Do you know the squeeze theorem ?whatlifeforme said:Homework Statement
Converge or diverge?
Homework Equations
[itex]\displaystyle a_{n} = \frac{(-1)^{n+1}}{2n-1}[/itex]
The Attempt at a Solution
all i know is to perhaps take ln of numerator and denominator to get the exponent down below?
SammyS said:Do you know the squeeze theorem ?
whatlifeforme said:I do.
whatlifeforme said:Homework Statement
Converge or diverge?Homework Equations
[itex]\displaystyle a_{n} = \frac{(-1)^{n+1}}{2n-1}[/itex]The Attempt at a Solution
all i know is to perhaps take ln of numerator and denominator to get the exponent down below?
Seq Convergence refers to the behavior of a sequence, which is a list of numbers that follow a specific pattern. In simple terms, it is the tendency of a sequence to approach a specific value as the number of terms increases.
The formula for a_n in this sequence is (-1) raised to the power of n+1, divided by 2n-1. This means that the first term (n=1) is equal to -1/1, the second term (n=2) is equal to 1/3, the third term (n=3) is equal to -1/5, and so on.
To determine if a sequence converges or diverges, you can use the limit test. If the limit of the sequence approaches a specific value as n approaches infinity, then the sequence converges. If the limit does not exist or approaches infinity, then the sequence diverges.
The limit of this sequence is 0, as n approaches infinity. This means that the sequence converges to 0.
The sequence (-1)^(n+1)/(2n-1) is convergent, as the limit of the sequence approaches 0 as n approaches infinity. This can also be confirmed by graphing the sequence, which will show that the terms get closer and closer to 0 as n increases.