How do you test the alternating series #Sigma (-1)^nsqrtn# from n is #[1,oo)# for convergence?

Answer 1

Samantha Adkison

As #lim_(n->oo) sqrtn = +oo# the series does not satisfy Cauchy's necessary condition and thus cannot be convergent.

Besides let #a_n = (-1)^n sqrtn#. Clearly #a_(2k) > 0# and #a_(2k+1) < 0# so the series oscillates indefinitely.

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Answer 2

Evelyn Agard

To test the alternating series ( \sum_{n=1}^{\infty} (-1)^n \sqrt{n} ) for convergence, you can use the Alternating Series Test. Here are the steps:

Check the sequence ( \sqrt{n} ) for monotonic decrease to ( 0 ) as ( n ) approaches infinity.
Verify that the terms of the series are alternating in sign.
Apply the Limit Test to verify that the limit of the absolute value of the terms as ( n ) approaches infinity is ( 0 ).

If all these conditions are met, then the alternating series ( \sum_{n=1}^{\infty} (-1)^n \sqrt{n} ) 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.