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Calculus IIChapter 4: Series and SequencesSection 4-9

Absolute Convergence and Divergence — Question 7

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Question 7

Consider ∑n=0∞(−1)n2nn!\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n2^n}{n!}.

  1. Apply the Ratio Test to the absolute-value series.

  2. Classify convergence.

  3. Identify the exact sum using the exponential series.

Original worksheet page 1: question and worked solution for 4-9-007
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Question 7 – Solution

Step 1: Test absolute values.

Let bn=2n/n!b_n=2^n/n!. Then bn+1bn=2n+1→0<1.\frac{b_{n+1}}{b_n}=\frac{2}{n+1}\longrightarrow0<1. The Ratio Test proves that ∑bn\sum b_n converges.

Step 2: Classify.

Therefore the signed series converges absolutely. More generally, replacing 22 by any fixed real base cc gives ratio |c|/(n+1)→0|c|/(n+1)\to0.

Step 3: Find the sum.

From ex=∑n=0∞xn/n!e^x=\sum_{n=0}^{\infty}x^n/n!, substitute x=−2x=-2: ∑n=0∞(−1)n2nn!=e−2.\sum_{n=0}^{\infty}\frac{(-1)^n2^n}{n!}=e^{-2}.

Original worksheet page 2: question and worked solution for 4-9-007

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