Question: > 
> If $a,b,c$ be real constants, analyze the convergence of $$\sum\_{n=1}^{\infty} (-1)^{n-1}\frac{n+a}{(n+b)(n+c)}$$
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**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.

Answer: 
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.
As Julian Aguirre pointed, we are not allowed to use a comparison with an alternating series. However, Dirichlet's test is enough to ensure convergence, since the partial sums of $(-1)^n$ are bounded and $\frac{(n+a)}{(n+b)(n+c)}$ is eventually decreasing to zero as $n\to +\infty$.

We may notice that:
$$\frac{(n+a)}{(n+b)(n+c)}=\frac{b-a}{b-c}\cdot\frac{1}{(n+b)}+\frac{c-a}{c-b}\cdot\frac{1}{(n+c)}$$
from which it follows that:
$$\begin{eqnarray\*} \sum\_{n\geq 1}(-1)^{n-1}\frac{(n+a)}{(n+b)(n+c)}&=&\int\_{0}^{1}\sum\_{n\geq 1}(-1)^{n-1}\left(\frac{b-a}{b-c}x^{n+b-1}+\frac{c-a}{c-b}x^{n+c-1}\right)\,dx\\&=&\frac{1}{b-c}\int\_{0}^{1}\frac{(b-a)x^b+(a-c)x^c}{1+x}\,dx.\end{eqnarray\*}$$