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

Question over a Series

  • #1

Homework Statement


I was watching a PatrickJMT video on ratio test and he gave this problem. I solved it before he did, and he got that it was divergent. He didn't simplify it initially, so our methods of approach are different. Did I do something wrong? I checked my calculator to make sure all my simplifications were correct. So I'm assuming I did something near the limit portion that was incorrect (if at all). Thanks


∑ n/31+3n
n=1

Homework Equations




The Attempt at a Solution



∑ n/31+3n = n/(3⋅27n)
n=1

lim |(n+1)/(3⋅27n+1) / (n/3 *27n)|
n→∞

= lim |(n+1)/n| * (1/27)
n→∞
= 1 * 1/27
=1/27 < 1

Convergent by Ratio Test
 

Answers and Replies

  • #2
Math_QED
Science Advisor
Homework Helper
2019 Award
1,392
515
Did you use the ratio test correctly?
 
  • #3
IMG_1809.jpg
Did you use the ratio test correctly?
Nope. Got it, thanks.

[edit]

Actually I typed my work incorrectly. I'll fix it. but to be sure...
 
  • #4
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728

Homework Statement


I was watching a PatrickJMT video on ratio test and he gave this problem. I solved it before he did, and he got that it was divergent. He didn't simplify it initially, so our methods of approach are different. Did I do something wrong? I checked my calculator to make sure all my simplifications were correct. So I'm assuming I did something near the limit portion that was incorrect (if at all). Thanks


∑ n/31+3n
n=1

Homework Equations




The Attempt at a Solution



∑ n/31+3n = n/(3⋅27n)
n=1

lim |(n+1)/(3⋅27n+1) / (n/3 *27n)|
n→∞

= lim |(n+1)/n| * (1/27)
n→∞
= 1 * 1/27
=1/27 < 1

Convergent by Ratio Test
It is convergent, as you have shown correctly using the ratio test.

You can actually find the sum ##\sum_{n=1}^N n/3^{3n+1}## explicitly (as a reasonably simple closed-from expression in ##N##), then take the limit as ##N \to \infty##. All you need to do is find a formula for ##f_N(x) =\sum_{n=1}^N n x^n##, and these are widely available, or you can work it out as ##f_N(x) = x (d/dx) g_N(x)##, where ##g_N(x) = \sum_{n=1}^N x^n##, which is elementary.
 
Last edited:

Related Threads on Question over a Series

Replies
4
Views
1K
Replies
1
Views
4K
  • Last Post
Replies
0
Views
976
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
7
Views
638
  • Last Post
Replies
3
Views
433
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
4
Views
965
Replies
4
Views
3K
Replies
2
Views
3K
Top