Do you know the squeeze theorem?sara_87 said:Homework Statement
Find the limit as n tends to [itex]\infty[/itex] of:
[itex]\frac{(-1)^n}{n}[/itex]
Homework Equations
The Attempt at a Solution
I know that (-1)^n alternates between 1 and -1.
I also know that the limit of 1/n is 0. But I don't know how to compute the above limit.
Any ideas will be appreciated. Thank you.
The limit of (-1)^n/n as n approaches infinity is equal to 0. This can be proven using the squeeze theorem or by observing that as n gets larger and larger, the value of (-1)^n alternates between 1 and -1, while n continues to increase, ultimately approaching infinity. Therefore, the fraction (-1)^n/n gets closer and closer to 0, resulting in a limit of 0.
To evaluate the limit of (-1)^n/n as n approaches infinity, you can use the concept of the squeeze theorem or direct substitution. The squeeze theorem states that if a function is always between two other functions as x approaches a certain value, then the limit of the function must also approach that value. In the case of (-1)^n/n, we can show that it is always between -1/n and 1/n, which both approach 0 as n approaches infinity. Therefore, the limit of (-1)^n/n must also be 0.
No, the limit of (-1)^n/n as n approaches infinity cannot be undefined. This is because as n gets larger and larger, the fraction (-1)^n/n will always approach 0. Even if n is an odd number, resulting in a negative value for (-1)^n, the overall fraction will still approach 0 as n approaches infinity.
Yes, the limit of (-1)^n/n as n approaches infinity is the same as the limit of (-1)^n+1/n as n approaches infinity. This is because when n approaches infinity, the value of n+1 is essentially the same as n. Therefore, the exponents of (-1) will still alternate between even and odd, resulting in the same limit of 0.
The limit of (-1)^n/n as n approaches infinity is an example of a convergent sequence in mathematics. A convergent sequence is a sequence of numbers that approaches a specific value as the number of terms in the sequence increases. In this case, as n approaches infinity, the values of the fraction (-1)^n/n get closer and closer to 0, resulting in a convergent sequence with a limit of 0.