Preliminary Test of Alternating Geometric Series

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
7 replies · 2K views
The-Mad-Lisper
Messages
12
Reaction score
1

Homework Statement


[tex]\lim_{n \to \infty}\frac{(-1)^{n+1} \cdot n^2}{n^2+1}[/tex]

Homework Equations


[tex]\lim_{n \to \infty}a_n \neq 0 \rightarrow S \ is \ divergent[/tex]

The Attempt at a Solution


I tried L'Hopital's rule, but I could not figure out how to find the limit of that pesky [itex](-1)^{n+1}[/itex].

Edit: This is not actually a geometric series, disregard that part of the title.
 
on Phys.org
The-Mad-Lisper said:

Homework Statement


[tex]\lim_{n \to \infty}\frac{(-1)^{n+1} \cdot n^2}{n^2+1}[/tex]

Homework Equations


[tex]\lim_{n \to \infty}a_n \neq 0 \rightarrow S \ is \ divergent[/tex]

The Attempt at a Solution


I tried L'Hopital's rule, but I could not figure out how to find the limit of that pesky [itex](-1)^{n+1}[/itex].

Edit: This is not actually a geometric series, disregard that part of the title.

Is there a question somewhere here? You seem to have arrived at the correct conclusion for the correct reason, so what are you unsure about?
 
Computing the derivative of an exponential function results in another exponential function, which doesn't really help when it comes to using l'hospital's rule.
 
Your post is very confusing! You title this "alternating geometric series" but appear to be asking about a "sequence" rather than a "series". Further this is not a "geometric" series or sequence. If the "problem statement" is to determine whether or not the series [itex]\sum_{n=0}^\infty \frac{(-1)^n n^2}{n^2+ 1}[/itex] converges then you should know that an alternating series, [itex]\sum_{n=0}^\infty a_n[/itex], converges if and only if the sequence [itex]a_n[/itex] converges to 0. Finally, you ask about the limit of [itex](-1)^n[/itex]. That sequence does not converge so has no limit but that is irrelevant to this problem.
 
  • Like
Likes   Reactions: The-Mad-Lisper
try dividing the denominator and the numerator by n^2. you'll be left out with ones and one/infinities which go to zero then you're left with a simple equation.
 
An alternating series is of the form ##\sum_n (-1)^na_n##. The limit in your test concerns ##a_n##, not ##(-1)^na_n##.
Capture.PNG