How do you determine whether the sequence #a_n=(-1)^n/sqrtn# converges, if so how do you find the limit?

Answer 1

Limit is 0

#0 <= abs((-1)^n/sqrt(n)) = 1/sqrt(n)#
But #lim_{n to oo} sqrt(n) = + oo#, therefore #lim_{n to oo} 1/sqrt(n) = 0#.

Now, use the theorem of comparison to conclude.

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Answer 2

To determine whether the sequence (a_n = \frac{(-1)^n}{\sqrt{n}}) converges, you can analyze its behavior as (n) approaches infinity.

Firstly, note that the sequence oscillates between positive and negative values as (n) increases.

Secondly, consider the behavior of (\frac{1}{\sqrt{n}}) as (n) approaches infinity. As (n) grows larger, the denominator (\sqrt{n}) also grows, which causes the fraction (\frac{1}{\sqrt{n}}) to approach zero.

However, because the sequence (a_n) alternates between positive and negative values, it does not converge to a single limit.

Therefore, the sequence (a_n = \frac{(-1)^n}{\sqrt{n}}) does not converge.

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Answer 3

To determine whether the sequence (a_n = \frac{(-1)^n}{\sqrt{n}}) converges, we can analyze its behavior as (n) approaches infinity.

Since (n) is approaching infinity, the oscillating term ((-1)^n) alternates between (-1) and (1) as (n) increases. However, the denominator (\sqrt{n}) grows without bound as (n) increases.

For even values of (n), (a_n = \frac{1}{\sqrt{n}}), and for odd values of (n), (a_n = \frac{-1}{\sqrt{n}}). As (n) approaches infinity, both (\frac{1}{\sqrt{n}}) and (\frac{-1}{\sqrt{n}}) tend towards (0).

Therefore, the sequence (a_n) converges to (0) as (n) approaches infinity.

The limit of the sequence (a_n) as (n) approaches infinity is (0).

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Answer 4

To determine whether the sequence (a_n = \frac{(-1)^n}{\sqrt{n}}) converges, we can analyze its behavior as (n) approaches infinity.

The sequence alternates between positive and negative terms as (n) changes. However, the terms (1/\sqrt{n}) decrease as (n) increases, approaching zero.

To find the limit, we need to consider two cases: when (n) is even and when (n) is odd.

  1. When (n) is even, (a_n = \frac{1}{\sqrt{n}}). As (n) approaches infinity, (1/\sqrt{n}) approaches zero.

  2. When (n) is odd, (a_n = \frac{-1}{\sqrt{n}}). As (n) approaches infinity, (-1/\sqrt{n}) also approaches zero.

Since both cases converge to zero, the sequence (a_n) converges to zero as (n) approaches infinity.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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