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<br />