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for questions about sequences and series, e.g. convergence, closed form expressions, etc. Note that there is a different tag for spectral sequences, and also note that MathOverflow is not for homework. Please consider consulting the online encyclopedia for integer sequences, if you are trying to identify a given sequence that you have found in your research.
10
votes
Accepted
Is the factorization of $a_m-a_n$ affected by the fact that $\Sigma \frac{1}{a_k}<+\infty$?
No, this is false. Define $a_1=1$, and for all $k \geq 2$ let $a_k = \big\lfloor \frac{k}{2}\big\rfloor^2$. Note that $\sum_{k=1}^\infty \frac{1}{a_k}$ converges since it is equal to $1+2\sum_{k=1}^ …
4
votes
Elementary proof of the equidistribution theorem
Not sure if this qualifies, but there is a short proof using Fourier Analysis. No hardcore stuff, just Fejér's Theorem. See Chapter 3 of Körner's Fourier Analysis book (there are 110 chapters in th …
5
votes
2
answers
1k
views
Degree sequences of multigraphs with bounded multiplicity
I got to thinking about this problem while sifting through the math puzzles for dinner thread. There's a fun puzzle by rgrig which asks the guests to prove that when they came to dinner two of them s …
10
votes
Accepted
A term for sequences whose mean is defined?
The standard term is Cesàro summable, named after Ernesto Cesàro. Note that a convergent sequence is also Cesàro summable (with the same limit), but the converse does not always hold.
Edit. I real …
1
vote
Two Equal Series?
This is just Sergei Ivanov's comment finished with Pietro Majer's answer. If all the terms are real and positive then rearrange both $a:=(a_n)$ and $b:=(b_n)$ so they are non-increasing. If $a \neq b …
16
votes
How many integers are of the form $n/d(n)$, where $d(n)$ is the number of divisors of $n$?
Here's an elementary argument proving that the set of numbers $I$ that fail the conjecture is infinite.
Claim. $p^{17} \in I$ for all primes $p > 19$.
Proof. Suppose that $p^{17}=\frac{N}{d(N)}$ …
2
votes
Probabilities in a riddle involving axiom of choice
I also like this version of the riddle. To answer the actual question though, I would say that it is not possible to guess incorrectly with probability only $\frac{1}{N}$, even for $N=2$. In order f …