Help power series

  • #1
at the serie [tex] \sum_0^{\infty} a_n (x - c)^n [/tex], the radius of convergency is:
.

[tex]R= \lim_{n \to \infty } |\frac{a_n}{a_{n+1}}|[/tex]

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


i don't understand mainly who is [tex]a_n[/tex] .

The answer is [tex]R \to \infty[/tex]
 

Answers and Replies

  • #2
22,129
3,297
Hi alejandrito29, :smile:

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

[tex]R= \lim_{n \to \infty } |\frac{a_n}{a_{n+1}}|[/tex]??

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).
 
  • #3
Hi alejandrito29, :smile:

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

[tex]R= \lim_{n \to \infty } |\frac{a_n}{a_{n+1}}|[/tex]??

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:

[tex] \lim_{n \to \infty} |\frac{c_{n+1}}{c_n}|<1 [/tex]

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

then

[tex]|x^2|<\lim _{n \to \infty}|\frac{a_n}{a_{n+1}}}| [/tex]

[tex]-\lim _{n \to \infty}|\frac{a_n}{a_{n+1}}}|<x^2<\lim _{n \to \infty}|\frac{a_n}{a_{n+1}}}|[/tex]

but [tex]x^2>0[/tex]....I dont understand

i don't see that [tex]\lim _{n \to \infty}|\frac{a_n}{a_{n+1}}}|[/tex] is a radius
 
Last edited:
  • #4
22,129
3,297
D'Lambert is:

[tex] \lim_{n \to \infty} |\frac{c_{n+1}}{c_n}|<1 [/tex]

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

then

[tex]|x^2|<\lim _{n \to \infty}|\frac{a_n}{a_{n+1}}}| [/tex]

Indeed, so the series converges if and only if

[tex]|x|<\sqrt{\lim _{n \to \infty}|\frac{a_n}{a_{n+1}}}|}[/tex]

So the convergence radius is [tex]\sqrt{\lim _{n \to \infty}|\frac{a_n}{a_{n+1}}}|}[/tex]
 
  • #5
Indeed, so the series converges if and only if

[tex]|x|<\sqrt{\lim _{n \to \infty}|\frac{a_n}{a_{n+1}}}|}[/tex]

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

very thanks you
 
  • #6
Here is your answer
[tex]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
[/tex]
 

Related Threads on Help power series

  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
4
Views
711
Replies
4
Views
2K
Replies
2
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
2
Views
1K
Top