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Results tagged with sequences-and-series
Search options answers only
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user 6101
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.
2
votes
Accepted
Equivalence of sequences related to A033264
Given the structure of these sequences, it seems convenient to describe them as integer valued
functions of binary strings rather than of natural numbers. This way, the recursions from $b$ and $c$ app …
- 60.6k
3
votes
Uniqueness of Neumann series
We can say a bit more: if a Neumann series converges point-wise, it also converges absolutely and in $\mathbb C[[z]]$, and the limit is an entire function.
Let $(a_n)_{n\ge0}$ be a sequence of complex …
- 60.6k
8
votes
Uniqueness of Neumann series
edit: This was meant to recall a first elementary but relevant fact that was not mentioned at all, that is orthogonality.
Bessel functions $\{J_n\}_{n\in\mathbb N_+}$ are orthogonal on $\mathbb R$ w.r …
- 60.6k
3
votes
Accepted
Difference between finite partial sums from two divergent series
A necessary and sufficient condition for $\{s_i\}_{i\in\mathbb N}\subset(0,1)$ to satisfy your property $(*)$, as it is, is simply that $$|s_i-r_i|\le\frac{r_i}2\, \text{ for all } i\in\mathbb N.$$
…
- 60.6k
2
votes
Lower bound in recurrence relation
From the definition $$v_{k-1}=\frac1{15}\prod_{j=1}^k(2^j+1)=$$$$= \frac1 {15}{2^{ {\frac12k(k+1)}}}\prod_{j=1}^k(1+2^{-j}),$$
and from the bound of the infinite product
$$\frac1 {15}\prod_{j=1}^\inft …
- 60.6k
3
votes
Is there another representation for the summation: $\sum_{j=1}^{N}\frac{a_j}{(c+a_j)(c+a_j+1...
A few words on existence and approximation. Write your equation as
$$ \sum_{n=1}^N \frac{c^2a_n}{(c+b_n)(c+b_n+1)}=1.$$
There is a nice partial fractions decomposition,
$$\frac{c^2 }{(c+b )(c+b +1)}=1 …
- 60.6k
2
votes
Finding $\sum_i x_i$ given $\{\sum_i x_i^{2n}\}_{n\in \mathbb{N}}$
[edit: completed] Assuming $x_i\ge0$ with $ \sum_i x_i <\infty$, we have that $\phi(t):=\sum_i(e^{x_it}-1)=\sum_{k\ge1} \big(\sum_i x_i^k\big)t^k/k!$ is an entire function (we can expand the expo …
- 60.6k
4
votes
Binomial series
As remarked by Iosif Pinelis, this is a matter of law of great numbers; we may also describe it in terms of Bernstein polynomials. Specifically, for $\alpha\ge0$ and $n\ge1$, let $p_n$ be the value of …
- 60.6k
1
vote
Another chicken or egg: sequence or series
Why shouldn’t they come together? Here’s a possible program. The first object to appear, in an elementary course of analysis, if one starts by an axiomatic presentation of $\mathbb R$, is the supremum …
- 60.6k
9
votes
Accepted
On the finite sum of reciprocal Fibonacci sequences
We need to prove, equivalently
$$\frac1{F_{2n-1}+1}<\sum_{k=n}^{2n}\frac1{F_{2k}}\le \frac1{F_{2n-1}}, $$
that is, by the above expression for $F_{k}$, since $\beta=-\alpha^{-1}$, we need to check the …
- 60.6k
5
votes
Accepted
A need for analytic continuation of a finite sum function
As for the sum $$\sum_{k\geq1}(-1)^k\binom{k}{\frac23}^{-1}$$
one can evaluate it by means of the Beta function integral, like in this recent computation.
$$\sum_{k\geq1}(-1)^k\binom{k}{\frac23}^{-1}= …
- 60.6k
22
votes
Closed form of an infinite series
Denote $c_n:={(-1)^n \frac{\Gamma(\frac{1}{3}+\frac{n}{3})}{\Gamma(1+\frac{n}{3})} \sin(\frac{2\pi n}{3})}$ the $n$-th term of the series. We have for all $k\ge0$
$$c_{3k}=0,$$
$$c_{3k+1}= (-1)^{k+1}\ …
- 60.6k
5
votes
A differential equation governing compositional inversion
I do not have a reference, but I’d say it is a plain instance of the Implicit Function Theorem for formal power series. I add the computation below, in case you had a different one.
We choose our inde …
- 60.6k
3
votes
Accepted
Modulo $2$ binomial transform of $m^n$
If $n\in \mathbb N$ has binary expansion $n=\sum_{k\in S}2^k$ for a finite subset $S\subset \mathbb N$, then
$$ \sum_{j=0}^n\Big[\big( {n \atop j}\big)\text{mod}\, 2 \Big]x^jy^{n-j}=\prod_{k\in S} (x^ …
- 60.6k
4
votes
Accepted
A restricted version of Riemann series theorem: rearrangements with alternating signs
Here you prescribe in addition the sequence of signs of the rearranged series in the Riemann-Dini theorem to be alternating, but note that any non-stationary binary sequence of signs does as well. Mo …
- 60.6k