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: 
Your argument is incorrect. The comparison principle is for series of non-negative terms. The series
$$
\sum\_{n=2}^\infty\frac{(-1)^n}{\sqrt n+(-1)^n}
$$
is alternating. Leibniz's test cannot be used because $1/(\sqrt n+(-1)^n)$ is not decreasing; in fact, the series diverges. But
$$
\lim\_{n\to\infty}\frac{\dfrac{1}{\sqrt n+(-1)^n}}{\dfrac{1}{\sqrt n}}=1\quad\text{and}\quad\sum\_{n=2}^\infty\frac{(-1)^n}{\sqrt n}\quad\text{converges.}
$$
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:
$$\begin{eqnarray\*} \sum\_{n\geq 1}(-1)^{n-1}\frac{(n+a)}{(n+b)(n+c)}&=&\int\_{0}^{1}\sum\_{n\geq 1}(-1)^{n-1}\left(\frac{b-a}{b-c}x^{n+b-1}+\frac{c-a}{c-b}x^{n+c-1}\right)\,dx\\&=&\frac{1}{b-c}\int\_{0}^{1}\frac{(b-a)x^b+(a-c)x^c}{1+x}\,dx.\end{eqnarray\*}$$