Mathematics

Mathematics

Problem

Consider the infinite series ∑n=0∞(−1)n+143n\displaystyle \sum_{n=0}^{\infty}(-1)^{n+1}\dfrac{4}{3^{n}}. When the series is written in the standard geometric form ∑n=0∞arn\sum_{n=0}^{\infty} ar^{n}, which of the following is the common ratio rr? a) −43-\dfrac43 b) \dfrac13 c) −13-\dfrac13 d) \dfrac43
Intuitive explanation

This is a geometric series in disguise—your job is to rewrite it in the standard form ∑n=0∞arn\sum_{n=0}^{\infty} ar^{n} so you can identify the common ratio rr. The key is to separate the (−1)n+1(-1)^{n+1} factor from the rest by recognizing that (−1)n+1=(−1)⋅(−1)n(-1)^{n+1} = (-1) \cdot (-1)^n, which lets you pull out a constant and absorb the alternating sign into the ratio itself. Once you've rewritten the series cleanly, the common ratio will be whatever multiplies by itself each step.

Intuitive explanation

This is a geometric series in disguise—your job is to rewrite it in the standard form ∑n=0∞arn\sum_{n=0}^{\infty} ar^{n} so you can identify the common ratio rr. The key is to separate the (−1)n+1(-1)^{n+1} factor from the rest by recognizing that (−1)n+1=(−1)⋅(−1)n(-1)^{n+1} = (-1) \cdot (-1)^n, which lets you pull out a constant and absorb the alternating sign into the ratio itself. Once you've rewritten the series cleanly, the common ratio will be whatever multiplies by itself each step.

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