Discussing the Convergence of a Series: Get My Opinion!

Click For Summary
The discussion centers on the convergence of a series represented by the ratio (n^2 + 3n + 1)/(n^2 + 5n + 7). The initial claim suggests it can be treated as a geometric series, but others argue that this comparison is invalid due to the nature of the function involved. To analyze convergence, it is recommended to evaluate the limit of (n^2 + 3n + 1)/(n^2 + 5n + 7) raised to the power of n^2, which has an indeterminate form. Using logarithms and L'Hopital's Rule is suggested as a method to resolve this limit. Ultimately, establishing that the limit does not approach zero indicates that the series diverges.
Amaelle
Messages
309
Reaction score
54
Homework Statement
studying the convergence of a serie (look at the image)
Relevant Equations
geometric serie, convergence
Good day
I want to study the connvergence of this serie

1612182366542.png

I already have the solution but I want to discuss my approach and get your opinion about it
it s clear that n^2+5n+7>n^2+3n+1 so 0<(n^2+3n+1)/(n^2+5n+7)<1 so we can consider this as a geometric serie that converge?
many thanks in advance
 
Last edited by a moderator:
Physics news on Phys.org
You can't compare to a geometric series; you have a function of n raised to a power which depends on n. That is similar to <br /> \left(1 - \frac 1n\right)^{n} \to e^{-1} &gt; 0. For that reason \sum_{n=1}^\infty \left(1 - \frac1n\right)^n does not converge.

I would write <br /> \frac{n^2 + 3n + 1}{n^2 + 5n +7} = 1 - \frac{2n + 6}{n^2 + 5n +7} and take logs.
 
Amaelle said:
Homework Statement:: studying the convergence of a serie (look at the image)
Relevant Equations:: geometric serie, convergence

Good day
I want to study the connvergence of this serie

View attachment 277247
I already have the solution but I want to discuss my approach and get your opinion about it
it s clear that n^2+5n+7>n^2+3n+1 so 0<(n^2+3n+1)/(n^2+5n+7)<1 so we can consider this as a geometric serie that converge?
many thanks in advance
If you can establish the fact that ##\lim_{n \to \infty}\left( \frac{n^2 + 3n + 1}{n^2 + 5n + 7}\right)^{n^2} \ne 0##, then you can conclude that the series diverges. This limit has the indeterminate form ##[1^\infty]##, so the best way of determining the limit is by the use of logarithms, and getting it to a form in which L'Hopital's Rule can be applied.
 
pasmith said:
You can't compare to a geometric series; you have a function of n raised to a power which depends on n. That is similar to <br /> \left(1 - \frac 1n\right)^{n} \to e^{-1} &gt; 0. For that reason \sum_{n=1}^\infty \left(1 - \frac1n\right)^n does not converge.

I would write <br /> \frac{n^2 + 3n + 1}{n^2 + 5n +7} = 1 - \frac{2n + 6}{n^2 + 5n +7} and take logs.
thanks you so much!
 
Mark44 said:
If you can establish the fact that ##\lim_{n \to \infty}\left( \frac{n^2 + 3n + 1}{n^2 + 5n + 7}\right)^{n^2} \ne 0##, then you can conclude that the series diverges. This limit has the indeterminate form ##[1^\infty]##, so the best way of determining the limit is by the use of logarithms, and getting it to a form in which L'Hopital's Rule can be applied.
thanks so much!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
29
Views
5K
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K