Calculus 4 - by Vlad Vasilescu (W)

From the formula $(\frac{1}{x+a})^{(k)}=\frac{(-1)^kk!}{(x+a)^{k+1}}$ we get $f^{(k)}(1)=\frac{(-1)^kk!}{2^{k+1}}$ . Then , $a_{n}$ will be $\displaystyle \sum_{k=0}^n$ $\frac{1}{2^{k+1}} = \frac{1}{2} \frac{(\frac{1}{2})^n - 1}{\frac{1}{2} -1 }$ . $\Rightarrow$ $lim_{n\rightarrow \infty}$ $a_{n}$ = $\boxed{1} .$

\(\begin{array} {} f^{(0)} (x) = f(x) = (1+x)^{-1} & = 0!(f(x))^1 \\ f^{(1)} (x) = (-1)(1+x)^{-2} & = -(1!)(f(x))^2 \\ f^{(2)} (x) = 2(1+x)^{-3} & = 2!(f(x))^3 \\ f^{(3)} (x) = (-6)(1+x)^{-4} & = -(3!)(f(x))^4 \\ ... & = \ ... \end{array} \)

$\implies f^{k} (x) = (-1)^kk!(f(x))^{k+1}$

Therefore, we have:

$\begin{aligned} a_n & = \sum_{k=0}^n \frac {(-1)^k}{k!} f^{(k)}(1) \\ & = \sum_{k=0}^n \frac {(-1)^k}{k!} (-1)^kk!(f(1))^{k+1} \\ & = \sum_{k=0}^n (f(1))^{k+1} \\ & = \sum_{k=0}^n \left(\frac 12 \right)^{k+1} \\ & = \frac 12 \times \frac 1{1-\frac 12} \\ & = \boxed{1} \end{aligned}$