Preliminary Test of Alternating Geometric Series

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SUMMARY

The discussion centers on the limit of the sequence defined by \lim_{n \to \infty}\frac{(-1)^{n+1} \cdot n^2}{n^2+1}. Participants clarify that this is not a geometric series, but rather a sequence. The conclusion drawn is that the sequence does not converge, as \lim_{n \to \infty} a_n \neq 0 indicates divergence. The use of L'Hôpital's rule is deemed ineffective for this problem, and the correct approach involves simplifying the expression by dividing the numerator and denominator by n^2.

PREREQUISITES
  • Understanding of limits in calculus
  • Familiarity with L'Hôpital's rule
  • Knowledge of alternating series convergence criteria
  • Ability to manipulate algebraic expressions involving limits
NEXT STEPS
  • Study the criteria for convergence of alternating series
  • Learn about the application of L'Hôpital's rule in different contexts
  • Explore the concept of sequences versus series in calculus
  • Practice simplifying limits by dividing by the highest power in the denominator
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Students and educators in calculus, particularly those focusing on sequences and series, as well as anyone seeking to clarify the distinction between geometric series and other types of sequences.

The-Mad-Lisper
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Homework Statement


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

Homework Equations


\lim_{n \to \infty}a_n \neq 0 \rightarrow S \ is \ divergent

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 (-1)^{n+1}.

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

Homework Statement


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

Homework Equations


\lim_{n \to \infty}a_n \neq 0 \rightarrow S \ is \ divergent

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 (-1)^{n+1}.

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.
 
The-Mad-Lisper said:
\lim_{n \to \infty}a_n \neq 0 \rightarrow S \ is \ divergent
This is a rule for series. Why is this relevant? There are no series in your post.
 
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 \sum_{n=0}^\infty \frac{(-1)^n n^2}{n^2+ 1} converges then you should know that an alternating series, \sum_{n=0}^\infty a_n, converges if and only if the sequence a_n converges to 0. Finally, you ask about the limit of (-1)^n. That sequence does not converge so has no limit but that is irrelevant to this problem.
 
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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
 
The question in the first post has been answered, and the OP hasn't been back for a couple of years, so I'm closing this thread.
 

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