Convergence and Divergence of Series — Question 5

Question 5 – Solution

Step 1: Simplify before selecting a test.

Use $(n+1)!=(n+1)n!$ to cancel the common factorial: $$\frac{n!}{(n+1)!}=\frac1{n+1}.$$

Step 2: Rewrite the partial sums.

Thus $$s_N=\sum_{n=1}^N\frac1{n+1}=\sum_{k=2}^{N+1}\frac1k=H_{N+1}-1,$$ where $H_m$ is the $m$th harmonic number.

Step 3: Apply the known harmonic behavior.

Because $H_{N+1}\to\infty$, subtracting the fixed number $1$ does not change the divergence. Therefore the partial sums are unbounded and $$\boxed{\displaystyle\sum_{n=1}^{\infty}\frac{n!}{(n+1)!}\text{ diverges}.}$$

Step 4: Interpret and verify.

The terms $1/(n+1)$ do approach zero, so the nth-term test alone is inconclusive. The factorial notation suggests rapid growth, but all of $n!$ cancels, leaving exactly a shifted harmonic series. Removing or shifting finitely many harmonic terms never changes convergence or divergence.