Convergence condition

Infinite geometric series appear frequently in HSC exam questions, testing your understanding of convergence conditions and sum formulas. It's a core part of the Advanced course and often appears in multi-part questions. In this problem, we'll work through a classic type of question where the common ratio contains an unknown parameter $m$, and we must find the values that make the series converge, then compute its limiting sum. This exercise consolidates your skills with absolute value inequalities and series notation.

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Problem Statement

Consider the geometric series

$$\sum_{n=0}^{\infty} 7\left(\frac{m-1}{3}\right)^n.$$

Find all real values of $m$ for which the series converges, and then state its sum.

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Hints

Use the condition $|r|<1$ for convergence of an infinite GP.

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Solutions

First, we identify the common ratio $r$ of the series. Since the series is geometric with first term $7$ (the $n=0$ term), the ratio is the factor raised to the $n$th power:

$$r=\frac{m-1}{3}.$$

For an infinite geometric series $\sum_{n=0}^\infty ar^n$, convergence occurs if and only if $|r|<1$. Applying this condition gives

$$\left|\frac{m-1}{3}\right|<1 \implies |m-1|<3 \implies -2<m<4.$$

Since we have established the convergence condition, we can safely use the sum to infinity formula. Now that we know the series converges for $-2<m<4$, we can find its sum to infinity using $S_\infty = \frac{a}{1-r}$, with $a=7$. Substituting $r$ yields

$$S_\infty=\frac{7}{1-\frac{m-1}{3}}=\frac{21}{4-m}.$$

Takeaways

• The convergence condition for an infinite geometric series is $|r|<1$ (strict inequality).
• The sum formula $S_\infty = \frac{a}{1-r}$ is only valid when the series converges.
• Solving $|m-1|<3$ yields an open interval $(-2,4)$; be careful with inequality signs.

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Further Readings

HSC Vectors, HSC Polynomials, HSC Integrals, HSC Mechanics