Ratio Test — Question 1

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Question 1

Consider ∑n=1∞n!3n\displaystyle\sum_{n=1}^{\infty}\frac{n!}{3^n}.

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

  2. Decide whether the series converges.

  3. Explain what the calculation says about factorial growth compared with 3n3^n.

Original worksheet page 1: question and worked solution for 4-10-001
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Question 1 – Solution

Step 1: Identify the term.

Let an=n!/3na_n=n!/3^n. Its terms are positive, so absolute values do not change the ratio.

Step 2: Simplify before taking the limit.

an+1an=(n+1)!3n+13nn!=n+13.\frac{a_{n+1}}{a_n}=\frac{(n+1)!}{3^{n+1}}\frac{3^n}{n!}=\frac{n+1}{3}.

Step 3: Interpret the ratio.

Thus L=lim⁡n→∞(n+1)/3=∞>1\displaystyle L=\lim_{n\to\infty}(n+1)/3=\infty>1. The terms eventually increase; indeed, an+1>ana_{n+1}>a_n for n>2n>2. Therefore ana_n cannot approach 00.

Conclusion.

The series diverges by the Ratio Test (and by the nth-term test). The unbounded ratio shows that n!n! eventually grows faster than 3n3^n.

Original worksheet page 2: question and worked solution for 4-10-001

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