to infty)-1/n^2=-1/infty^2=-1/infty=0`

Since the limit is the same in both cases, the sequence is convergent and its limit is equal to zero.

Image below shows first 15 terms of the sequence. We can see that the odd-numbered terms are negative, while the even-numbered terms are positive, but they are both approaching the `x`-axis implying convergence to zero.

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