Answer
$a_{n} = 2.5$ if $n=1$
$a_{n} = \frac{n-2}{2^{n}}$ if $n \geq 2$
and
$\Sigma^{\infty}_{n=1} a_{n} = 3$
Work Step by Step
When $n=1$
$a_{1} = s_{1} = 3-1\times 2^{-1} =3 - 0.5 = 2.5$
When $n \geq 2$
$a_{n} = s_{n} - s_{n-1} = [3-n2^{-n}] - [3- (n-1)2^{-n+1}]$
$=(n-1)2^{-n+1} -n2^{-n}$
$=\frac{(n-1)}{2^{n-1}}-\frac{n}{2^{n}}$
$=\frac{2n-2}{2^{n}}-\frac{n}{2^{n}}$
$=\frac{n-2}{2^{n}}$
Now, $\Sigma^{\infty}_{n=1} a_{n} = \lim\limits_{n \to \infty}s_{n}$
$=\lim\limits_{n \to \infty} 3-n2^{-n}$
$=3-\lim\limits_{n \to \infty} \frac{n}{2^{n}}$
Let us apply L'Hospital's Rule.
$=3-\lim\limits_{n \to \infty}\frac{1}{2^{n}\ln2}$
$=3-\frac{1}{\infty}$
$=3-0=3$
Hence, $a_{n} = 2.5$ if $n=1$
$a_{n} = \frac{n-2}{2^{n}}$ if $n \geq 2$
and
$\Sigma^{\infty}_{n=1} a_{n} = 3$
