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Results tagged with sequences-and-series
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user 10744
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.
5
votes
5 different ways to define the same family of integer sequences
Here is a proof that $a_1(n, p, q) = a_2(n,p,q)$.
The generating function for the Stirling numbers of the second kind is
$$\sum_{n=i}^\infty {n\brace i}\frac{x^n}{n!}= \frac{(e^x-1)^i}{i!}.$$
Also
$$\ …
- 17k
6
votes
Accepted
Closed form for the A110501 (unsigned Genocchi numbers (of first kind) of even index)
This is a known result. To quote from Richard Stanley's Enumerative Combinatorics, Volume 2, second edition, solution to problem 8(e) of Chapter 5, page 115: This is equivalent to a conjecture of J. M …
- 17k
7
votes
Accepted
$R$-recursion for the A143017
We proceed by "guessing" a generating function for $R(n,q)$ and verifying that it has the right properties.
According to https://oeis.org/A143017, the generating function $G= \sum_{n=1}^\infty a(n) x^ …
- 17k
4
votes
Accepted
$R$-recursion for the A307389
Let
\begin{equation*}
A(x,q) = e^{qx}A(x)=e^{(q+1)x+(e^x-1)^2/2}
\end{equation*}
and define $a(n,q)$ by
\begin{equation*}
A(x,q) = \sum_{n=0}^\infty a(n,q) \frac{x^n}{n!}.
\end{equation*}
Then $a(n,0) …
- 17k
4
votes
Accepted
General case of the some $R$-recursions
Let
$$
A(x,q)=\sum_{i=0}^{\infty}\frac{x^i}{\prod\limits_{j=0}^{i}(1-f(q+j)x)},
$$
so that $A(x) = A(x,0)$. Define $a(n,q)$ by
$$A(x,q) = \sum_{n=0}^\infty a(n,q) x^n,$$
so that $a(n) = a(n,0)$.
I wil …
- 17k
7
votes
Recreation with Catalan
Here's a sketch of a generating function proof. Recall that
\begin{equation*}
\sum_{m=0}^\infty \binom{2m+k}{m} x^m =\frac{c(x)^k}{\sqrt{1-4x}},
\end{equation*}
where $c(x) = \sum_{n=0}^\infty C_n x^n …
- 17k
1
vote
Evaluating a sinusoidal series
Here's a formula for $A_{nk}$:
$$A_{nk} = (-1)^{n-k}2^{k-2n} \frac{(2n-k)!}{k!\,(n-k)!}.$$
This can be proved by standard methods for proving binomial coefficient identities.
- 17k
22
votes
Fibonacci series captures Euler $e=2.718\dots$
More generally,
$$\sum_{k=0}^\infty F_{n+k} \frac{x^k}{k!} = e^x\sum_{k=0}^\infty F_{n-k}\frac{x^k}{k!},\tag{1}$$
which is equivalent to Will Sawin's identity.
Similarly,
$$e^x\sum_{k=0}^\infty F_{n+k …
- 4,713
8
votes
Accepted
Infinite sum of Laguerre polynomials: is $\sum_{k=0}^{\infty}\frac{n!}{(k+n)!}x^kL_n^k(x)^2=...
Everything becomes simpler if add some parameters and start the sum at $k=-n$ instead of $k=0$. Note that if $k$ is a negative integer with $-n\le k \le -1$ then $L_n^k(x)$ is a polynomial in $x$ divi …
- 17k
4
votes
Integrality of a sequence formed by sums
Here is another proof, inspired by Tewodros Amdeberhan's. We represent the sum as a constant term in a power series.
To represent $(7k+8) \frac{(3k+1)!}{k!\,(2k+3)!}$ as a constant term, we need to ex …
- 17k
32
votes
Accepted
Integrality of a sequence formed by sums
Let $A(x) = \sum_{n=1}^\infty a_n x^n$ and let
$$S(x) = \sum_{k=0}^\infty (7k+8)\frac{(3k+1)!}{k!\,(2k+3)!} x^k.$$
Then the formula for $a_n$ gives
$A(x) = R(x)S(x)$,
where
$$R(x) = \frac{1}{3}\biggl( …
- 43.2k
2
votes
Upper bound for an infinite series of Pochhammer Symbol
The sum is $r/(1-\alpha)^{(1+r/\alpha)}$ by the binomial theorem.
- 17k
9
votes
Method to evaluate an infinite sum of ratio of Gamma functions (how does Mathematica do it?)
This is a special case of Watson's Theorem
$$\def\h{\frac{1}{2}}
\def\g#1{\Gamma(#1)\,}
{}_3F_2\left({a,\ b,\ c\atop\h a+\h b+\h, 2c }\biggm| 1 \right)
=\frac{\g\h\g{c+\h}\g{\h a+\h b +\h}\g{c+\h -\ …
- 17k
5
votes
Non-arithmetic proof of the integrality of a rational expression
The integrality of the coefficients of $(1-k^2x)^{-1/k}$ follows from the integrality of the coefficients of the generalized Catalan number generating function
$$c_k(x)=\frac{1-(1-k^2 x)^{1/k}}{kx},$$ …
- 17k
2
votes
Bilinear recurrence relation between even Bernoulli numbers
Here's a simple proof of the identity. Let
$$B(x) = \frac{x}{e^x-1} = \sum_{n} B_n \frac{x^n}{n!}.$$ Then
$$\sum_n (2^n-1) B_n \frac{x^n}{n!} = B(2x) - B(x) = -\frac{x}{e^x+1}$$
and
$$\bigl(B(2x) -B …
- 17k