The given series is: $$6+3*2^{1-n}$$
Prove that this series is geometric progression. What is $a_n$ in this series? And also show if the series is convergent.
I tried to:
So we know that the sum of n terms in geometric progression is: $$\frac{a_1(q^n-1)}{q-1}$$ I created the equation: $$\frac{a_1(q^n-1)}{q-1} = 6+3*2^{1-n} $$
The thing here is, that there is more than one variable. So I don't know what $q$ is and also what $a_1$ is. This is when things get difficult. To show that the series is convergent I know that I would have to use: $$-1<q<1$$, and for the $a_n$ term I would use $$a_n=a_1*q^{n-1}$$ But how to get $a_1$ and $q$ ?
$$S_n = 6+3\cdot 2^{1-n}$$ $$S_{n-1} = 6+3\cdot 2^{2-n}$$
$$a_n = S_n - S_{n-1}=3\cdot (2^{1-n}-2^{2-n})=3(2^{1-n})(1-2)=(-3)\cdot 2^{1-n}=-6\cdot 2^{-n}$$