I am looking at the following:
$$\lim_{n \rightarrow \infty} \frac{\sum_{m=1}^{n} \sin m}{\sum_{m=1}^{n} \cos m}$$
I first thought the limit would be clearly $1$, but now I don't think it exists. The limits of the individual sums of course do not exist, but I am not sure about the ratio. I have looked for bounds on the individuals sums, but I don't see any that are helpful here.
First note that $\sum_{m=1}^n\sin m,\cos m$ are the imaginary,real part of $\frac{1-e^{(n+1)i}}{1-e^i}=e^i+...e^{ni}$, resp.
Thus, that the limit of your formula exists implies that the argument of $1-e^{(n+1)i}$ is convergent.
Just by comparing three adjacent $n$, you can check that this is not convergent.
If you multiply both the numerator and denominator by $\sin \frac{1}{2}$ and apply formulas of the product of sines and cosines:
$$\sin k \sin \frac{1}{2} = \frac{1}{2}\Big[-\cos(k+\frac{1}{2}) + \cos(k-\frac{1}{2})\Big]$$ $$\cos k \sin \frac{1}{2} = \frac{1}{2}\Big[\sin(k+\frac{1}{2}) - \sin(k-\frac{1}{2})\Big]$$ and next use the chain rule, you'll get $$\frac{-\cos (n+\frac{1}{2}) + \cos \frac{1}{2}}{\sin (n+\frac{1}{2}) - \sin \frac{1}{2}} = \tan\frac{n+1}{2}$$ if my calculations are correct. Since the tangent function is periodic, there exists no limit.
You have $$ \sum_{m=1}^n \sin m = {\rm Im} \Big(\sum_{m=1}^n e^{im} \Big) = {\rm Im} \Big(\frac{e^{i} - e^{i(n+1)}}{1-e^{i}} \Big) = {\rm Im}\Big(\frac{e^{i(n+2)/2}}{e^{i/2}} \frac{e^{-in/2}-e^{in/2}}{e^{-i/2}-e^{i/2}} \Big) = \\ = {\rm Im}\Big(e^{i(n+1)/2} \frac{\sin\frac n2}{\sin \frac12}\Big) = \frac{\sin\frac{n+1}{2}\sin\frac n2}{\sin \frac12}$$ Similar calculation gives you $$ \sum_{m=1}^n \cos m = {\rm Re} \Big(\sum_{m=1}^n e^{im} \Big) = \frac{\cos\frac{n+1}{2}\sin\frac n2}{\sin \frac12}$$ You have then $$ \frac{\sum_{m=1}^n \sin m}{\sum_{m=1}^n \cos m} = \tan\frac{n+1}{2}$$ and it's easy to check that it does not have a limit.
