About the gaps between consecutive primes If my question is inappropriate, non-uniform or ill-shaped; please comment (or edit maybe).
Let $t(n)$ be the nth triangle number, so $t(n) = {(n)(n+1) \over 2}$.
And let $p(n)$ be the nth prime number.
I want to show that $t(n) > p(n)$ for all $n>2$. I tried this:
Let $t(n) > p(n)$ and $p(n+1) > t(n+1)$. If I can get a contradiction, I'm done.
With this I have $p(n+1) - p(n) > t(n+1) - t(n) = n+1$.
I tried to show the last part and couldn't. I checked with computer and if my code is correct, this last part is probably correct.
Then, I looked at Wikipedia's Prime Gap article and get suspicious.
Thanks any help for proving last part and checking first part.
 A: Rosser and Schoenfeld (1962), page 69, Theorem 3, Formula (3.11), 
$$   \mbox{for} \; \;  n \geq 20, \; \; \; \; p(n) < n (\log n + \log \log n - \frac{1}{2})    $$
Here we go, as corollary, (3.13)
$$   \mbox{for} \; \;  n \geq 6, \; \; \; \; p(n) < n (\log n + \log \log n )    $$
This is quite a good bound, as (3.12)
$$   \mbox{for} \; \;  n \geq 1, \; \; \; \; p(n) > n \log n     $$
A: It is known that $p_n \lt n (\ln n + \ln \ln n)$ for $n \ge 6$, 
and also $(\ln n + \ln \ln n) \le (n+1)/2$ for $n \ge 6$. So you have the result for $n \ge 6$ and you can check the rest by hand. 
The result I quote is a bit complicated to obtain, though. 
A: You're trying to show that
$$\pi(n(n+1)/2)\ge n$$
where $\pi(x)$ is the usual prime-counting function.  Since $\pi(x)$ is nondecreasing, it suffices to let $u=n^2/2$, find where
$$\pi(x)\gt\sqrt{2x}$$
for all $x\ge u$, and then check any smaller cases that might remain.  There ought to be some fairly simple proof that finds a fairly small $u$ past which the desired inequality holds. (The Prime Number Theorem, of course, says that $\pi(x)\approx x/\log x$ for large $x$, which is quite a bit larger than $\sqrt{2x}$.  Since there's so much room to spare, there should be a proof that doesn't rely on PNT.)
