Finding the Taylor polynomial of $f(x) = \frac{1}{x}$ with induction So I am asked to find the Taylor polynomial of $f(x) = \frac{1}{x}$ about the point $a=1$ for ever n$\in{N}$, and then use induction to justify the answer. 
I got the Taylor polynomial which was simple enough:
$$T_{n}(x)=\sum_{n=0}^{\infty} \frac{f^n(x)(a)}{n!}(x-a)^n$$
$$f(x) = 1-(x-1)+(x-1)^2-(x-1)^3+(x-1)^4+...$$
$$T_{n}(x)=\sum_{n=0}^{\infty} (-1)^n(x-1)^n$$
That wasn't too bad. How do I justify this with induction though? I am a little confused as to how I would start this.
I tried writing out the terms as such:
$$1-(x-1)+(x-1)^2-(x-1)^3+(x-1)^4+...+(-1)^k(x-1)^k = \frac{1}{x}$$
I am not really sure how to go about doing this though... Is my step valid? I would appreciate is someone could guide me in the right direction.
 A: We will prove that if $f:x \mapsto \dfrac{1}{x}$ then $f^{(n)}(x)=\dfrac{(-1)^n n!}{x^{n+1}}$, for every $n \in \mathbb{N}^*$.
For $n=1$ we have $f'(x)=\dfrac{-1}{x^2}=\dfrac{(-1)^1 \cdot 1!}{x^{1+1}}$.
If we asume tht $f^{n}(x)=\dfrac{(-1)^nn!}{n^{1+n}}$ then $$f^{(n+1)}(x)=\left[ \dfrac{(-1)^n \cdot n!}{x^{n+1}}\right]^´=(-1)^nn! \cdot \dfrac{-(n+1)}{x^{n+2}}=\dfrac{(-1)^{n+1}(n+1)!}{x^{(1+n)+1}},$$ hence by induction the result follows. 
Finally, the Taylor series of $x \mapsto 1/x$ about the point $a=1$ is $$T_n(x)=\sum_{k=0}^{\infty} \dfrac{\frac{(-1)^kk!}{1^{k+1}}}{k!}(x-1)^k=\sum_{k=0}^{\infty}(-1)^k(x-1)^k.$$
A: The known form of your Taylor polynomial is $P_n(x)=\sum^{n}_{k=0} a_k(x-1)^k$ where the coefficients satisfy $a_k = \frac{f^{(k)}(1)}{k!}$
You managed to find the coefficients $a_k = (-1)^k$. What the question is probably asking is to prove that this is correct using induction.
You would take the base case and show that $a_0=\frac{f(1)}{0!}=(-1)^0$
Then you would show that if $a_n=(-1)^n$ then it follows that $a_{n+1} = (-1)^{n+1}$
Doing these two steps would show that the formula holds for all $n$. 
