
Re: Induction proof
Posted:
Aug 20, 2006 2:12 PM


On 20 Aug 2006 10:29:37 0700, <emailtgs@gmail.com> wrote in <news:1156094977.426428.218100@i42g2000cwa.googlegroups.com> in alt.math.undergrad:
> okay, my question is, just how do you decide to go from here ... to > here.
> < (1  1/n) + 1/(n+1)^2 (by induction hypotheses) > < (1  1/n) + 1/(n*(n+1))
> I know you've made a note of what you did but I don't > understand how you can do that.
The 'how' is trivial. 1/(n+1)^2 < 1/(n*(n+1)), so of course (1  1/n) + 1/(n+1)^2 < (1  1/n) + 1/(n*(n+1)): you're just adding 1  1/n to both sides.
To see why 1/(n+1)^2 < 1/(n*(n+1)), if it isn't already obvious, observe that (n+1)^2 > n*(n+1). In even more detail: n+1 > n, so multiplying both sides by n+1 yields (n+1)^2 > n*(n+1) if n+1 > 0, i.e., if n > 1. Here you know that n > 1, so certainly n > 1.
The 'why' is so that you can split 1/(n*(n+1)) into partial fractions, one of which is 1/n that cancels the 1/n that you already have.
[...]
Brian

