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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
Japanese Olympiad
parkjungmin   4
N 2 hours ago by parkjungmin
It's about the Japanese Olympiad

I can't solve it no matter how much I think about it.

If there are people who are good at math

Please help me.
4 replies
parkjungmin
Saturday at 6:51 PM
parkjungmin
2 hours ago
ISI UGB 2025 P3
SomeonecoolLovesMaths   11
N 3 hours ago by Levieee
Source: ISI UGB 2025 P3
Suppose $f : [0,1] \longrightarrow \mathbb{R}$ is differentiable with $f(0) = 0$. If $|f'(x) | \leq f(x)$ for all $x \in [0,1]$, then show that $f(x) = 0$ for all $x$.
11 replies
SomeonecoolLovesMaths
Yesterday at 11:32 AM
Levieee
3 hours ago
D1020 : Special functional equation
Dattier   4
N 3 hours ago by Dattier
Source: les dattes à Dattier
1) Are there any $(f,g) \in C(\mathbb R,\mathbb R_+)$ increasing with
$$\forall x \in \mathbb R, f(x)(\cos(x)+3/2)+g(x)(\sin(x)+3/2)=\exp(x)$$?

2) Are there any $(f,g) \in C(\mathbb R,\mathbb R_+)$ increasing with
$$\forall x \in \mathbb R, f(x)(\cos(x)+3/2)+g(x)(\sin(x)+3/2)=\exp(x/2)$$?
4 replies
Dattier
Apr 24, 2025
Dattier
3 hours ago
UC Berkeley Integration Bee 2025 Bracket Rounds
Silver08   24
N 4 hours ago by quasar_lord
Regular Round

Quarterfinals

Semifinals

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24 replies
Silver08
May 9, 2025
quasar_lord
4 hours ago
No more topics!
fractional binomial limit sum
Levieee   3
N Apr 20, 2025 by Levieee
this was given to me by a friend

$\lim_{n \to \infty} \sum_{k=1}^{n}{\frac{1}{\binom{n}{k}}}$

a nice solution using sandwich is
$\frac{1}{n}   + \frac{1}{n} + 1 + \frac{n-3}{\binom{n}{2}} \ge \frac{1}{n} +  \sum_{k=2}^{n-2}{\frac{1}{\binom{n}{k}}}+ \frac{1}{n} + 1 \ge \frac{1}{n} +  + \frac{1}{n} + 1$

therefore $\lim_{n \to \infty} \sum_{k=1}^{n}{\frac{1}{\binom{n}{k}}}$ = $1$

ALSO ANOTHER SOLUTION WHICH I WAS THINKING OF WITHOUT SANDWICH BUT I CANT COMPLETE WAS TO USE THE GAMMA FUNCTION

we know

$B(x, y) = \int_0^1 t^{x - 1} (1 - t)^{y - 1} \, dt$

$B(x, y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x + y)}$

and $\Gamma(n) = (n-1)!$ for integers,

$\frac{1}{\binom{n}{k}}$ = $\frac{k! (n-k)!}{n!}$

therefore from the gamma function we get

$ (n+1) \int_{0}^{1}  x^k (1-x)^{n-k} dx$ = $\frac{1}{\binom{n}{k}}$ = $\frac{k! (n-k)!}{n!}$
$\Rightarrow$ $\lim_{n \to \infty} (n+1) \int_{0}^{1} \sum_{k=1}^{n} x^k (1-x)^{n-k} dx$ $=\lim_{n \to \infty} \sum_{k=1}^{n}{\frac{1}{\binom{n}{k}}}$

somehow im supposed to show that

$\lim_{n \to \infty} (n+1) \int_{0}^{1} \sum_{k=1}^{n} x^k (1-x)^{n-k} dx$ $= 1$

all i could observe was if we do L'hopital (which i hate to do as much as you do)

i get $\frac{ \int_{0}^{1} \sum_{k=1}^{n} x^k (1-x)^{n-k} dx}{1/n+1}$

now since $x \in (0,1)$ , as $n \to \infty$ the $(1-x)^{n-k} \to 0$ which gets us the $\frac{0}{0}$ form therefore L'hopital came to my mind , which might be a completely wrong intuition, anyway what should i do to find that limit

:noo: :pilot:
3 replies
Levieee
Apr 19, 2025
Levieee
Apr 20, 2025
fractional binomial limit sum
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Levieee
236 posts
#1
Y by
this was given to me by a friend

$\lim_{n \to \infty} \sum_{k=1}^{n}{\frac{1}{\binom{n}{k}}}$

a nice solution using sandwich is
$\frac{1}{n}   + \frac{1}{n} + 1 + \frac{n-3}{\binom{n}{2}} \ge \frac{1}{n} +  \sum_{k=2}^{n-2}{\frac{1}{\binom{n}{k}}}+ \frac{1}{n} + 1 \ge \frac{1}{n} +  + \frac{1}{n} + 1$

therefore $\lim_{n \to \infty} \sum_{k=1}^{n}{\frac{1}{\binom{n}{k}}}$ = $1$

ALSO ANOTHER SOLUTION WHICH I WAS THINKING OF WITHOUT SANDWICH BUT I CANT COMPLETE WAS TO USE THE GAMMA FUNCTION

we know

$B(x, y) = \int_0^1 t^{x - 1} (1 - t)^{y - 1} \, dt$

$B(x, y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x + y)}$

and $\Gamma(n) = (n-1)!$ for integers,

$\frac{1}{\binom{n}{k}}$ = $\frac{k! (n-k)!}{n!}$

therefore from the gamma function we get

$ (n+1) \int_{0}^{1}  x^k (1-x)^{n-k} dx$ = $\frac{1}{\binom{n}{k}}$ = $\frac{k! (n-k)!}{n!}$
$\Rightarrow$ $\lim_{n \to \infty} (n+1) \int_{0}^{1} \sum_{k=1}^{n} x^k (1-x)^{n-k} dx$ $=\lim_{n \to \infty} \sum_{k=1}^{n}{\frac{1}{\binom{n}{k}}}$

somehow im supposed to show that

$\lim_{n \to \infty} (n+1) \int_{0}^{1} \sum_{k=1}^{n} x^k (1-x)^{n-k} dx$ $= 1$

all i could observe was if we do L'hopital (which i hate to do as much as you do)

i get $\frac{ \int_{0}^{1} \sum_{k=1}^{n} x^k (1-x)^{n-k} dx}{1/n+1}$

now since $x \in (0,1)$ , as $n \to \infty$ the $(1-x)^{n-k} \to 0$ which gets us the $\frac{0}{0}$ form therefore L'hopital came to my mind , which might be a completely wrong intuition, anyway what should i do to find that limit

:noo: :pilot:
This post has been edited 3 times. Last edited by Levieee, Apr 20, 2025, 10:14 AM
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KAME06
159 posts
#2
Y by
Check Putnam 1958 B1, you might find something useful there :D
This post has been edited 1 time. Last edited by KAME06, Apr 19, 2025, 11:51 PM
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anudeep
199 posts
#3
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btw the limit must be at least $2$ (and turns out it is precisely $2$) as the first and the last term are both $1$ i.e, $1/\binom{n}{0}+1/\binom{n}{n}=2$.
This post has been edited 1 time. Last edited by anudeep, Apr 20, 2025, 7:52 AM
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Levieee
236 posts
#4
Y by
anudeep wrote:
btw the limit must be at least $2$ (and turns out it is precisely $2$) as the first and the last term are both $1$ i.e, $1/\binom{n}{0}+1/\binom{n}{n}=2$.

oh my bad edited the question
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