Question

Can someone help me with PMI proofs?

Answer 1

With what?

Answer 2

Prove that this is true for all
\[n \in \mathbb{N}\]
\[1/2!+2/3!+...+n/(n+1)!=1-(1/(n+1)!)\]

Answer 3

Proof by induction
Base: True for n=1: 1/2! = 1 - 1/2!
1/2 = 1/2 True

Now assume true for N. Show it holds for N+1:
\[\sum_{n=1}^{N+1}\frac{n}{(n+1)!}= 1-\frac{1}{(N+2)!}\]
\[\sum_{n=1}^{N+1}\frac{n}{(n+1)!} = \sum_{n=1}^{N}\frac{n}{(n+1)!}+\frac{N+1}{(N+2)!}\]
By assumption our equality for the summation to N holds, so re-write the RHS as:
\[= 1 -\frac{1}{(N+1)!}+\frac{N+1}{(N+2)!}\]
This simplifies to
\[1-\frac{1}{(N+2)!}\]
so the equation is true for all N>=1