This is a periodic function of $t$.
At $t = 2m \pi $, it diverges. But for other values, we know by the Abel criterion that it converges.
The problem is, what is the asymptotic behavior of $f$ as $t\rightarrow 0^+ $?
As a similar question, how about the function
$$g (t ) = \sum_{n\geq 1 } \sin (nt) /\sqrt{n} $$
It should have a jump at $t = 0 $.
You can analyze $$g(t) = \sum_{n=1}^\infty n^{-1/2} e^{-nt},\qquad \Re(t) \ge 0, t \not \in 2i \pi \mathbb{Z}$$ from its Mellin transform $$G(s)=\int_0^\infty t^{s-1}g(t)dt=\Gamma(s) \zeta(s+1/2)$$ obtaining that for $t \in (0,1)$ $$g(t) = \frac{1}{2i\pi } \int_{1-i\infty}^{1+i\infty} G(s) t^{-s}ds=\sum Res(G(s) t^{-s})=\Gamma(1/2) t^{-1/2}+ \sum_{k=0}^\infty \frac{(-1)^k}{k!} \zeta(1/2-k) t^k $$ which stays true for $|t|< 1,\Re(t) \ge 0$ by analytic continuation, as well as for $|t|< 2\pi,\Re(t) \ge 0$ by the decay property of $\frac{\zeta(1/2-k)}{k!}$ found from the functional equation for $\zeta(s)$