Finding the limit $\lim_{n\to\infty}{\sum_{i=0}^n \frac{(-1)^i}{i!}f^{(i)}(1)}$. Let
$$f(x)=\frac{1}{x^2+3x+2}$$
I must find $$\lim_{n\to\infty}\sum_{i=0}^n \frac{(-1)^i}{i!}f^{(i)}(1)$$
How should I proceed?
 A: Let $f(x)=\frac{1}{x^2+3x+2}$.  Then, using partial fraction expansion, we find that
$$f(x)=\frac{1}{x+1}-\frac{1}{x+2}$$
The $i$'th derivative of $f(x)$ is given by 
$$f^{(i)}(x)=(-1)^i\, i!\left(\frac{1}{(x+1)^{i+1}}-\frac{1}{(x+2)^{i+1}}\right)$$
Therefore, we find that
$$\begin{align}
\sum_{i=0}^n \frac{(-1)^i}{i!}f^{(i)}(1)&=\sum_{i=0}^n\left(\frac{1}{2^{i+1}}-\frac{1}{3^{i+1}}\right)\\\\
&=\left(1-\frac{1}{2^{n+1}}\right)-\left(\frac12-\frac{1/2}{3^{n+1}}\right)\\\\
&\to \frac12\,\,\text{as}\,\,n\to \infty
\end{align}$$
A: I figured it out, here is the solution:
by decomposing f(x) and applying some derivates you get (note that I have skipped induction)
 $$f^{(n)}(1)=\frac{(-1)^nn!}{2^{n+1}}-\frac{(-1)^nn!}{3^{n+1}}$$
after some simple algebraic manipulation the limit becomes
$$\lim_{n->\infty}{\sum^n_{k=0}{\frac{1}{2^{k+1}}-\frac{1}{3^{k+1}}}}$$
those are 2 geometric series with ratios of q = $\frac{1}{2}$ and $\frac{1}{3}$
using the formula for the geometric series
$$\sum^n_{k=0}{b_k}=b_1\frac{q^n-1}{q-1}$$
we can rewrite the limit as
$$\lim_{n->\infty}{1-\frac{1}{2^{n+1}}} - \lim_{n->\infty}{\frac{1}{2}(1-\frac{1}{3^n})} = \frac{1}{2}$$
