Recent content by dot.hack

The idea is almost exactly the same as equality. Starting from where you left off you have \sum_{i=1}^{n}\frac{1}{i^2} + \frac{1}{n+1}≤ 2- \frac{1}{n+1} See what happens if you combine like terms and use what you already know about \sum_{i=1}^{n}\frac{1}{i^2} to help you out.

First off be careful using F(t) as an arbitrary function as F(t) generally denotes Force as a function of time. Your issue here is that x is being used for two different purposes on the LHS and RHS. On the RHS x denotes the distance traveled due to time (x(t)) and c is the initial starting...

As best as I can figure out (don't know anything about statistical mechanics) this is just a simple derivative. If the LHS reads -1/Z(beta) dZ(beta)/d beta (which I imagine it should) then this is just an application of the chain rule. To make it easier to think about consider LHS and RHS as...

Actually, the last equation I wrote have n!/2 thanks

I'm a little confused because the original equation was an inequality (n! > n^2) not an equality (n! = n^2). Assuming you mean the original n! > n^2 it is just a matter of arranging the terms we already know and multiplying. We know that k! > k*(k-1) (by the definition of k!) and we showed that...

Yea, absolutely. I'm just trying to get as much practice as I can so thank you for the advice.

I'm an analysis student as well and I'm running around these boards answering proofs as best as I know how. I'm not sure if the last step in this proof is rigorous but i'll try anyways. Starting from where you left off, divide both sides by (k+1) this gives k! > k+1. We know that k! is...

Homework Statement Prove that n^n > 2^n * n! when n > 6 using the Binomial theorem. I just proved the Binomial theorem using induction which was not that difficult but in applying what I learned through it's proof I am having difficulty. Homework Equations Binomial theorem = (x+y)^n =...

This is coming from another Real Analysis student so I may be making a mistake, but the practice for me is definitely worth it so please point out if I did something wrong. Taking k=1 we see that the equality holds, so we assume the equality holds for k and attempt to prove for k+1. We then...