Infimum of $(\frac{1}{n}+\cos(\pi n))$ Question I have to prove the following:

$\inf\left(\frac{1}{n}+cos(\pi n)|n\in \Bbb N\right)=-1$

I thought that if I prove that $\inf(A+B)=\inf(A)+\inf(B)$ I can solve the question, since $\inf\left(\frac{1}{n}\right)=0$ and $\inf(\pi n)=-1$, so the sum is $-1$.
Can I use this approach solving the question?
Thanks,
Alan

If you fix $n$ odd, then you have the subsequence $-1 + \frac 1n$ and clearly 
$\inf \left(-1 + \frac 1n\right) = -1 + \inf \left(\frac 1n\right)$.
If you fix $n$ even, then the subsequence is $1 + \frac 1n$ which is always bigger than the other one so we don't really care about it.
