## Calculus 8th Edition

$a_{n} = 2.5$ if $n=1$ $a_{n} = \frac{n-2}{2^{n}}$ if $n \geq 2$ and $\Sigma^{\infty}_{n=1} a_{n} = 3$
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 $\Sigma^{\infty}_{n=1} a_{n} = 3$