polynomial approximations put into sigma notation (just for fun) We're doing polynomial approximations in my Calc 2 class and because I've been told by virtually every person who has taken calc 2 before me that Taylor series is the most difficult part of the class, I've been trying to put as many of the Taylor polynomials I get into sigma notation. It's pretty fun actually! It blows my mind that you can put these long series into a short compact form that gives you any term you want so I've been doing quite a few today.
I'm wondering how you'd do $f(x) = \ln(2+x)$, however.
I calculated the following derivatives:
$$\begin{align*}
f^{(1)}(x) &= \frac{1}{(2+x)}\\
f^{(2)}(x) &= \frac{-1}{(2+x)^2}\\
f^{(3)}(x) &= \frac{4+2x}{(2+x)^3}
\end{align*}$$
and then solved for $f^*(0)$ for each derivative, which gave me the values:
$$\begin{align*}
f(0) &= \ln(2)\\
f^{(1)}(0) &= \frac{1}{2}\\
f^{(2)}(0) &= -\frac{1}{4}\\
f^{(3)}(0) &= \frac{1}{4}
\end{align*}$$
So with these Taylor coefficients, I can finally get the Taylor polynomial approximation (for $n=3$ in this case):
$$y = \ln(2) + \frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{24}$$
Now, how do I put this into sigma notation? I can see a pattern with the numbers and the alternating signs, but I don't see how I'd possibly get a $\ln(2)$ to pop out as the first term. 
Thanks in advance!
 A: You'll just have to handle the constant term specially. Since we know that $\log(x+1)$ has a nice Taylor series, we can write
$$\log(2+x)=\log\left(2(\frac{x}{2} + 1)\right)=\log 2 + \log\left(\frac{x}{2}+1\right) =\log2+\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k2^k}x^k$$
So one shouldn't expect the Taylor series to match up nicely going from the constant term to the rest.
A: You seem to be using the quotient rule to get the successive derivatives; this is obscuring what is going on here. Note that if $g(x) = f'(x) = \frac{1}{x+2} = (x+2)^{-1}$, after that you can just use the Chain Rule and you have:
$$\begin{align*}
g(x) &= (x+2)^{-1}\\
g'(x) &= -(x+2)^{-2}\\
g''(x) &= (-1)(-2)(x+2)^{-3}\\
g^{(3)}(x) &= (-1)(-2)(-3)(x+2)^{-4}\\
g^{(4)}(x) &= (-1)(-2)(-3)(-4)(x+2)^{-5}\\
&\vdots\\
g^{(n)}(x) &= (-1)^{n}(n!)(x+2)^{-n-1}\\
&\vdots
\end{align*}$$
and after a few steps one can see the general formula for $g^{(n)}(x)$, hence for $f^{(n+1)}(x)$, hence for $f^{(n+1)}(0)$.
That said, you can't really make the first term fit (that is, the constant term $f(0)=\ln(2)$). This is often the case: the first (or first few) terms don't quite follow an obvious rule, but after a while the terms "settle down" into a nice pattern (not always the case: sometimes we can't spot a pattern in any case). So we just write the first few terms separately, and then the rest. So here
$$ y = \ln(2) + \sum_{n=0}^{\infty}\left(\frac{(-1)^nn!}{2^{n+1}}\cdot\frac{x^{n+1}}{(n+1)!}\right) = \ln(2) + \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n2^n}x^n.$$
