Alternating Series Test — Question 6

Question 6 – Solution

Step 1: Check the domain and AST template.

At $n=1$ the denominator is $1-1=0$, so the series must start at $2$. Magnitudes satisfy $b_{2k}=1/(2k+1)$ and $b_{2k+1}=1/(2k)$; hence $b_{2k+1}>b_{2k}$ infinitely often, so the standard monotonic AST does not apply.

Step 2: Pair terms.

$$a_{2k}+a_{2k+1}=\frac1{2k+1}-\frac1{2k}=-\frac1{2k(2k+1)}.$$ The paired series converges absolutely by comparison with $1/(4k^2)$. Since the unpaired final term tends to zero, both even and odd partial-sum subsequences approach the same limit; the original series converges.

Step 3: Test absolute convergence.

$$|a_{2k}|+|a_{2k+1}|=\frac1{2k+1}+\frac1{2k}\ge\frac1{2k}.$$ Thus the absolute series diverges, so convergence is conditional.