Question

Derive the formula (1.4) for the sum \(S_{n}\) of the geometric progression \(S_{n}=a+a r+\) \(a r^{2}+\cdots+a r^{n-1} .\) Hint: Multiply \(S_{n}\) by \(r\) and subtract the result from \(S_{n} ;\) then solve for \(S_{n} .\) Show that the geometric series (1.6) converges if and only if \(|r|<1\); also show that if \(|r|<1,\) the sum is given by equation (1.8).

Short answer

Sum: \(S_n = \frac{a(1 - r^n)}{1 - r}\). For \(|r|<1\): Sum converges to \(\frac{a}{1 - r}\).

Step-by-step solution

Write the sum of the geometric progression

Start by writing the sum of the geometric progression: \[ S_n = a + ar + ar^2 + \ ar^3 + ... + ar^{n-1} \]

Multiply both sides by r

Multiply both sides of the equation by the common ratio, r: \[ rS_n = ar + ar^2 + ar^3 + ... + ar^n \]

Subtract the second equation from the first

Subtract the equation obtained by multiplication from the original sum: \[ S_n - rS_n = (a + ar + ar^2 + ... + ar^{n-1}) - (ar + ar^2 + ... + ar^n) \]

Simplify the left-hand side

Factor out \(S_n\) on the left-hand side: \[ S_n(1 - r) = a - ar^n \]

Solve for \(S_n\)

Isolate \(S_n\) to get the sum formula: \[ S_n = \frac{a(1 - r^n)}{1 - r} \]

Determine the condition for convergence

To determine the convergence of the geometric series, consider the condition \(|r|<1\). As \(n\) approaches infinity, the term \(r^n\) approaches 0. Therefore, if \(|r|<1\), \[ \lim_{n \to \infty} S_n = \frac{a}{1 - r} \]

Key concepts

Sum of Geometric Series

The geometric series is a sum of terms where each term after the first is found by multiplying the previous term by a constant, called the common ratio (r).The formula for the sum of a geometric series can be derived easily.Step 1 illustrates the sum of such series. As follows:

• Step 2: Multiply both sides by the common ratio, r, transforming the series into one that aligns neatly under the original when subtracted.
• Step 3: Subtract these two series to eliminate most terms, making further simplification possible.
• Step 4: Factorizing and simplifying these equations allows for isolating the sum, \( S_n \).
• Step 5: Solve the simplified equation to obtain the formula: \[ S_n = \frac{a(1 - r^n)}{1 - r} \]

This formula helps us find the sum of any geometric series without needing to add all terms manually. Each variable plays a key role: 'a' is the first term, 'r' is the common ratio, and 'n' is the number of terms.

Convergence Criteria

Understanding when a geometric series converges is crucial. Convergence tells us if adding up infinitely many terms will produce a finite sum. A series converges when it approaches a specific value as the number of terms goes to infinity. For a geometric series to converge, the common ratio \( r \) must satisfy the condition \( |r| < 1 \). This ensures that the terms get smaller and smaller, eventually approaching zero. This convergence condition can be mathematically shown:

• For \( |r| < 1 \), as 'n' (the number of terms) approaches infinity, \( r^n \) approaches zero.
• This results in the series sum being simplified further as most terms become negligible.
• Therefore, the sum of an infinite geometric series is: \[ \text{Sum} = \frac{a}{1 - r} \ \text{if} \|r\| < 1 \]

When the value of the common ratio r is outside this range, the series does not converge, meaning it doesn't settle towards a distinctive finite value.

Formula Derivation

Deriving the formula for the sum of a geometric progression involves elementary algebraic methods, making it understandable and approachable. Reviewing each step:

• Start with the general form: \[ S_n = a + ar + ar^2 + ... + ar^{n-1} \]
• Multiply the whole series by 'r': \[ rS_n = ar + ar^2 + ar^3 + ... + ar^n \]
• Subtract the second equation from the first: \[ S_n - rS_n = a - ar^n \]
• Simplify and isolate 'S_n' by factoring and rearranging terms: \[ S_n(1 - r) = a(1 - r^n) \]
• Solve for 'S_n': \[ S_n = \frac{a(1 - r^n)}{1 - r} \]

This formula assists us efficiently in calculating the sum of any finite geometric series, giving a structured method to achieve the desired result. This specific derivation also highlights the power of algebraic manipulation in simplifying seemingly complex series.