For i)
Let $K$ be a compact of $\mathbb{C}$. Then there exists $\delta > 0$ such that $\forall z \in K$ $d( z , \mathbb{N}) > \delta$ and since $K$ is bounded the sequence of function $ z \mapsto \frac{1}{1 +\frac{z}{n}}$ converges uniformily to $1$ on $K$ (Ask me if you want details on the proof of this statment)
Now remind us that to show the power series converges uniformly on $K$ we only have to demonstrate that it is unifomrly Cauchy on $K$. Put $B_k(z) = \frac{1}{z+k+1} - \frac{1} {z+k}$ and $a_k = 0$ if $k$ is even and $1$ if $k$ is odd.
Let's write $\sum_p^{q}{\frac{(-1)^n}{z +n}} = \sum_p^q{a_kB_k(z)} + a_{p-1}\frac{1}{z+q} - a_q\frac{1}{z+q+1}$
$B_k(z) = \frac{-1}{(z+k+1)(z+k)} = \frac{-1}{k^2}\frac{1}{(1+\frac{z}{k})(1+\frac{z+1}{k})}$
Since the sequence of function $ z \mapsto \frac{1}{1 +\frac{z}{n}}$ converges uniformily to $1$ on $K$, for all $\epsilon > 0$ there exists $p$ such that $\forall k \geq p$ and $\forall z \in K$ we have $|B_k(z)| \leq \frac{2}{k^2}$ and $|\frac{1}{z+k}| \leq \epsilon$
Then $ |\sum_p^q{ \frac{(-1)^n}{z +n}}| \leq 2\sum_p^q{\frac{1}{n^2}} + 2\epsilon $
Since it doesn't depends on $z$, the convergence is uniform.
(The method used in this proof is a particular case of the Abel Transformation)