Sum of series using complex numbers 
In this I just know $e^{im}=\cos m +i\sin m$
but in this none of the terms are cancelling .
The answer of this question is 1008
 A: Let $u_k=e^{2\pi  i/k}$ and show that for $1\leq \lambda \leq k$
$$\frac{1}{1-u_{k+1}^{\lambda }}+\frac{1}{1-u_{k+1}^{k+1-\lambda }}=1 $$ 
To see this, just multiply top and bottom of the second fraction by $u_{k+1}^{\lambda }$ and use the fact that $u_{k+1}^{k+1}=1$
Next, let
$$g(k)=\sum _{\lambda =1}^k \frac{1}{1-u_{k+1}^{\lambda }}$$
By a change of index $\lambda \to  k+1-\lambda $ we see that
$$g(k)=\sum _{\lambda =1}^k \frac{1}{1-u_{k+1}^{k+1-\lambda }}$$
Therefore
$$2g(k)=\sum _{\lambda =1}^k \left(\frac{1}{1-u_{k+1}^{\lambda }}+\frac{1}{1-u_{k+1}^{k+1-\lambda }}\right)=\sum _{\lambda =1}^k 1=k$$
so
$$g(k)=\frac{k}{2}$$
and 
$$g(2016)=1008$$
A: I would write this as
$$S=\sum_{j=1}^{2016}\frac1{1-\zeta^j}$$
where $\zeta=\exp(2\pi i/2017)$.
Then $S=\lim_{t\to 1^-}F(t)$ where
$$F(t)=\sum_{j=1}^{2016}\frac1{1-t\zeta^j}.$$
For $|t|<1$,
$$F(t)=\sum_{j=1}^{2016}\sum_{n=0}^\infty t^n\zeta^{jn}
=\sum_{n=0}^\infty a_nt^n$$
where
$$a_n=\sum_{j=1}^{2016}\zeta^{nj}=-1+\sum_{j=0}^{2016}\zeta^{nj}.$$
Now $\sum_{j=0}^{2016}\zeta^{nj}=0$ unless $2017\mid n$ (why?),
so $a_n=-1$ if $2017\nmid n$. I'll leave you to work out $a_n$
when $2017\mid n$. You easily get an explicit rational function for $F(t)$ and then working out its limit as $t\to1^-$ should be OK.
Of course, all this works for other values of 2017.
