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Results tagged with sequences-and-series
Search options answers only
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user 11919
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.
77
votes
Accepted
A mysterious connection between primes and $\pi$
Here is a proof of the conjecture. We shall use Hecke's theorem that the angles of the lattice points $(x_p,y_p)$ are asymptotically equidistributed in $[\pi/4,\pi/2]$, cf. this MO post.
Let $t_p\in[ …
- 105k
45
votes
Accepted
Is this a rational function?
The function
$$ f(z)=\sum_{n=1}^{\infty} \frac{z^n}{2^n-1} $$
defines a holomorphic function for $|z|<2$, and it satisfies
$$ f(2z) = f(z)+\frac{z}{1-z} $$
for $|z|<1$. Based on this identity, it is e …
- 105k
37
votes
Accepted
Identity for an infinite product
The identity is true. To see this, we consider
$$f(q):=q\prod_{k\geq 1}(1-q^k)^{24},\qquad |q|<1.$$
With this notation,
$$\frac{qf(q)}{f(q^2)}=\prod_{\substack{k\geq 1\\\text{$k$ odd}}}(1-q^k)^{24}\qq …
- 105k
32
votes
Alternating sum of square roots of binomial coefficients
The following proof of $c_n>0$ is based on Gjergji Zaimi's response to this related question. In particular the positive answer follows for that question, too. Moreover, the argument below should also …
- 105k
27
votes
Accepted
An infinite series that converges to $\frac{\sqrt{3}\pi}{24}$
Here is an elementary proof. We rewrite the series as
$$\frac{1}{4}\int_0^1\frac{1-x^4}{1-x^6}\,dx=\frac{1}{8}\int_0^1\frac{dx}{1-x+x^2}+\frac{1}{8}\int_0^1\frac{dx}{1+x+x^2}.$$
It is straightforward …
- 105k
24
votes
Is $ \sum\limits_{n=0}^\infty x^n / \sqrt{n!} $ positive?
The affirmative answer follows from my response to this related question.
EDIT. Noam Elkies gave a nicer and more general argument here.
- 105k
23
votes
Accepted
Evaluating an infinite sum related to $\sinh$
Here is a proof, surely not the simplest. Using
$$\frac1{\sinh(n\pi)}=\frac{2e^{-\pi n}}{1-e^{-2\pi n}}=2\sum_{m\text{ odd}}e^{-mn\pi}$$
and
$$2\sum_{n\text{ odd}}\frac{x^n}{n}=-2\ln(1-x)+\ln(1-x^2 …
- 105k
22
votes
Accepted
Order of magnitude of $\sum \frac{1}{\log{p}}$
The contribution of the primes $p\leq n/\log ^3 n$ is clearly $O(n/\log^3 n)$. For the remaining primes we have
$$ \log n-3\log\log n <\log p<\log n,$$
$$ \frac{1}{\log p}=\frac{1}{\log n}+O\left(\fra …
- 105k
18
votes
Accepted
An inequality involving square roots and sums
For every $n\in\{1,\dotsc,N\}$, we have
$$2\sqrt{\sum_{i\leq n} a_i}-2\sqrt{\sum_{i\leq n-1} a_i}=\frac{2a_n}{\sqrt{\sum_{i\leq n} a_i}+\sqrt{\sum_{i\leq n-1} a_i}}>\frac{a_n}{\sqrt{\sum_{i\leq n} a_i …
- 105k
18
votes
Does $x_0=1/3$ lead to periodicity in the logistic map $x_{k+1}=4x_k(1-x_k)$?
It is straightforward to show that $(2\pi)^{-1}\arcsin(\sqrt{1/3})$ is irrational. Indeed, if this equals a rational number $r\in\mathbb{Q}$, then $2\sin(2\pi r)=\sqrt{4/3}$. However, $2\sin(2\pi r)$ …
- 105k
16
votes
Series whose convergence is not known
Assume that $A\subset\mathbb{N}$ contains no 3-term arithmetic progression. It was conjectured by Erdős and Turán that $\sum_{a\in A}\frac{1}{a}$ converges. As far as I know this is open, although we …
- 105k
12
votes
How to show that $\log 2(1/2\log 2\log 4 + 1/3\log 3\log 6 + \dotsb) + 1/2\log 2 - 1/3\log 3...
Observe that the left-hand side is the sum of two convergent series. Let $N\geq 2$ be an integer tending to infinity. Truncate the first series at the $N$-th term and the second series at the $2N$-th …
- 105k
12
votes
Is this formula expressing $\operatorname{erf} \left(\tfrac 1 {\sqrt 2}\right)$ as an infini...
WolframAlpha returns this result promptly, so it is probably known or straightforward to show. See here.
- 105k
10
votes
iterated harmonic numbers vs Riemann zeta
The identity in Question 1 is a special case of Theorem 2 in Ohno: A Generalization of the Duality and Sum Formulas on the Multiple Zeta Values (see here). Indeed, putting $k=1$ and $n=m+1$ in this th …
- 105k
10
votes
Accepted
Growth of a linear recurrent sequence
First note that $\frac{1+i\sqrt{2}}{1-i\sqrt{2}}\in\mathbb{Q}(\sqrt{-2})$ is not a root of unity, because it does not equal $\pm 1$. Therefore Baker's famous theorem shows that, for some effectively c …
- 105k