Can $$\sum_{n=1}^{\infty} \frac{2 + (-1)^n }{1.25^n} $$ be split into two so that it could be solved without the comparison test?
I am thinking of splitting the sum into to: the first one will be a convergent series, whereas the second one could be solved using the ratio test.
If $$\sum{f}=\sum a\pm\sum b$$ where $\sum a$ and $\sum b$ are convergent, then $\sum f$ is convergent too. Apply this to your sereis by splitting it to: $$\sum_{n=1}^{\infty}{\frac{2+(-1)^n}{1.25^n}}=\sum_{n=1}^{\infty}{\frac{2}{1.25^n}}+\sum_{n=1}^{\infty}{\frac{(-1)}{1.25^n}}$$
$1.25^n>2$ when $n\ge 4$, and $1.25^{n+1}>1.25^n \forall n\ge1$.
Thus Ratio Test makes$$\sum_{n=1}^{\infty}\frac{2}{1.25^n}$$ convergent.
With regards to the other half$$\sum_{n=1}^{\infty}\frac{(-1)}{1.25^n}=\sum_{n=1}^{\infty}\frac{1}{1.25^{2n}}-\sum_{n=1}^{\infty}\frac{1}{1.25^{2n-1}}$$
Showing they are both convergent will resolve your question.