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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
Convergent series with weight becomes divergent
P_Fazioli   0
an hour ago
Initially, my problem was : is it true that if we fix $(b_n)$ positive such that $b_n\underset{{n}\longrightarrow{+\infty}}\longrightarrow +\infty$, then there exists $(a_n)$ positive such that $\displaystyle\sum_{n\geq 0}a_n$ converges and $\displaystyle\sum_{n\geq 0}a_nb_n$ diverges ?

Thinking about the continuous case : if $g:\mathbb{R}_+\longrightarrow\mathbb{R}$ is continuous, positive with $g(x)\underset{{x}\longrightarrow{+\infty}}\longrightarrow +\infty$, does $f$ continuous and positive exist on $\mathbb{R}_+$ such that $\displaystyle\int_0^{+\infty}f(x)\text{d}x$ converges and $\displaystyle\int_0^{+\infty}f(x)g(x)\text{d}x$ diverges ?

To the last question, the answer seems to be yes if $g$ is in the $\mathcal{C}^1$ class, increasing : I chose $f=\dfrac{g'}{g^2}$. With this idea, I had the idea to define $a_n=\dfrac{b_{n+1}-b_n}{b_n^2}$ but it is not clear that it is ok, even if $(b_n)$ is increasing.

Now I have some questions !

1) The main problem : is it true that if we fix $(b_n)$ positive such that $b_n\underset{{n}\longrightarrow{+\infty}}\longrightarrow +\infty$, then there exists $(a_n)$ positive such that $\displaystyle\sum_{n\geq 0}a_n$ converges and $\displaystyle\sum_{n\geq 0}a_nb_n$ diverges ? And if $(b_n)$ is increasing ?
2) is it true that if we fix $(b_n)$ positive increasing such that $b_n\underset{{n}\longrightarrow{+\infty}}\longrightarrow +\infty$
and $\frac{b_{n+1}}{b_n}\underset{{n}\longrightarrow{+\infty}}\longrightarrow 1$, then $\displaystyle\sum_{n\geq 0}\left(\frac{b_{n+1}}{b_n}-1\right)$ diverges ?
3) is it true that if we fix $(b_n)$ positive increasing such that $b_n\underset{{n}\longrightarrow{+\infty}}\longrightarrow +\infty$
and $\frac{b_{n+1}}{b_n}\underset{{n}\longrightarrow{+\infty}}\longrightarrow 1$, then $\displaystyle\sum_{n\geq 0}\frac{b_{n+1}-b_n}{b_n^2}$ converges ?
4) if $(b_n)$ is positive increasing and such that $b_n\underset{{n}\longrightarrow{+\infty}}\longrightarrow +\infty$
and $\frac{b_{n+1}}{b_n}$ does not converge to $1$, can $\displaystyle\sum_{n\geq 0}\frac{b_{n+1}-b_n}{b_n^2}$ diverge ?
5) for the continuous case, is it true if we suppose $g$ only to be continuous ?

0 replies
P_Fazioli
an hour ago
0 replies
D1026 : An equivalent
Dattier   4
N an hour ago by 3ch03s
Source: les dattes à Dattier
Let $u_0=1$ and $\forall n \in \mathbb N, u_{2n+1}=\ln(1+u_{2n}), u_{2n+2}=\sin(u_{2n+1})$.

Find an equivalent of $u_n$.
4 replies
Dattier
Saturday at 1:39 PM
3ch03s
an hour ago
Rolles theorem
sasu1ke   4
N Yesterday at 6:13 PM by sasu1ke

Let \( f: \mathbb{R} \to \mathbb{R} \) be a twice differentiable function such that
\[
f(0) = 2, \quad f'(0) = -2, \quad \text{and} \quad f(1) = 1.
\]Prove that there exists a point \( \xi \in (0, 1) \) such that
\[
f(\xi) \cdot f'(\xi) + f''(\xi) = 0.
\]

4 replies
sasu1ke
Saturday at 9:00 PM
sasu1ke
Yesterday at 6:13 PM
Putnam 1947 B1
sqrtX   3
N Yesterday at 5:38 PM by Levieee
Source: Putnam 1947
Let $f(x)$ be a function such that $f(1)=1$ and for $x \geq 1$
$$f'(x)= \frac{1}{x^2 +f(x)^{2}}.$$Prove that
$$\lim_{x\to \infty} f(x)$$exists and is less than $1+ \frac{\pi}{4}.$
3 replies
sqrtX
Apr 3, 2022
Levieee
Yesterday at 5:38 PM
No more topics!
A Typical Determinant Problem
Saucepan_man02   4
N Apr 26, 2025 by loup blanc
Source: Romania Contest, 2010
Let $A, B \in M_n(\mathbb R)$ with $B^2 = O_n$. Show that: $\det(AB+BA+I_n) \ge 0$.
4 replies
Saucepan_man02
Apr 17, 2025
loup blanc
Apr 26, 2025
A Typical Determinant Problem
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G H BBookmark kLocked kLocked NReply
Source: Romania Contest, 2010
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Saucepan_man02
1331 posts
#1
Y by
Let $A, B \in M_n(\mathbb R)$ with $B^2 = O_n$. Show that: $\det(AB+BA+I_n) \ge 0$.
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loup blanc
3591 posts
#2 • 1 Y
Y by Tip_pay
$B$ is similar, over $\mathbb{R}$, to $diag(0_{n-2r},J_1,\cdots,J_r)$, where $J_i=J=\begin{pmatrix}0&1\\0&0\end{pmatrix}$.
Then we may assume that $B$ is in Jordan form.
$\bullet$ Here, we consider only the case $n=2r$.
Then $C=AB+BA$ is a $r\times r$ block-matrix where the entries are $2\times 2$ matrices in the form $C_{i,j}=p_{i,j}I_2+q_{i,j}J$.
Let $t\in\mathbb{R}$; since the $C_{i,j}$ pairwise commute, $\det(AB+BA+tI_n)=\det(u_0I_2+u_1J+u_2J^2+\cdots)$, where
$\sum_i u_iJ^i$ is obtained by developing the determinant by considering the $C_{i,j}$ as elements.
Finally $\det(AB+BA+tI)=\det(u_0I_2)={u_0}^2$, where $u_0$ is a polynomial of degree $r$ in $t$. $\square$
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loup blanc
3591 posts
#3
Y by
$\bullet$ The general case -the sequel of post #2-
We may assume (see above) that $B=diag(0_{n-2r},K_{2r})$ and $A=\begin{pmatrix}P_{n-2r}&Q\\R&S_{2r}\end{pmatrix}$.
Then $AB+BA+I=\begin{pmatrix}I&QK\\KR&SK+KS+I\end{pmatrix}$ and
$\det(AB+BA+I)=\det(SK+KS+I-KTK)$, where $T=RQ\in M_{2r}$.
Note that $SK+KS$ is the matrix $C$ in post #2 and that $KTK$ is a $r\times r$ block-matrix where the entries are
$2\times 2$ matrices in the form $(KTK)_{i,j}=q_{i,j}J$ (then nilpotent).
As in post #2, $\det(SK+KS+I-KTK)=\det(v_0I_2+v_1J+v_2J^2+\cdots)=\det(v_0 I_2)$ but here $v_0$ does not depend on
the matrix $KTK$, that is, $\det(AB+BA+I)=\det(SK+KS+I)=\det(v_0 I_2)={v_0}^2$. $\square$
PS. If $t>0$, then $\det(AB+BA+tI)=t^n\det(A'B+BA'+I)$, where $A'=\dfrac{1}{t}A$.
Then $\det(AB+BA+tI)\geq 0$ and, by continuity in $0^+$, $\det(AB+BA)\geq 0$.
This post has been edited 1 time. Last edited by loup blanc, Apr 17, 2025, 8:59 PM
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removablesingularity
568 posts
#4 • 3 Y
Y by loup blanc, GreenKeeper, RobertRogo
Attempt for shorter answer : Using fact $B^2 = 0$ we have $AB+BA + I = AB^2A + AB + BA + I = AB.BA + AB + BA + I = \left(AB+I\right).\left(BA+I\right)$.
We have $\det\left(AB+BA+I\right) =  \det\left(AB+I\right) \det\left(BA+I\right).$ Meanwhile, recall that $\det \left(I+AB\right)=\det \left(I+BA\right)$.Proceed.
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loup blanc
3591 posts
#5
Y by
@ removablesingul ; well done.
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