matematikk.net

[tex]$$\sum\limits_{n = 1}^\infty {{{{2^n}{x^n}} \over {\left( {2n - 1} \right)!}}} $$[/tex]


Forholdstesten:

[tex]$$R = {\lim }\limits_{n \to \infty } \left | \frac{2^{n+1}x^{n+1}}{\left ( 2\left ( n+1 \right )-1 \right )!}\cdot \frac{\left ( 2n-1 \right)!}{2^nx^n} \right | $$[/tex]

[tex]$$R = 2\left| x \right| \cdot {\lim }\limits_{n \to \infty } \frac{\left ( 2n-1 \right )!}{\left ( 2n+1 \right )!} $$[/tex]

der vi har to triks:

1. [tex]$$\left( {2n - 1} \right)! = {{\left( {2n} \right)!} \over {2n}}$$[/tex]

2. [tex]$$\left( {2n + 1} \right)! = \left( {2n + 1} \right) \cdot \left( {2n} \right)!$$[/tex]

[tex]$$R = 2\left| x \right| \cdot {\lim }\limits_{n \to \infty} \frac{\frac{\left ( 2n \right )!}{2n}}{\left ( 2n+1 \right )\left ( 2n \right )!} $$[/tex]

[tex]$$R = 2\left| x \right| \cdot {\lim }\limits_{n \to \infty} \frac{2}{2+\frac{1}{n}} $$[/tex]

[tex]R= 2\left| x \right|[/tex]


Er dere enige så langt? Jeg mener i følge Wolfram her:

http://www.wolframalpha.com/input/?i=Sum[%282^%28n%29x^%28n%29%29%2F%28%282n-1%29!%29%2Cn%3D1%2C+infinity+]


At jeg burde fått en grenseverdi som gikk mot null, slik at den kunne konvergert for alle verdier av x slik Wolfram mener.


Ser dere noen feil?

Du tuller litt med brøkreglene: [tex]\frac{\frac{(2n)!}{2n}}{(2n+1)(2n)!} = \frac{1}{(2n+1)(2n)} \to 0[/tex].

Vektormannen wrote:Du tuller litt med brøkreglene: [tex]\frac{\frac{(2n)!}{2n}}{(2n+1)(2n)!} = \frac{1}{(2n+1)(2n)} \to 0[/tex].

Jeg ser det - takk vektormannen!

(godt du er våken)