be positive integers. Bob has 
stones with weights 
.
be an integer greater than 
and let 
be the number of non empty subsets 
of 
with the property that the average of the elements of 
is always even.
be a function verifying :
Assume that 
.
,
Assume that 
,
,
.
prove that :
be a prime number. Show that : ![\[ \binom{np}{0} - \binom{np}{p} + \binom{np}{2p} - \binom{np}{3p} + \ldots + (-1)^n\binom{np}{np} \equiv 0 \pmod{p^{\max(3,n)}} \]](http://latex.artofproblemsolving.com/6/d/6/6d6b4c7b5c5c760b6ef3670896a8b130aa1ce4db.png)
be a group and let 
be two distinct elements. Let 
.![\[ E = \{a^kb^{n-k} | k \in \{0, 1, \ldots, n\}\} \]](http://latex.artofproblemsolving.com/6/3/f/63f8b31fa845bdabfd2fe44b3559180b805eecbb.png)
Show that the subgroup of 
,
,
and 
,![\[ w = x_1 x_2 \ldots x_n \text{ where each } x_i \in \{a,b\} \]](http://latex.artofproblemsolving.com/2/7/a/27af4bc7835db198f7ba81488bccd173645ceec6.png)
![$f,g : [0,1] \to \mathbb{R}_{>0}$](http://latex.artofproblemsolving.com/3/3/d/33da9c7226a1d4478c4c90997d4178f8d7ee6915.png)
be two continuous and strictly positive functions. We assume that 
.![$I \subset [0,1]$](http://latex.artofproblemsolving.com/0/0/8/008310cd7e1f41303a88a7173c0b9cd719f98be8.png)
such that 
![\[
f(x)=
\begin{cases}
\displaystyle\int_{x}^{\infty}\! u^{-1}e^{-u}\,du, & x>0,\\[6pt]
0, & x\le 0
\end{cases}
\]](http://latex.artofproblemsolving.com/e/a/7/ea723fe708579bd6e3e140f3ad705f7207bd2097.png)
is a probability density. Determine its characteristic function and give its power–series expansion.
![$\left[ {{\left( \prod\limits_{k=1}^{n}{\frac{2k}{2k-1}} \right)}^{2}}-\pi n \right]$](http://latex.artofproblemsolving.com/5/9/7/597b891775ad35de89721c1207ad1164cf8e4b82.png)
,![$[a]$](http://latex.artofproblemsolving.com/d/e/3/de38c39460667f504b7b57c3daa08d8d13cdaa9c.png)
it is the integer part of the number 
teams lasted for 
days, as follows. On each day, every team played one game against another team, with one team winning and one team losing in each of the 
,
,
,
be chosen uniformly at random in the range ![$[0,1]$](http://latex.artofproblemsolving.com/e/8/6/e861e10e1c19918756b9c8b7717684593c63aeb8.png)
.![\[\mathbb{E}(\min(X)\cdot \max(X)) = \dfrac{1}{n+2}.\]](http://latex.artofproblemsolving.com/b/8/4/b84d5581f617beda3c068283978a8c49e93613e0.png)
I'm pretty sure this requires calculus (although would love a non-calculus based proof), hence why I posted in College Math.
is a random variable that takes on only nonnegative integer values, with ![$E[X]=1,$](http://latex.artofproblemsolving.com/e/8/2/e826f5b79a2adeacdaccd3884658e94a3b1f8e22.png)
![$E[X^2]=2,$](http://latex.artofproblemsolving.com/5/9/1/5915ee4a36f8615ddc8075a197f3b44a902936c6.png)
and ![$E[X^3]=5.$](http://latex.artofproblemsolving.com/d/1/1/d11c925af9f188f0140a89a0fbd6c83e1dd4fc12.png)
(Here ![$E[Y]$](http://latex.artofproblemsolving.com/a/2/9/a29ea940a5d54ad2a5d668a596f9db0c88fa1123.png)
denotes the expectation of the random variable 
)
a subgroup of 
a character of a 
.
.
Y by KAME06
Could anybody possibly prove that the limit of 
is 
, maybe even with rather elementary means? As usual, 
denotes the 
-th prime number. The problem of that limit came up in my partial solution of this problem: https://artofproblemsolving.com/community/c7h3495516.
Thank you for your efforts.
is 
, maybe even with rather elementary means? As usual, 
denotes the 
-th prime number. The problem of that limit came up in my partial solution of this problem: https://artofproblemsolving.com/community/c7h3495516.Thank you for your efforts.
and this can be done using the Prime Number Theorem: 
is the "official" statement of the prime number theorem, with the prime-counting function 
.
Although I admit that the second statement expresses the same "idea" as the first, I cannot see that it follows immediately from the first. As I have never tried to understand the proof of the first statement, I never bothered to look for a proof of the second, but I assume that it still takes some real work to make the step from the first to the second statement. Do you agree?
,
Since 
we get 
i.e. 
Taking logarithm yields 
This implies 
so 
Since 
it follows that 
which combined with 
then 
and 
(recall that 
means 
)
and 
and 
does not exist, which illustrates that "additively" the sequences might be very different.
so
which is entirely new to me. And your other remarks were also very helpful!
