All Questions
8
questions
2
votes
2
answers
141
views
How to choose which stronger claim to prove when proving $\sum_{i=1}^n \frac{1}{i^2} \le 2$?
I am studying an inductive proof of the inequality $\sum_{i=1}^n \frac{1}{i^2} \le 2$. In the proof, it was decided to prove the stronger claim $\sum_{i=1}^n \frac{1}{i^2} \le 2-\frac{1}{n}$, as this ...
- 3,019
6
votes
2
answers
201
views
How prove this inequality $\sum_{i=1}^{n}\sum_{j=1}^{n}\text{lcm}(i,j)\le\frac{n^3}{5}(n+4)$?
Let $n$ be postive integers. Show that
$$\sum_{i=1}^{n}\sum_{j=1}^{n}[i,j]\le\dfrac{n^3}{5}(n+4)\,,$$ where $[a,b]$ denote the least common multiple of $a$ and $b$.
$S_1=1=\dfrac{1^3}{5}(4+1)=1$
...
- 93.6k
1
vote
1
answer
99
views
How to prove this inequality: $\sum_{i=1}^{n}\sum_{j=1}^{n}\text{lcm}(i,j)\ge\frac{7}{8}n^3$?
Let $n$ be postive integers. Show that
$$\sum_{i=1}^{n}\sum_{j=1}^{n}[i,j]\ge\dfrac{7}{8}n^3\,,$$ where $[a,b]$ denote the least common multiple of $a$ and $b$.
For $n=1,2,3 $, it is clear. How ...
- 93.6k
5
votes
3
answers
65
views
Prove that $\sum_{i=1}^{n} \frac{n+i}{i+1} \le 1 + n(n-1) \ \forall n \in \Bbb{N}$ (without calculus)
I'm trying to prove the following proposition (I'm not supposed to use calculus):
$$\sum_{i=1}^{n} \frac{n+i}{i+1} \le 1 + n(n-1) \ \forall n \in \Bbb{N}$$
(I'm assuming that $0 \notin \Bbb{N}$)
...
- 751
1
vote
0
answers
46
views
An inequality involving Möbius function [duplicate]
For any positive integer $n$ show the inequality holds :
$$\left|\sum_{i=1}^{n}\frac{\mu(i)}{i}\right|\le 1$$
I tried induction. when $\mu(n+1)=0$ it is trivial. But what if $\mu(n+1)\ne 0$? I am ...
- 5,636
2
votes
1
answer
56
views
Is $\frac{k!!}{j!!(k-j)!!}\leq\frac{k!}{j!(k-j)!}$ for all integers $j$ and $k$, where $0\leq j\leq k$?
For all integers $j$ and $k$, where $0\leq j\leq k$, is the inequality
$\frac{k!!}{j!!(k-j)!!}\leq\frac{k!}{j!(k-j)!}$
true?
I have a feeling that it is and it would be helpful to me if it is, ...
- 51
4
votes
2
answers
647
views
Simple Divisor Summation Inequality (with Moebius function)
Show that
$$\left| \sum_{k=1}^{n} \frac {\mu(k)}{k} \right| \le 1 $$
where $\mu$ is Moebius function and n is a positive integer.
The hard thing here is that the sum is not directly divisor sum; it's ...
- 1,502
3
votes
1
answer
439
views
Floor Inequalities
Proving the integrality of an fractions of factorials can be done through De Polignac formula for the exponent of factorials, reducing the question to an floored inequality.
Some of those inequalities ...
- 453