Alternating Series Test — Question 3

Section 11-8 · Calculus pages 2085–2086

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*Question 3 Consider ∑n=1∞(−1)nn+1n\displaystyle\sum_{n=1}^{\infty}(-1)^n\frac{n+1}{n}.

  1. Evaluate the limit of the term magnitude.

  2. Apply the nth-term test before considering the AST.

  3. Explain exactly which AST condition fails.

*Question 3 – Solution

bn=n+1n=1+1n→1.b_n=\frac{n+1}{n}=1+\frac1n\longrightarrow1. Thus the series terms an=(−1)nbna_n=(-1)^nb_n alternate near ±1\pm1 and do not approach zero.

Every convergent series must satisfy an→0a_n\to0. Here the even subsequence tends to 11 and the odd subsequence tends to −1-1, so the term sequence has no limit and the series diverges.

Although bnb_n is positive and decreasing, the required condition bn→0b_n\to0 fails. Alternating signs cannot compensate for terms of persistent size.