Which comparison correctly proves that ∑_n=1^∞13^n-1 converges?

13^n-1≤32·13^n for n≥1; the geometric series converges

13^n-1≤13^n for n≥1; the geometric series converges

13^n-1≥13^n for n≥1, hence the series converges

13^n-1≤1n for n≥1; the harmonic series converges

Exercise · ZW-COMPARISON-TEST-008 · ★★☆ Intermediate

Which comparison correctly proves that \(\sum_{n=1}^{\infty}\frac1{3^n-1}\) converges?

Choose one option.

Show answer

C. \(\dfrac1{3^n-1}\le\dfrac32\cdot\dfrac1{3^n}\) for \(n\ge1\); the geometric series converges

Explanation

Since \(3^n-1\ge\tfrac23\,3^n\) for \(n\ge1\), the bound with factor \(\tfrac32\) holds. Without that factor the inequality is false; with \(\ge\) the direction is useless; and the harmonic series diverges.

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Review the concept: Comparison test

A test that decides whether a series \(\sum a_n\) of nonnegative terms converges by comparing it with another series \(\sum b_n\) such that \(a_n\le b_n\).

\[0\le a_n\le b_n\ \Longrightarrow\ \Big(\sum b_n\text{ converges}\Rightarrow\sum a_n\text{ converges}\Big)\]

Which comparison correctly proves that \(\sum_{n=1}^{\infty}\frac1{3^n-1}\) converges?

Explanation

Review the concept: Comparison test

More practice problems: comparison test