# Convergence or divergence of $\sum _{k=1}^{\infty} \left [ \cos(\frac {1} {k^{2}})-\cos(\frac {1} {(k+1)^{2}}) \right ]$

I am working on a two part question. It asks me to determine whether two different infinite series are divergent or convergent. Here are the two infinite series:

(a) $\sum _{i=1}^{\infty} \cos(\frac {1} {i^{2}})$

(b) $\sum _{k=1}^{\infty} \left [ \cos(\frac {1} {k^{2}})-\cos(\frac {1} {(k+1)^{2}}) \right ]$

For part (a) I have determined it is divergent by finding $\lim_{i \to \infty} cos(\frac {1} {i^{2}})=1$, which shows that the series must be divergent. For part (b) I've found that both terms of the series approach 1 as $k \to \infty$. This means each term is divergent on its own. This doesn't tell me whether the series is convergent or divergent, it just tells me that the series could be convergent, because $\lim_{k \to \infty}a_{k}=0$. I am wondering what I can do to get a proper answer.

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For part $(b)$ write down the the $N$th partial sum and notice that it is telescoping, i.e. $$\sum_{k=1}^N\big[(\cos(\frac{1}{k^2}) - \cos\big(\frac{1}{(k+1)^2}\big)\big] = \cos(1) -\cos\big(\frac{1}{(N+1)^2}\big)$$
Now the series is just the limit of the $N$th partial sum as $N$ goes to infinity, and thus the sum is $$\lim_{N\to \infty} \big[\cos(1) -\cos\big(\frac{1}{(N+1)^2}\big)\big] = \cos(1) -\cos(0) = \cos(1) - 1$$
$\cos(a)-\cos(b)=-2\sin(\dfrac{(a+b)}{2})\sin(\dfrac{(a-b)}{2})$, maybe what you need is this.