Divergence of a trigonometric series

[Nono713]

Show that this series diverges:
$$\sum_{n = 0}^\infty \cos \left ( n^2 \right )$$
(in the sense that it takes arbitrarily large values as $n \to \infty$)

 

[jvicens]

To show that this series diverges, we can use the comparison test. First, let's consider the series $\sum_{n = 0}^\infty \cos \left ( n^2 \right )$. We know that $n^2$ grows faster than $n$, so we can write $n^2 > n$ for all $n \geq 2$. This means that $\cos \left ( n^2 \right ) > \cos \left ( n \right )$ for all $n \geq 2$.

Now, we can compare our series to the series $\sum_{n = 0}^\infty \cos \left ( n \right )$. This series is known to diverge, as it is the alternating harmonic series. Since $\cos \left ( n^2 \right ) > \cos \left ( n \right )$ for all $n \geq 2$, our original series must also diverge.

Therefore, we have shown that the series $\sum_{n = 0}^\infty \cos \left ( n^2 \right )$ diverges in the sense that it takes arbitrarily large values as $n \to \infty$. This means that as we continue to add terms to the series, the sum will continue to increase without bound.