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Calculus IIChapter 4: Series and SequencesSection 4-10

Ratio Test — Question 8

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Question 8

Analyze ∑n=1∞2n23n2\displaystyle\sum_{n=1}^{\infty}\frac{2^{n^2}}{3^{n^2}}. Rewrite its term using a single base, compute the consecutive-term ratio, and classify the series.

Original worksheet page 1: question and worked solution for 4-10-008
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Question 8 – Solution

Step 1: Combine the powers.

an=(2/3)n2\displaystyle a_n=(2/3)^{n^2}.

Step 2: Compute the ratio.

an+1an=(23)(n+1)2−n2=(23)2n+1.\frac{a_{n+1}}{a_n}=\left(\frac23\right)^{(n+1)^2-n^2}=\left(\frac23\right)^{2n+1}. Because 0<2/3<10<2/3<1, this expression tends to 00.

Conclusion.

The ratio limit is L=0<1L=0<1, so the series converges. Its quadratic exponent makes the terms decay faster than those of ∑(2/3)n\sum(2/3)^n.

Original worksheet page 2: question and worked solution for 4-10-008

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