How do you test the alternating series Sigma (-1)^(n+1)n/(10n+5) from n is [1,oo) for convergence?

1 Answer
Apr 16, 2018

Diverges by the Divergence Test.

Explanation:

The Alternating Series Test tells us that if we have a series sum(-1)^nb_n, where b_n is a sequence of positive terms, the series converges if

a) b_n>=b_(n+1), IE, the sequence is ultimately decreasing for all n.

b) lim_(n->oo)b_n=0

We should note that we don't need to have (-1)^(n+1) -- any term that causes alternating signs, such as cos(npi), (-1)^(n-1), (-1)^(n+1), is okay.

Here, we see b_n=n/(10n+5). Taking the limit,

lim_(n->oo)n/(10n+5)=1/10 ne 0 -- the Alternating Series Test is inconclusive here.

However, take the limit of the overall sequence:

lim_(n->oo)(-1)^(n+1)n/(10n+5) ne 0 -- we can say this because although the limit does not truly exist, we can convince ourselves that it alternates signs and gets closer to 1/10, causing divergence by the Divergence Test.