Home > Calculus > Alternating Series Test (Leibniz's Theorem) for Convergence of an Infinite Series

How do you determine the convergence or divergence of #Sigma ((-1)^(n))/(sqrtn)# from #[1,oo)#?

Answer 1

Paisley Johnson

The sum is convergent.

#sum_(k=1)^oo(-1)^k/sqrt(k)#

is an alternate series with #abs(a_k) < abs(a_(k-1))# so it converges.

Here #a_k = 1/sqrt(k)#.

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Answer 2

Emilia Aguilar

To determine the convergence or divergence of the series (\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}}), you can use the Alternating Series Test. This test states that if a series is alternating (meaning the signs of the terms alternate) and the absolute value of the terms decreases monotonically to zero, then the series converges. In this case, the terms (\frac{1}{\sqrt{n}}) decrease monotonically to zero as (n) increases, and the series alternates in sign. Therefore, the series converges.

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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.