• Support PF! Buy your school textbooks, materials and every day products Here!

Ratio Test

  • Thread starter fauboca
  • Start date
  • #1
158
0
[tex]\displaystyle\sum_{n = 1}^{\infty}\frac{(n!)^3}{(3n)!}z^n , \ z\in\mathbb{C}[/tex]

By the ratio test,
[tex]
\displaystyle L = \lim_{n\to\infty}\left|\frac{[(n + 1)!]^3 z^{n + 1} (3n)!}{[3(n + 1)]! (n!)^3 z^n}\right| = \lim_{n\to\infty}\left|\frac{z (n + 1)^2}{3}\right| = \infty.[/tex]

Therefore, the series diverges and there is no radius of convergence.

Correct?
 
Last edited:

Answers and Replies

  • #2
lanedance
Homework Helper
3,304
2
i don't think your cancellation is quite correct 3(n+1)=3n+3
 
  • #3
158
0
i don't think your cancellation is quite correct 3(n+1)=3n+3
Thanks, I see the problem.

[tex]|z|<27[/tex]

So [itex]R = 27[/itex] then
 
Last edited:
  • #4
lanedance
Homework Helper
3,304
2
can you show how you got there?
 
Last edited:
  • #5
158
0
can you show how you got there?
[tex]\lim_{n\to\infty}\left|\frac{z (n + 1)^3}{(3n + 3)(3n + 2)(3n + 1)}\right|[/tex]


[tex]\Rightarrow \lim_{n\to\infty}|z|\left(\frac{n^3}{27n^3}\right) = \frac{1}{27}|z| < 1 \Rightarrow |z| < 27[/tex]
 
  • #6
lanedance
Homework Helper
3,304
2
yeah looks good

you can just just take the result below if you like , rather than carrying the z through though its good to understand why
[tex] r = \lim_{n \to \infty} \left|\frac{c_n}{c_{n+1}}\right| [/tex]
 

Related Threads for: Ratio Test

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
11
Views
685
  • Last Post
Replies
1
Views
688
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
942
  • Last Post
Replies
7
Views
782
  • Last Post
Replies
17
Views
3K
Top