Simplificar el cálculo de la suma de esta serie

Question

Necesito simplificar el último producto de esta suma. Si es posible u otra forma de encontrar esta suma\

$\displaystyle\sum_{n=0}^{\infty}\left((-1)^n\frac{500n^3+750n^2+360n+50}{625n^4+1125n^3+800n^2+190n+24}\right)=\frac{1}{20}\left\{(2-2\sqrt{5})\log(2)+2\sqrt{10-2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}+1}{\sqrt{10-2\sqrt{5}}}\right)-6\sqrt{10+2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}\right)+2(2\sqrt{10+2\sqrt{5}}-2\sqrt{10-2\sqrt{5}})\tan^{-1}\left(\frac{3+\sqrt{5}}{\sqrt{10+2\sqrt{5}}}\right)-2\sqrt{10-2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}-3}{\sqrt{10-2\sqrt{5}}}\right)\right\}$ \

Mi solución:\

$(A)$ ....\

$(500n^3+750n^2+360n+50)=(250n^3+425n^2+225n+36)+(250n^3+325n^2+135n+14)=10n(25n^3+35n+12)+3(25n^2+35n+12)+10n(25n^2+15n+2)+7(25n^2+15n+2)=(10n+3)(5n+4)(5n+3)+(10n+7)(5n+1)(5n+2)$ \

$(B)$ ..\

$(625n^4+1125n^3+800n^2+190n+24)=25n^2(25n^2+35n+12)+10n(25n^2+35n+12)+2(25n^2+35n+12)=(25n^2+10n+2)(25n^2+35n+12)=(5n(5n+2)+(5n+2))(5n(5n+4)+3(5n+4))=(5n+1)(5n+2)(5n+3)(5n+4)$ \

Así que..:
$\left(\frac{500n^3+750n^2+360n+50}{625n^4+1125n^3+800n^2+190n+24}\right)=\left(\frac{(10n+3)(5n+4)(5n+3)+(10n+7)(5n+1)(5n+2)}{(5n+1)(5n+2)(5n+3)(5n+4)}\right)$
$=\left(\frac{10n+3}{(5n+1)(5n+2)}+\frac{10n+7}{(5n+3)(5n+4)}\right)=\left(\frac{1}{5n+1}+\frac{1}{5n+2}+\frac{1}{5n+3}+\frac{1}{5n+4}\right)$
encontramos: $$\displaystyle\sum_{n=0}^{\infty}(-1)^n\left(\frac{500n^3+750n^2+360n+50}{625n^4+1125n^3+800n^2+190n+24}\right)$$ \ $$=\displaystyle\sum_{n=0}^{\infty}(-1)^n\left(\frac{1}{5n+1}+\frac{1}{5n+2}+\frac{1}{5n+3}+\frac{1}{5n+4}\right) =\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{5n+1}+\sum_{n=0}^{\infty}\frac{(-1)^n}{5n+2}+\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{5n+3}+\sum_{n=0}^{\infty}\frac{(-1)^n}{5n+4}$$ y tenemos :\

(1)
$\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{5n+1}=\frac{1}{20}\left[2\sqrt{5}\log(3+\sqrt{5})+\log(2)(4-2\sqrt{5})-2\sqrt{10-2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}-3}{\sqrt{10-2\sqrt{5}}}\right)+2\sqrt{10+2\sqrt{5}}\tan^{-1}\left(\frac{3+\sqrt{5}}{\sqrt{10+2\sqrt{5}}}\right)+2\sqrt{10-2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}+1}{\sqrt{10-2\sqrt{5}}}\right)-2\sqrt{10+2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}\right)\right]$
preuve (1):
$\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{5n+1}=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{5n+1}x^{5n+1}|_{x=1}=S(1)$ \ donde $S(x)=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^nx^{5n+1}}{5n+1}$
Así que..: $S^{'}(x)=\displaystyle\sum_{n=0}^{\infty}(-1)^nx^{5n}=\frac{1}{1+x^5}$
a continuación, busque : $S(x)=\int S^{'}(x)dx=\int_{}^{}\frac{1}{1+x^5}dx$
$S(x)=\frac{1}{20}\left[(\sqrt{5}-1)\log(x^2+\frac{\sqrt{5}-1}{2}x+1)-(1+\sqrt{5})\log(x^2-\frac{1+\sqrt{5}}{2}x+1)+4\log(x+1)-2\sqrt{10-2\sqrt{5}}\tan^{-1}\left(\frac{-4x+\sqrt{5}+1}{\sqrt{10-2\sqrt{5}}}\right)+2\sqrt{10+2\sqrt{5}}\tan^{-1}\left(\frac{4x+\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}\right)\right]$
tenemos :
$S(0)=0=\frac{1}{20}\left[-2\sqrt{10-2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}+1}{\sqrt{10-2\sqrt{5}}}\right)+2\sqrt{10+2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}\right)\right]+c$
$\implies c=-\frac{1}{20}\left[-2\sqrt{10-2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}+1}{\sqrt{10-2\sqrt{5}}}\right)+2\sqrt{10+2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}\right)\right]$
Así que encontramos :
$S(1)=\frac{1}{20}\left[2\sqrt{5}\log(3+\sqrt{5})+\log(2)(4-2\sqrt{5})-2\sqrt{10-2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}-3}{\sqrt{10-2\sqrt{5}}}\right)+2\sqrt{10+2\sqrt{5}}\tan^{-1}\left(\frac{3+\sqrt{5}}{\sqrt{10+2\sqrt{5}}}\right)+2\sqrt{10-2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}+1}{\sqrt{10-2\sqrt{5}}}\right)-2\sqrt{10+2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}\right)\right]$
$\implies S(1)=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{5n+1}=\frac{1}{20}\left[2\sqrt{5}\log(3+\sqrt{5})+\log(2)(4-2\sqrt{5})-2\sqrt{10-2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}-3}{\sqrt{10-2\sqrt{5}}}\right)+2\sqrt{10+2\sqrt{5}}\tan^{-1}\left(\frac{3+\sqrt{5}}{\sqrt{10+2\sqrt{5}}}\right)+2\sqrt{10-2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}+1}{\sqrt{10-2\sqrt{5}}}\right)-2\sqrt{10+2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}\right)\right]$
(2)
$\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{5n+2}=\frac{1}{20}\left[\sqrt{5}\log\left(\frac{3+\sqrt{5}}{3-\sqrt{5}}\right)-4\log(2)-2\sqrt{10+2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}-3}{\sqrt{10-2\sqrt{5}}}\right)-2\sqrt{10-2\sqrt{5}}\tan^{-1}\left(\frac{3+\sqrt{5}}{\sqrt{10+2\sqrt{5}}}\right)+2\sqrt{10+2\sqrt{5}}\tan^{-1}\left(\frac{1+\sqrt{5}}{\sqrt{10-2\sqrt{5}}}\right)+2\sqrt{10-2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}\right)\right]$
preuve(2):
$\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{5n+2}=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^nx^{5n+2}}{5n+1}|_{x=1}=S(1)$ donde
$S(x)=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^nx^{5n+2}}{5n+1}$ \

Así que..:
$S(x)=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^nx^{5n+2}}{5n+2}\implies S^{'}(x)=\displaystyle \sum_{n=0}^{\infty}(-1)^nx^{5n+1}=x\displaystyle\sum_{n=0}^{\infty}(-x^5)^{n}=\frac{x}{1+x^5}$
entonces:
$S(x)=\int S^{'}(x)dx=\int\frac{x}{1+x^5}dx$ \ $=\frac{1}{20}\left[(1+\sqrt{5})\log(x^2+\frac{\sqrt{5}-1}{2}x+1)-(\sqrt{5}-1)\log(x^2-\frac{1+\sqrt{5}}{2}x+1)-4\log(x+1)-2\sqrt{10+2\sqrt{5}}\tan^{-1}\left(\frac{-4x+\sqrt{5}+1}{\sqrt{10-2\sqrt{5}}}\right)-2\sqrt{10-2\sqrt{5}}\tan^{-1}\left(\frac{4x+\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}\right)\right]+c$
tenemos :
$S(0)=c+\frac{1}{20}\left[-2\sqrt{10+2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}+1}{\sqrt{10-2\sqrt{5}}}\right)-2\sqrt{10-2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}\right)\right]$
$\implies c=\frac{1}{20}\left[2\sqrt{10+2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}+1}{\sqrt{10-2\sqrt{5}}}\right)+2\sqrt{10-2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}\right)\right]$
Finalmente encontramos :
$S(1)=\frac{1}{20}\left[\sqrt{5}\log\left(\frac{3+\sqrt{5}}{3-\sqrt{5}}\right)-4\log(2)-2\sqrt{10+2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}-3}{\sqrt{10-2\sqrt{5}}}\right)-2\sqrt{10-2\sqrt{5}}\tan^{-1}\left(\frac{3+\sqrt{5}}{\sqrt{10+2\sqrt{5}}}\right)+2\sqrt{10+2\sqrt{5}}\tan^{-1}\left(\frac{1+\sqrt{5}}{\sqrt{10-2\sqrt{5}}}\right)+2\sqrt{10-2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}\right)\right]=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{5n+2}$
(3)
$\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{5n+3}=\frac{1}{20}\left[\sqrt{5}\log\left(\frac{3-\sqrt{5}}{3+\sqrt{5}}\right)+4\log(2)-2\sqrt{10+2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}+1}{\sqrt{10-2\sqrt{5}}}\right)-2\sqrt{10-2\sqrt{5}}\tan^{-1}\left(\frac{3+\sqrt{5}}{\sqrt{10+2\sqrt{5}}}\right)+2\sqrt{10+2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}+1}{\sqrt{10-2\sqrt{5}}}\right)+2\sqrt{10-2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}\right)\right]$
Preuve(3):
$\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{5n+3}=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^nx^{5n+3}}{5n+3}|_{x=1}=S(1)$
donde
$S(x)=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^nx^{5n+3}}{5n+3}$
Así que tenemos :
$ S^{'}(x)=\displaystyle\sum_{n=0}^{\infty}(-1)^nx^{5n+2}=x^2\displaystyle\sum_{n=0}^{\infty}(-x^5)^n=\frac{x^2}{1+x^5}$
$\implies S(x)=\int\frac{x^2}{1+x^5}dx=\frac{1}{20}\left[-(1+\sqrt{5})\log(x^2+\frac{(\sqrt{5}-1)x}{2}+1)+(\sqrt{5}-1)\log(x^2-\frac{(1+\sqrt{5})x}{2}+1)+4\log(x+1)-2\sqrt{10+2\sqrt{5}}\tan^{-1}\left(\frac{-4x+\sqrt{5}+1}{\sqrt{10-2\sqrt{5}}}\right)-2\sqrt{10-2\sqrt{5}}\tan^{-1}\left(\frac{4x+\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}\right)\right] +c$
tenemos :
$S(0)=c+\frac{1}{20}\left[-2\sqrt{10+2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}+1}{\sqrt{10-2\sqrt{5}}}\right)-2\sqrt{10-2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}\right)\right]=0$
$\implies c=\frac{1}{20}[2\sqrt{10+2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}+1}{\sqrt{10-2\sqrt{5}}}\right)+2\sqrt{10-2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}\right)$
Por fin:
$S(1)=\frac{1}{20}\left[\sqrt{5}\log\left(\frac{3-\sqrt{5}}{3+\sqrt{5}}\right)+4\log(2)-2\sqrt{10+2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}+1}{\sqrt{10-2\sqrt{5}}}\right)-2\sqrt{10-2\sqrt{5}}\tan^{-1}\left(\frac{3+\sqrt{5}}{\sqrt{10+2\sqrt{5}}}\right)+2\sqrt{10+2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}+1}{\sqrt{10-2\sqrt{5}}}\right)+2\sqrt{10-2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}\right)\right]= \displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{5n+3}$
(4)
$\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{5+4}=\frac{1}{20}\left[\sqrt{5}\log\left(\frac{3-\sqrt{5}}{3+\sqrt{5}}\right)-4\log(2)-2\sqrt{10-2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}-3}{\sqrt{10-2\sqrt{5}}}\right)+2\sqrt{10+2\sqrt{5}}\tan^{-1}\left(\frac{3+\sqrt{5}}{\sqrt{10+2\sqrt{5}}}\right)-2\sqrt{10-2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}+1}{\sqrt{10-2\sqrt{5}}}\right)-2\sqrt{10+2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}\right)\right]$
preuve (4):
$\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{5n+4}=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^nx^{5n+4}}{5n+4}|_{x=1}=S(1)$
donde
$S(x)=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^nx^{5n+4}}{5n+4}$
Así que..:
$S^{'}(x)=\displaystyle\sum_{n=0}^{\infty}(-1)^nx^{5n+3}=\frac{x^3}{1+x^5}$ \ $\implies S(x)=\int\frac{x^3}{1+x^5}dx=\frac{1}{20}\left[-(\sqrt{5}-1)\log(x^2+\frac{(\sqrt{5}-1)x}{2}+1)+(\sqrt{5}+1)\log(x^2-\frac{(1+\sqrt{5})x}{2}+1)-4\log(x+1)-2\sqrt{10-2\sqrt{5}}\tan^{-1}\left(\frac{-4x+\sqrt{5}+1}{\sqrt{10-2\sqrt{5}}}\right)+2\sqrt{10+2\sqrt{5}}\tan^{-1}\left(\frac{4x+\sqrt{5}+1}{\sqrt{10+2\sqrt{5}}}\right)\right]+c$
tenemos :
$S(0)=c+\frac{1}{20}\left[2\sqrt{10-2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}+1}{\sqrt{10-2\sqrt{5}}}\right)+2\sqrt{10+2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}\right)\right]$
$\implies c=-\frac{1}{20}\left[2\sqrt{10-2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}+1}{\sqrt{10-2\sqrt{5}}}\right)+2\sqrt{10+2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}\right)\right]$
Finalmente encontramos:
$S(1)=\frac{1}{20}\left[\sqrt{5}\log\left(\frac{3-\sqrt{5}}{3+\sqrt{5}}\right)-4\log(2)-2\sqrt{10-2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}-3}{\sqrt{10-2\sqrt{5}}}\right)+2\sqrt{10+2\sqrt{5}}\tan^{-1}\left(\frac{3+\sqrt{5}}{\sqrt{10+2\sqrt{5}}}\right)-2\sqrt{10-2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}+1}{\sqrt{10-2\sqrt{5}}}\right)-2\sqrt{10+2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}\right)\right]= \displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{5n+4}$
tenemos :\

$\displaystyle\sum_{n=0}^{\infty}(\frac{(-1)^n}{5n+1}+\frac{(-1)^n}{5n+2})=\frac{1}{20}\left[3\sqrt{5}\log(3+\sqrt{5})-2\sqrt{5}\log(2)-\sqrt{5}\log(3-\sqrt{5})+4\sqrt{10-2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}+1}{\sqrt{10-2\sqrt{5}}}\right)-4\sqrt{10+2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}\right)\right]$
y :\

$\displaystyle\sum_{n=0}^{\infty}(\frac{(-1)^n}{5n+3}+\frac{(-1)^n}{5n+4})=\frac{1}{20}\left[2\sqrt{5}\log\left(\frac{3-\sqrt{5}}{3+\sqrt{5}}\right)+\tan^{-1}\left(\frac{3+\sqrt{5}}{\sqrt{10+2\sqrt{5}}}\right)(-2\sqrt{10-2\sqrt{5}}+2\sqrt{10+2\sqrt{5}})+\tan^{-1}\left(\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}\right)(2\sqrt{10-2\sqrt{5}}-2\sqrt{10+2\sqrt{5}})-2\sqrt{10-2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}-3}{\sqrt{10-2\sqrt{5}}}\right)-2\sqrt{10-2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}+1}{\sqrt{10-2\sqrt{5}}}\right)\right]$ \

Por fin:
$\displaystyle\sum_{n=0}^{\infty}\left((-1)^n\frac{500n^3+750n^2+360n+50}{625n^4+1125n^3+800n^2+190n+24}\right)=\frac{1}{20}\left[(2-2\sqrt{5})\log(2)+2\sqrt{10-2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}+1}{\sqrt{10-2\sqrt{5}}}\right)-6\sqrt{10+2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}\right)+2(2\sqrt{10+2\sqrt{5}}-2\sqrt{10-2\sqrt{5}})\tan^{-1}\left(\frac{3+\sqrt{5}}{\sqrt{10+2\sqrt{5}}}\right)-2\sqrt{10-2\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{5}-3}{\sqrt{10-2\sqrt{5}}}\right)\right]$ \

¿Hay alguna ayuda para simplificar el resultado o encontrar la suma de esta serie de otra manera?

Answer 1

Esta es una respuesta parcial.

Hay algún punto que no entiendo. Para mi $$\frac{1}{5n+1}+\frac{1}{5n+2}+\frac{1}{5n+3}+\frac{1}{5n+4}=\frac{500 n^3+750 n^2+350 n+50 }{625 n^4+1250 n^3+875 n^2+250 n+24 }$$ que no es la expresión que estás trabajando.

Volviendo al problema planteado en el post $$625n^4+1125n^3+800n^2+190n+24=0$$ no muestra ninguna raíz real. Así que escribe la expresión como $$\frac{500n^3+750n^2+360n+50}{625n^4+1125n^3+800n^2+190n+24}=\frac{500n^3+750n^2+360n+50}{(5n-a)(5n-b)(5n-c)(5n-d)}$$ donde $a,b,c,d$ son números complejos y utilizar la descomposición en fracciones parciales para obtener $$\frac{500n^3+750n^2+360n+50}{625n^4+1125n^3+800n^2+190n+24}=\frac {k_1}{5n-a}+\frac {k_2}{5n-b}+\frac {k_3}{5n-c}+\frac {k_4}{5n-d}$$ El problema empieza a ser complejo ya que la suma parcial $$S_p=\sum_{n=0}^p \frac {(-1)^n}{5n-a}=\frac{1}{5} \left((-1)^p \,\Phi \left(-1,1,p+1-\frac{a}{5}\right)+\Phi \left(-1,1,-\frac{a}{5}\right)\right)$$ donde aparece la función trascendente de Lerch (para la que las asintóticas no son las más agradables - echa un vistazo aquí ).

Por el momento (estén seguros de que continuaré), el valor de la suma infinita es $$1.68876362556729891404128608773597318553351101190557646863328498735\cdots$$ que no es reconocido por las calculadoras simbólicas inversas. No parece estar de acuerdo con su último resultado.

Eche un vistazo aquí .

Editar

• Si el problema era $$S=\sum_{n=0}^\infty (-1)^n \left(\frac{1}{5n+1}+\frac{1}{5n+2}+\frac{1}{5n+3}+\frac{1}{5n+4} \right)$$ el resultado sería simplemente $$S=\frac{2}{5} \sqrt{1+\frac{2}{\sqrt{5}}} \pi \approx 1.72961$$

• Esperando no haberme equivocado al copiar la última expresión del post, se podría simplificar como $$\frac{1}{100} \left(\left(3 \sqrt{10+2 \sqrt{5}}-2 \sqrt{10-2 \sqrt{5}}\right) \pi -10 \left(\sqrt{5}-1\right) \log (2)\right)\approx 0.125136$$

Answer 2

Hice una prueba con el software y muestra que esta serie diverge. Te sugiero que utilices una prueba de comparación para demostrarlo. Por ejemplo, observa que $$\frac{500n^3}{625n^4+1125n^3+800n^2+190n+24}<\frac{500n^3+750n^2+360n+50}{625n^4+1125n^3+800n^2+190n+24}$$ y si $\frac{500n^3}{625n^4+1125n^3+800n^2+190n+24}$ diverge, también lo hará el $\frac{500n^3+750n^2+360n+50}{625n^4+1125n^3+800n^2+190n+24}$ . Puede que entonces sea más sencillo.