# Help power series

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 $$R \to \infty$$

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).

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 dont understand

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

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

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

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