I have a two quick questions on mathematical induction. I'm new to it, so my answers seem a little shaky. I wanted to see what some of the you guys thought.

Question 1

We're asked to prove using mathematical induction that for all natural numbers $\displaystyle n$,

$\displaystyle \frac{1}{2!} + \frac{2}{3!}+\dots+\frac{n}{(n+1)!} = 1 - \frac{1}{(n+1)!}$

Proof (?):

Let $\displaystyle k\in\mathbb{N}$

1. Base case: $\displaystyle k=3$:

$\displaystyle \begin{align*} \frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!}&=\frac{1}{2}+\frac{2}{6}+\frac{3}{24} \\&=\frac{23}{24}\\&=1-\frac{1}{(3+1)!}\end{align*}$

CHECK

2. Assume $\displaystyle \forall k\in\mathbb{N}$,

$\displaystyle \frac{1}{2!} + \frac{2}{3!}+\dots+\frac{k}{(k+1)!} = 1 - \frac{1}{(k+1)!}$.

Then,

$\displaystyle \begin{align*}\frac{1}{2!} + \frac{2}{3!}+\dots+\frac{k}{(k+1)!} + \frac{k+1}{(k+2)} &= 1 - \frac{1}{(k+1)!} + \frac{1}{(k+2)!}\\&=1 - \frac{1}{(k+1)!}\left(\frac{k+2}{k+2}\right) + \frac{k+1}{(k+2)!}\\&=1 - \frac{k+2}{(k+2)!} + \frac{k+1}{(k+2)!}\\&=1-\frac{1}{(k+2)!}\end{align*}$

3. Thus, by P.M.I. blah blah QED.

Is this legit? It just seems strange to me since I'm working completely on the right side. Yet any other way I try to attempt seems to lead me down some crazy algebra path (I could very well be doing it wrong).

Question 2

UsegeneralizedPMI principles of mathematical induction to prove

$\displaystyle n!>3n\quad\text{for}\quad n\geq 4$

Proof(?):

First, we know that since we're taking a factorial that n must be an integer (this was not originally given to us, so I guess it's implied?).

Let $\displaystyle k\in\mathbb{Z},\quad k\geq 4$

(I guess that since k is greater than 4, I might rotate that Z pi radians and define k as a natural number?)

1. Base case: k = 4

$\displaystyle 4!=24>3\cdot4 = 12$

Good.

2. Mathematical induction:

Assume that for some integer $\displaystyle k\geq 4$ that $\displaystyle k!>3k$

Note,

$\displaystyle 3(k+1)=3k+3>3k$

Also note,

$\displaystyle k!\geq k(3!)>3(k+1)=3k+3$

And

$\displaystyle (k+1)!>k!$

(since k is always positive.)

So,

$\displaystyle (k+1)!>k!>3(k+1)>3k$

3. by general PMI, i'm done?

Again, this feels a bit shaky to me. Perhaps because I dealt with so many pieces individually before putting it all back together.

So again, my question: Am I using mathematical induction properly here? Do my proofs prove what they are supposed to?