<s>[INST] Whether $\sum \limits_{n=1}^{\infty} \dfrac{\left(-1\right)^n n^2}{n^2+5} =\sum \limits_{n=1}^{\infty} \left(-1\right)^n -\dfrac{5}{n^2+5} $ $$\sum \limits_{n=1}^{\infty} \dfrac{\left(-1\right)^n n^2}{n^2+5} =\sum \limits_{n=1}^{\infty} \left(-1\right)^n -\dfrac{5}{n^2+5} $$
I don't trust my solution, mostly because I don't have much experience with using $\left(-1\right)^n$. 
Because both summations have the same starting point ($1$) and ending point ($\infty$). Thus $a_n = b_n$.
$$\dfrac{\left(-1\right)^n n^2}{n^2+5} = \left(-1\right)^n - \dfrac{5}{n^2+5} $$
$$\dfrac{\left(-1\right)^n n^2}{n^2+5} + \dfrac{5}{n^2+5} = \left(-1\right)^n  $$
$$ \left(-1\right)^n\dfrac{n^2+5}{n^2+5} = \left(-1\right)^n$$
$$ \dfrac{n^2+5}{n^2+5} = \dfrac{\left(-1\right)^n}{\left(-1\right)^n} $$ 
$$ 1 = 1 $$
 [/INST] $$ \dfrac{\left(-1\right)^n n^2}{n^2+5} =  \dfrac{\left(-1\right)^n ((n^2+5)-5)}{n^2+5} = \dfrac{\left(-1\right)^n (n^2+5)}{(n^2+5)} -  \dfrac{\left(-1\right)^n 5}{n^2+5} $$ $$=(-1)^n- \dfrac{\left(-1\right)^n 5}{n^2+5} $$
You missed $(-1)^n$ on R.H.S. with $5$ in numerator.
</s>