Induction Proof for a series expansion of a function I have done induction proofs of many different types, but trying to prove by induction that a derivative from the Taylor series expansion of a function has me stumped in terms of how to get the final step to work.
In most induction proofs we start with a base case of $n=1$, then assume $N=n$ works, then apply
$N+1$, and usually make use of the assumption to get a statement that is TRUE to prove it.
Here is the function:
$$f(x)= \ln (1+x).$$
And here is the equation we wish to apply Induction to:
$$f^{(n)}(x)= (-1)^{n-1}\frac{(n-1)!}{(1+x)^{n}}.$$
The approach I took was to apply $N+1$ where ever I see $n$, and the other thing that I did was
to just differentiate the equation which was straight forward, and then both were the same.
BUT i don't feel that this was a good proof because in Induction you usually make use of
step 2 the equation itself.
Hope someone can guide me or explain to me how one goes about doing an induction proof that
has derivatives involved of this sort.
 A: The way you've described induction is a bit muddled, and I blame the fact that proofs using induction are often written very informally.
The conceptually simplest form of the principle of mathematical induction is this:
Let $S$ be a set of natural numbers with the following two properties:


*

*$0\in S$

*For all $n\in \Bbb N$, if $n\in S$ then $n+1 \in S$.
Then $S = \Bbb N$.
This principle can be recast in logical terms (as a "schema" as follows):
Let $P$ be a propositional function in $\Bbb N$. That is, for each $n\in \Bbb N$, let $P(n)$ be a statement.
Suppose that the following hold:


*

*$P(0)$ is true.

*For each $n\in \Bbb N$, if $P(n)$ is true then $P(n+1)$ is true.
Then we can conclude that for each $n\in \Bbb N$, $P(n)$ is true.
In fact, it's easy to see that either of these approaches can be extended to allow any initial value, not just $0$, at the cost of having the proposition hold (or the number be an element of the set) only for numbers greater than some value (in your case $1$).
Using one of these somewhat more formal approaches to induction should help you keep a clear idea in your mind of how you need to approach the proof.
Let $P(n)$ be the statement that $$f^{(n)}(x) = (-1)^{n-1}\frac{(n-1)!}{(1+x)^n}$$. Prove first that $P(1)$ is true. Then prove that for each $n\in\Bbb N$, if $P(n)$ is true then so is $P(n+1)$.
A: I suppose you had no problem to check that the formula wich we have to proof is true for $n=1$. Wich is step 1 of our induction proof. So we go to step 2 : 
We show : if $ \ \ \ \ f^{(n)}(x)= (-1)^{n-1}\dfrac{(n-1)!}{(1+x)^{n}}$ $\ \ \ \ \ \ \ \ \ \ \ \ (1)$
is valid for $n$ then it is valid for $n+1$. Wich is step 2 of te induction proof.
Now knowing that $(1)$ holds we defferintiate both sides of the equation : 
$f^{(n+1)}(x)= -n \cdot (-1)^{n-1}\dfrac{(n-1)!}{(1+x)^{n+1}}= (-1)^{n}\dfrac{(n)!}{(1+x)^{n+1}}$ 
As you can see $(1)$ is valid for $n+1$. 
Indeed as you claimed "in induction we make use of step 2 the equation itself" we actually used $(1)$  wich is  "the equation" and manipulated it to show that it also holds for $n+1$. As far as i understood you made the same steps. I hope this helps to convince you that you were right.
