sum of reciprocals’ problem

Nuestro objetivo es estudiar las funciones generadoras cuyas series de Taylor en el origen generan una secuencia creciente de coeficientes y se informa acerca de la convergencia o no, de esa serie evaluada en uno…

En otras palabras estudiamos $f(x)$ tal que $f(x)=\sum_{n=0}^{\infty}\frac{x^n}{a_n}$ en algun radio de convergencia mayor que uno, pero que $f(1)=\frac{1}{a_0}+\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+...$ determine convergencia o no. Esto lo vemos e investigamos para algunas secuencia creciente $a_n$ de enteros.

SCN: para sucesión creciente natural

SOR: sum of reciprocals

GF: generating function

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SCN: 1, 2, 3, 4, 5, 6, 7 …

SOR: 1/1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7+ … = diverge

GF:

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SCN: 1, 1, 2, 6, 24, 120, 720, …

SOR: 1/1 + 1/1 + 1/2 + 1/6 + 1 / 24 + 1/120 + …. = e

GF: $e^x= 1+x+x^2/2+x^3/6+x^4/24+\cdots$

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SCN: 1, 4 , 9, 16, 25, 36, ….

SOR: 1/1 + 1/4 + 1/9 + 1/16 + 1/25+ $\cdots=\frac{\pi^2}{6}$

GF:

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sum of the reciprocals primes $\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots=\infty$

Euler
zeta de Riemann
Sprugnoli, sum of reciprocals problem for Central Binomial Coefficients $B_n={2n \choose n}$
central binomials: ${2n\choose n}$ : 1, 2, 6, 20 , 70, 252, … A000984. The gf for is

$\frac{4\,\left({\sqrt{4-x}}+{\sqrt{x}}\,\arcsin(\frac{{\sqrt{x}}}{2})\right) }{{\sqrt{(4-x)^3}}}$

the following formulas are for testing at, say, W|A for example. First, the gf of CBC_n inverses, second multiplied by $x$ and third differentiated to get the gf of Calatans inverses:

4(\sqrt{4 – x} + \sqrt{x}\arcsin (\sqrt{x}/ 2))/\sqrt{(4 – x)^3}

4x(\sqrt{4 – x} + \sqrt{x}\arcsin (\sqrt{x}/ 2))/\sqrt{(4 – x)^3}

d/dx[4x(\sqrt{4 – x} + \sqrt{x}\arcsin (\sqrt{x}/ 2))/\sqrt{(4 – x)^3}]

You can see in another way in OeisWiki.

Catalan: $\frac{{2n\choose n}}{n+1}$ : 1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, … meanwhile this is the gf story:

is seed, and this other:

$\frac{2\left(\sqrt{4-x}(8+x)+12\sqrt{x}\arctan{\frac{\sqrt{x}}{\sqrt{4-x}}}\right)}{\sqrt{(4-x)^5}}$

is the generating function. That give us

$2+\frac{4\sqrt{3}\pi}{27}=1+1+\frac{1}{2}+\frac{1}{5}+\frac{1}{14}+\frac{1}{42}+\cdots$

that you can confer with the entry A121839 of OEIS.

Tryyourself by entering

2( sqrt(4-x)(8+x)+12 sqrt(x) arctan((sqrt(x))/(sqrt(4-x))))/(sqrt((4-x)^5))

in Wolfram-alpha. Or directly

Series[(2 (Sqrt[4 – x] (8 + x) + 12 Sqrt[x] ArcTan[Sqrt[x]/Sqrt[4 – x]]))/Sqrt[(4 – x)^5], {x, 0, 11}]

copy-paste in your mathematica.

$\frac{d}{dx}\left(\frac{4x\left({\sqrt{4 - x}} + {\sqrt{x}}\arcsin (\frac{{\sqrt{x}}}{2}) \right) }{{\sqrt{(4 - x)^3}}}\right)=\frac{2\left(\sqrt{4-x}(8+x)+12\sqrt{x}\arctan{\frac{\sqrt{x}}{\sqrt{4-x}}}\right)}{\sqrt{(4-x)^5}}$

Siehler: numbers $\mathbf{s_n}$ : 1, 3, 7, 15, 23, 39, 47, 71, 87, 111, … has nice intrincacies … does there exist a gf zumbeispiel?… This corresponds to A171503 at Oeis. It has $s_n=s_{n-1}+4\varphi(n)$ where $\varphi$ is the Totient de Euler given by: $\varphi(m)=\#{{\mathbb{Z}}_m}^*$ , the amount of coprimes with $m$ ….

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para la A000127: 1, 2, 4, 8, 16, 31, 57, 99,… tenemos función generadora y relación de recurrencia, lo cual permitiría quizá, que la suma

$1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{31}+\frac{1}{57}+\frac{1}{99}+\cdots$

se le pueda encontrar una función generadora correspondiente

Para sumas finitas o sumas parciales: www.math.upenn.edu/~wilf/AeqB.html

interesante también es estudiar las ecuaciones diferenciales entre estas GF’s
utilizando software, sure!

OEISes

For $B_n={2n\choose n}$

BZ A000984 $1,2,6,20,70,252,924,...$
BW A005430 $0,2,12,60,280,1260,5544,...$
BT A002736 $0,2,24,180,1120,6300,33264,168168,...$
BA A002457 $1,6,30,140,630,2772,12012,...$
BM 2 A000917 $6,40,210,1008,4620,20592,...$
BN Axxxxxx $6,80,630,4032,23100,123552,..$
BP A037965 $1,4,18,80,350,1512,6468,...$

For $C_n=\frac{{2n\choose n}}{n+1}$

CZ A000108 $1,1,2,5,14,42,132,1430,...$
CW A001791 $0,1,4,15,56,210,792,3003,...$
CT A110609 $1,8,45,224,1050,4752,21021,...$
CA A001700 $1,3,10,35,126,462,1716,6435,...$
CM A024483 $0,2,10,42,168,660,2574,10010,...$
CN A145885 $0,1,10,63,336,1650,7722,35035,...$
CP A000984 $1,2,6,20,70,252,924,...$

For $D_n=\frac{{2n\choose n}}{2n-1}$

DZ A002420 $-1,2,2,4,10,28,84,...$
DW A028329 $2,4,12,40,140,504,1848,...$
DT A037965 $2,8,36,160,700,3024,...$
DA A078718 $3,-5,14,14,-45,154,-546,1980,-7293,...$
DM A001791 $1,4,15,56,210,792,...$
DN A110609 $1,8,45,224,1050,4752,21021,...$
DP Axxxxxx $-1,4,6,16,50,168,588,2112,7722,28600,...$

For $U_n={n+2\choose 2}$

TRIANGULAR NUMBERS

series (2(x Li_2(x)-x+(x-1)log(1-x))/x) at origin

1, 2, 3, 4, 5, 7, 9, 12, 16, 21, 28, 37, 49, 65, 86, 114, 151, 200, 265, 351, 465, 616, 816, 1081, 1432, 1897, 2513, 3329, 4410, 5842, 7739, 10252, 13581, 17991, 23833, 31572, 41824, 55405, 73396, 97229, 128801, 170625, 226030, 299426