A geometric series like
[tex] \displaystyle \sum_{n=1}^\infty \dfrac{1}{\alpha^n} [/tex]
Converges if and only if [tex] |\alpha|>1 [/tex]. If this is the case, the sum equals
[tex] \displaystyle \sum_{n=1}^\infty \dfrac{1}{\alpha^n} = \dfrac{1}{\alpha-1} [/tex]
So, in your case, you have convergence if and only if
[tex] |2+a|>1 \iff 2+a>1 \lor 2+a<-1 \iff a>-1 \lor a<3 [/tex]
And if this is the case, the sum equals
[tex] \displaystyle \sum_{n=1}^\infty \dfrac{1}{(2+a)^n} = \dfrac{1}{2+a-1} = \dfrac{1}{a+1} [/tex]
