Is the argument I used to evaluate the convergence of the series $\sum_{n=1}^{\infty} (-1)^{n-1}\frac{n+a}{(n+b)(n+c)}$ right? 
If $a,b,c$ be real constants, analyze the convergence of $$\sum_{n=1}^{\infty} (-1)^{n-1}\frac{n+a}{(n+b)(n+c)}$$

What I tried to:
I compared the general term of my series to $\frac{1}{n}$: $$\lim \limits_{n \to \infty} \frac{\frac{n+a}{(n+b)(n+c)}}{\frac{1}{n}}= $$ $$\lim\limits_{n \to \infty} \frac{n^2 (1 + \frac{a}{n})}{n^2 \left ( 1 + \frac{b+c}{n} + \frac{bc}{n^2} \right ) } = 1 $$
So, as $\frac{n+a}{(n+b)(n+c)} \sim \frac{1}{n}$ when $n \to \infty$, the series  (conditionally) converges because the alternating harmonic series converges.

Since if $b\neq c$ then there exists $A,B\in\mathbb{R}$ such that $$\frac{n+a}{(n+b) (n+c)}=\frac{A}{n+b} +\frac{B}{n+c} $$
the series is convergent by Leibniz test.
If $b=c$ then $$\frac{n+a}{(n+b) (n+c)}=\frac{1}{n+b} +\frac{a-b}{(n+b)^2}$$
and the series is also convergent by Leibniz test.
