How do you test for convergence given #Sigma (-1)^n n^(-1/n)# from #n=[1,oo)#?

Answer 1

The series:

#sum_(=1)^oo (-1)^n n^(-1/n)#

is not convergent.

As the series:

#sum_(=1)^oo (-1)^n n^(-1/n)#

is alternating, we can test it for convergence using Leibniz's criteria stating that:

#sum_(n=1)^oo (-1)^n a_n# is convergent if:
# lim_(n->oo) a_n = 0# # a_(n+1) < a_n# at least for #n> N_0#

We have that:

# lim_(n->oo) n^(-1/n) = lim_(n->oo) (e^lnn)^(-1/n)= lim_(n->oo) e^(-lnn/n) = 1#

so the series is not convergent.

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Answer 2

To test for convergence of the series ∑(-1)^n * n^(-1/n) from n=1 to infinity, you can use the Alternating Series Test. This test states that if a series has alternating signs and its terms decrease in absolute value monotonically to zero, then the series converges.

Here, the series has alternating signs and the terms decrease as n increases to infinity. To show this, you can observe that as n increases, n^(-1/n) decreases since the exponent (-1/n) becomes smaller, leading to larger values for n. Additionally, n^(-1/n) approaches 1 as n approaches infinity.

Therefore, the series ∑(-1)^n * n^(-1/n) converges by the Alternating Series Test.

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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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