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

1 Answer
John D.
Jun 10, 2017

#sum_(n=1)^oo(-1)^(n+1)(1+1/n)# diverges.

Explanation:

The alternating series test basically says this:

If a series is alternating, then as long as the ABSOLUTE VALUE of each term is less than the ABSOLUTE VALUE of the previous term, the series will converge.

BUT, we must first check that #lim_(n->oo)a_n = 0# (the nth term test).

#lim_(n->oo)(-1)^(n+1)(1+1/n) = +-1 != 0#

Not only do the terms of this series NOT approach #0# as #n -> oo#, but also they approach the series #sum_(n=1)^oo(-1)^n# which is known to be divergent.

Therefore, the series diverges.

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NOTE: If you forget to check using the n-th term test before using the Alt. Series Test, you will usually get the wrong answer. For example, we could have proven that each term of this series is closer to #0# than the last term, so the series meets the criteria for the Alt. Series Test. We would then falsely assume convergence even though the series approaches a divergent sum.

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