What is the Radius of Convergence for the Power Series Given?

[alejandrito29]

at the serie $$\sum_0^{\infty} a_n (x - c)^n$$, the radius of convergency is:
.

$$R= \lim_{n \to \infty } |\frac{a_n}{a_{n+1}}|$$

My problem is : Find the radius of convergency when:
$$\sum_0^{\infty} \frac{(-1)^n}{(2n+1)!} \cdot x^{2n+1}$$


i don't understand mainly who is $$a_n$$ .

The answer is $$R \to \infty$$

 

[micromass]

Hi alejandrito29,

The problem here is of course that you will divide by zero, and this is not allowed. Thus the formula is not immediately applicable.

Now, how did you prove

$$R= \lim_{n \to \infty } |\frac{a_n}{a_{n+1}}|$$??

Can you adjust the prove such that it also applies to your series now? Essentialy, the proof uses a criterion for series that can be applied to this one (I used to call it the criterion of d'Alembert).

 

[alejandrito29]

Hi alejandrito29,

The problem here is of course that you will divide by zero, and this is not allowed. Thus the formula is not immediately applicable.

Now, how did you prove

$$R= \lim_{n \to \infty } |\frac{a_n}{a_{n+1}}|$$??

Can you adjust the prove such that it also applies to your series now? Essentialy, the proof uses a criterion for series that can be applied to this one (I used to call it the criterion of d'Alembert).

D'Lambert is:

$$\lim_{n \to \infty} |\frac{c_{n+1}}{c_n}|<1$$

i know that
$$\lim_{n \to \infty} |\frac{a_{n+1}}{a_n} \cdot \frac{x^{2n+3}}{x^{2n+1}}|<1$$

then

$$|x^2|<\lim _{n \to \infty}|\frac{a_n}{a_{n+1}}}|$$

$$-\lim _{n \to \infty}|\frac{a_n}{a_{n+1}}}|<x^2<\lim _{n \to \infty}|\frac{a_n}{a_{n+1}}}|$$

but $$x^2>0$$...I don't understand

i don't see that $$\lim _{n \to \infty}|\frac{a_n}{a_{n+1}}}|$$ is a radius

 

[micromass]

D'Lambert is:

$$\lim_{n \to \infty} |\frac{c_{n+1}}{c_n}|<1$$

i know that
$$\lim_{n \to \infty} |\frac{a_{n+1}}{a_n} \cdot \frac{x^{2n+3}}{x^{2n+1}}|<1$$

then

$$|x^2|<\lim _{n \to \infty}|\frac{a_n}{a_{n+1}}}|$$

Indeed, so the series converges if and only if

$$|x|<\sqrt{\lim _{n \to \infty}|\frac{a_n}{a_{n+1}}}|}$$

So the convergence radius is $$\sqrt{\lim _{n \to \infty}|\frac{a_n}{a_{n+1}}}|}$$

 

[alejandrito29]

Indeed, so the series converges if and only if

$$|x|<\sqrt{\lim _{n \to \infty}|\frac{a_n}{a_{n+1}}}|}$$

So the convergence radius is $$\sqrt{\lim _{n \to \infty}|\frac{a_n}{a_{n+1}}}|}$$

very thanks you

 

[3.1415926535]

Here is your answer
$$R= \lim_{n \to \infty } \left |\frac{a_n}{a_{n+1}} \right |=\lim_{n \to \infty } \left|\frac{\frac{(-1)^n}{(2n+1)!}}{\frac{(-1)^{n+1}}{(2n+3)!}}\right |=\lim_{n \to \infty } \left |\frac{(-1)^n(2n+3)!}{(-1)^{n+1}(2n+1)!}\right |=\lim_{n \to \infty } \left |\frac{(2n+3)(2n+2)}{(-1)}\right |=\lim_{n \to \infty }(2n+3)(2n+2)=\infty$$