Ratio Test — Question 7

Section 11-10 · Calculus pages 2133–2134

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*Question 7 Determine whether ∑n=1∞n!n10\displaystyle\sum_{n=1}^{\infty}\frac{n!}{n^{10}} converges or diverges.

  1. Simplify an+1/ana_{n+1}/a_n.

  2. Use its limiting behavior to justify the conclusion.

  3. Explain why the nth-term test also applies.

*Question 7 – Solution

an+1an=(n+1)!(n+1)10n10n!=(n+1)(nn+1)10=n(1+1n)−9.\begin{align*} \frac{a_{n+1}}{a_n}&=\frac{(n+1)!}{(n+1)^{10}}\frac{n^{10}}{n!}=(n+1)\left(\frac{n}{n+1}\right)^{10}\\&=n\left(1+\frac1n\right)^{-9}. \end{align*}

The factor (1+1/n)−9→1(1+1/n)^{-9}\to1, while n→∞n\to\infty. Hence L=∞>1L=\infty>1.

The series diverges. Its terms eventually grow rather than approach zero, so the nth-term divergence test confirms the result. A factorial eventually dominates every fixed power n10n^{10}.