# Finding the limit $\lim_{n\to\infty}{\sum_{i=0}^n \frac{(-1)^i}{i!}f^{(i)}(1)}$. [closed]

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?

• I'm sorry, where's the $n$? – KR136 Apr 5 '16 at 17:15
• What have you tried so far? Do you know any formulas that involve sums over the derivatives of a function, suitably scaled? – Steven Stadnicki Apr 5 '16 at 17:18
• I fixed it , it's the upper argument of the sum – oren revenge Apr 5 '16 at 17:19

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}

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

$$\lim_{n->\infty}{1-\frac{1}{2^{n+1}}} - \lim_{n->\infty}{\frac{1}{2}(1-\frac{1}{3^n})} = \frac{1}{2}$$

• Not sure how this became closed when you actually provided a solution. +1. – Mark Viola Apr 5 '16 at 19:22