multiple choice
(1) Find the sum:
$\displaystyle
S = x + \frac{x^2}{2\cdot 2}+ \frac{x^3}{3\cdot 2^2}+ \frac{x^4}{4\cdot 2^3} + ...
$
Consider this as a function of $\displaystyle x$ (note the negative sign in front of the term in $\displaystyle x^4$
I presume is a mistake), the derivative of this is:
$\displaystyle
U(x) = 1 + \frac{x}{ 2}+ \frac{x^2}{ 2^2}+ \frac{x^3}{ 2^3} + ...
$
which is a geometric series and it converges for $\displaystyle |x|<2$.
So write down the sum of this last series, then integrate with the constant
of integration set so that the integral is zero for $\displaystyle x=0$, and that
is your answer (it looks like (d) to me but you will need to check).
RonL
There are many verions, I use the the following one.
Taylor's Theorem (Lagrange): Let $\displaystyle f$ be $\displaystyle n$ differenciable on $\displaystyle (a,b)$ (with $\displaystyle a<0<b$) and for $\displaystyle 0\not =x\in (a,b)$ the remainder is given by $\displaystyle R_n(x) = \frac{f^{(n)}(y)}{n!}\cdot x^n$ for some $\displaystyle y$ between $\displaystyle 0$ and $\displaystyle x$. Where the remainder is the difference between the function and its $\displaystyle n-1$ degree Taylor polynomial $\displaystyle T_{n-1}(x)$, defined as $\displaystyle T_{n-1}(x) = \sum_{k=0}^{n-1} \frac{f^{(k)}(0)}{k!}\cdot x^k$ and $\displaystyle R_n(x) = f(x) - T_{n-1}(x)$.
So, if $\displaystyle n=5$ then $\displaystyle T_4(x) = 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}$ on $\displaystyle (-.1,.1)$
Then, $\displaystyle R_5(x) = \frac{f^{(5)}(y)}{5!}\cdot x^5 = \frac{e^y}{5!} \leq \frac{e^{.1}}{5!}\approx .003$
So it is 2 decimal points.