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

  • —————————————–

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:

  • —————————————–

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

  • ——————————————

SCN: 1, 4 , 9, 16, 25, 36, ….

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

GF:

  • —————————————–

\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….

—————————————-                                               ————————————————–

  • 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}

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

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

For U_n={n+2\choose 2}

  • UZ
  • UW
  • UT
  • UA
  • UM
  • UN
  • UP

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

3 responses to “sum of reciprocals’ problem

  1. \cdot para n{2n\choose n}

    http://oeis.org/A003506

    enter sum[(n+1)/(nBinomial[2n,n])x^n] at wolfram-alpha

    \cdot para (2n+1){2n\choose n}

    http://oeis.org/A002457

    \cdot para n^2{2n\choose n}

    http://oeis.org/A002736

    sum[(n+1)/((n^2)Binomial[2n,n])x^(n)]

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