Using Dirichlet's test I am trying to find out if the following series converges or diverges. If it converges I then want to find out if it converges absolutely or conditionally.
$$\sum _{n=1}^{\infty}\frac{\sin(n)}{\sqrt n}.$$
For the first part I used Dirichlet's law. 
I set $a_k = \frac {1}{\sqrt k }$ and $b_k = sink(k)$. 
So $a_k= \frac {1}{\sqrt k}$ converges to $0$ as n approaches infinity.
The second condition is to show that $b_k$ is bounded. 
$$\sum _{k=1}^{m}\sin(k)$$ using the trig identity $2\sin(a)\sin(b)=\cos(a-b)-\cos(a+b)$. Then,
$$2\sin(1) \sum _{k=1}^{m}\sin(k) =\sum _{k=1}^{m}2\sin(1)\sin(k) = \sum _{k=1}^{m}(\cos(1-k)-\cos(1+k)) $$ as $$\cos(1-k) = \cos(-1(k-1))$$ then $$=\sum _{k=1}^{m}(\cos(k-1)-\cos(k+1))$$
$$=(\cos0-\cos2)+(\cos1-\cos3)+(\cos2-\cos4)+(\cos3-\cos5)+(\cos4-\cos6)+...+(\cos(m-2)-\cos(m))+(\cos(m-1)-\cos(m+1))$$
$$=1+\cos1-\cos(m)-\cos(m+1)$$ Therefore, dividing through by the quantity $2\sin(1)$
$$\Biggl\lvert\sum _{k=1}^{m}\sin(k)\Biggr\rvert = \biggl\lvert\frac {1+\cos(1)-\cos(m)-\cos(m+1)}{2\sin(1)}\biggr\rvert \leqslant \biggl\lvert\frac {4}{\sin(1)}\biggr\rvert = \frac{2}{\sin(1)}= M.$$ Therefore this converges by Dirichlet's test. If this is correct, how do I show that it converges conditionally or absolutely?
 A: As stated in the comments, your proof of convergence using Dirichlet's test is basically correct. There are however a few minor issues.

So $a_k= \frac {1}{\sqrt k}$ converges to $0$ as n approaches infinity.

Dirichlet's test requires that the sequence is monotonic, so it would be better to explicitly say that $(a_k)$ is monotonic here. Also, you use the index $k$ but wrote "as $n$ approaches infinity".

The second condition is to show that $b_k$ is bounded.

That should say that the sequence of partial sums, $\sum_{k = 1}^m b_k$ is bounded, not the sequence $(b_k)$.
And at the end, you have a typo, $\bigl\lvert \frac{4}{\sin (1)}\bigr\rvert$ should be $\bigl\lvert\frac{4}{2\sin (1)}\bigr\rvert$.
None of that is a big issue, but these points would probably lead to a small deduction in a graded exercise.
To see that the series is not absolutely convergent, note that for any three consecutive integers, at least one has a sine of absolute value $\geqslant \frac{1}{2}$, for $\lvert\sin x \rvert < \frac{1}{2} \iff (\exists k\in \mathbb{Z})\bigl(\lvert x-k\pi\rvert < \frac{\pi}{6}\bigr)$, so the length of the intervals where $\lvert \sin x\rvert$ is small ($<\frac{1}{2}$) is shorter than $2$, thus such an interval can contain at most two integers, and the length of the intervals where $\lvert\sin x\rvert$ is large ($\geqslant \frac{1}{2}$) is longer than $1$, so one can't step over a whole such interval. That gives
$$\frac{\lvert\sin (3m-2)\rvert}{\sqrt{3m-2}} + \frac{\lvert\sin (3m-1)\rvert}{\sqrt{3m-1}} + \frac{\lvert\sin (3m)\rvert}{\sqrt{3m}} \geqslant \frac{1}{2\sqrt{3m}},$$
and from that we obtain
$$\sum_{k = 1}^{3n} \frac{\lvert\sin k\rvert}{\sqrt{k}} \geqslant \frac{1}{2\sqrt{3}}\sum_{m = 1}^n \frac{1}{\sqrt{m}},$$
which shows the divergence of
$$\sum_{k = 1}^{\infty} \frac{\lvert\sin k\rvert}{\sqrt{k}}.$$
