factorial proof by induction

Question

So I have an induction proof that, for some reason, doesn't work after a certain point when I keep trying it. Likely I'm not adding the next term correctly but I don't know for sure. Question is: If n∈N, then 1/(2!)+2/(3)!+3/(4)!+...+n/(n+1)!=1-(1/(n+1)!) P(1) works. Then assume P(k). Then P(k+1) is where, as I understand, you add the next term onto the left so LHS gets an additional "(k+1)/(k+2)!"... then the right gets that tacked on as well? Or do you just do 1-(1/(k+1+1))! and that's it (as in replace the (k+1)! in the lone denominator part with a (k+1+1)!? Algebraically I can never get the RHS to equal the left if I do that. Any thoughts would be great.

Answer 1

Your RHS is $1-\frac{1}{(k+1)!} + \frac{k+1}{(k+2)!}$.
But $(k+2)! = (k+2) \times (k+1)!$, so let us factor out: $1 + \frac{1}{(k+1)!} (\frac{k+1}{k+2}-1)$.
Then $\frac{k+1}{k+2} - 1 = - \frac{1}{k+2}$, and we get the new RHS: $1 - \frac{1}{(k+1)!} \frac{1}{k+2} = 1 - \frac{1}{(k+2)!}$.

QED

Answer 2

If you can reduce $P(k+1)$ to $P(k)$ then that is sufficient You should try and show $$P(k+1) = P(k) + (1-\frac{1}{(k+2)!})$$

Answer 3

It looks very simple to me. You assume that, for some number k, $\frac{1}{2}+ \cdot\cdot\cdot+ \frac{k}{(k+1)!}= 1- \frac{1}{(k+1)!}$ and you want to use that to prove that $\frac{1}{2}+ \cdot\cdot\cdot+ \frac{k}{(k+1)!}+ \frac{k+1}{(k+2)/1}= 1- \frac{1}{k+ 2}$.

Like any "sum" induction problem, the key is to replace the first part of that sum using the "induction hypothesis": $\frac{1}{2}+ \cdot\cdot\cdot+ \frac{k}{(k+1)!}+ \frac{k+ 1}{(k+ 2)!}= 1- \frac{1}{(k+1)!}+ \frac{k+1}{(k+ 2)!}= 1- (\frac{1}{(k+1)!}- \frac{k+1}{(k+2)!}$

You can, I hope, recognize that we can get the "common denominator" to subtract the last two fractions just by multiplying the numerator and denominator of the first fraction by k+ 2: $1- (\frac{1}{(k+1)!}- \frac{k+1}{(k+2)!}= 1- (\frac{k+2}{(k+2)!}- \frac{k+1}{(k+2)!}$ $= 1- \frac{(k+2)- (k+1)}{(k+ 2)!}= 1- \frac{1}{(k+2)!}$