{ "slug": "geometric-series-summation", "coreConcept": "A **geometric series** is the sum of the terms of a geometric progression. The sum of the first `$n$` terms is given by `$S_n = \\frac{a(1-r^n)}{1-r}$`, where `$a$` is the first term and `$r$` is the common ratio.\n\nFor a **convergent** geometric series, where the magnitude of the common ratio is less than one (`$|r|<1$`), the sum of an infinite number of terms approaches a finite value, known as the **sum to infinity**, given by `$S_\\infty = \\frac{a}{1-r}$`.", "mainText": "### Introduction to Geometric Series\n\nA **geometric progression** is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the **common ratio**, denoted by `$r$`. For a sequence with first term `$a$`, the terms are `$a, ar, ar^2, ar^3, \\dots$`, and the `$n^{th}$` term is given by the formula `$u_n = ar^{n-1}$`.\n\nA **geometric series** is formed by adding the terms of a geometric progression together. For instance, the sequence `$3, 6, 12, 24, \\dots$` is a geometric progression with `$a=3$` and `$r=2$`. The corresponding geometric series is `$3 + 6 + 12 + 24 + \\dots$`.\n\n### The Sum of a Finite Geometric Series\n\nTo find the sum of the first `$n$` terms of a geometric series, denoted `$S_n$`, we can derive a general formula. This avoids the need to add up each term individually, which would be inefficient for a large number of terms. The derivation itself is a valuable piece of mathematical reasoning.\n\nLet the sum be `$S_n = a + ar + ar^2 + \\dots + ar^{n-2} + ar^{n-1}$`. If we multiply this entire equation by the common ratio `$r$`, we obtain `$rS_n = ar + ar^2 + ar^3 + \\dots + ar^{n-1} + ar^n$`. Now, observe that most of the terms in the expressions for `$S_n$` and `$rS_n$` are identical. Subtracting the second equation from the first yields:\n`$S_n - rS_n = (a + ar + \\dots + ar^{n-1}) - (ar + ar^2 + \\dots + ar^n)$`\n`$S_n(1-r) = a - ar^n$`\n`$S_n(1-r) = a(1-r^n)$`\n\nProvided that `$r \\neq 1$`, we can divide by `$(1-r)$` to arrive at the formula for the sum of the first `$n$` terms:\n

\n`$S_n = \\frac{a(1-r^n)}{1-r}$`\n

\nAn alternative form, derived by multiplying the numerator and denominator by -1, is `$S_n = \\frac{a(r^n-1)}{r-1}$`. This version is often more convenient for calculations when `$|r|>1$` as it avoids working with negative numbers.\n\n**Example 1: Calculating a Finite Sum**\n\n**Question:** Find the sum of the first 8 terms of the geometric series `$2 + 6 + 18 + \\dots$`.\n\n**Solution:**\n1. **Identify the parameters.** The first term is `$a=2$`. The common ratio is `$r = 6/2 = 3$`. The number of terms is `$n=8$`. \n\n2. **Select the appropriate formula.** Since `$r=3$`, which is greater than 1, using the form `$S_n = \\frac{a(r^n-1)}{r-1}$` will be simpler.\n\n3. **Substitute and calculate.**\n `$S_8 = \\frac{2(3^8 - 1)}{3 - 1}$`\n `$S_8 = \\frac{2(6561 - 1)}{2}$`\n `$S_8 = 6560$`\n The sum of the first 8 terms is 6560.\n\n### The Sum to Infinity of a Convergent Geometric Series\n\nAn intriguing question arises when we consider adding up an infinite number of terms. Can an infinite series have a finite sum? The answer depends on the common ratio `$r$`. Let us examine the term `$r^n$` within the finite sum formula `$S_n = \\frac{a(1-r^n)}{1-r}$`.\n\nIf the magnitude of the common ratio is less than 1, i.e., `$-1 < r < 1$` or `$|r|<1$`, the term `$r^n$` approaches 0 as `$n$` becomes infinitely large (`$n \\to \\infty$`). For example, if `$r=1/2$`, the sequence `$r^n$` is `$1/2, 1/4, 1/8, 1/16, \\dots$`, which clearly tends to zero. In this case, the series is said to be **convergent**.\n\nAs `$n \\to \\infty$`, if `$|r|<1$`, then `$r^n \\to 0$`. The finite sum formula `$S_n = \\frac{a(1-r^n)}{1-r}$` simplifies to:\n`$S_\\infty = \\lim_{n\\to\\infty} \\frac{a(1-r^n)}{1-r} = \\frac{a(1-0)}{1-r}$`\n

\n`$S_\\infty = \\frac{a}{1-r}$`\n

\nThis is the formula for the **sum to infinity** of a convergent geometric series. If `$|r| \\ge 1$`, the terms do not decrease towards zero, and the sum will grow indefinitely. Such a series is called **divergent** and does not have a finite sum to infinity.\n\n**Example 2: Calculating the Sum to Infinity**\n\n**Question:** Find the sum to infinity of the series `$10 + 5 + 2.5 + 1.25 + \\dots$`.\n\n**Solution:**\n1. **Identify the parameters.** The first term is `$a=10$`. The common ratio is `$r = 5/10 = 1/2$`. \n\n2. **Check for convergence.** We must verify that the series converges by checking the common ratio. Here, `$|r|=|1/2|=0.5$`, which is less than 1. Therefore, the series is convergent, and its sum to infinity exists.\n\n3. **Substitute into the formula and calculate.**\n `$S_\\infty = \\frac{a}{1-r} = \\frac{10}{1 - 1/2}$`\n `$S_\\infty = \\frac{10}{1/2} = 20$`\n The sum to infinity of the series is 20.\n\n**Example 3: Problem-Solving with Geometric Series**\n\n**Question:** The sum of the first two terms of a convergent geometric series is 9, and its sum to infinity is 25. Find the possible values for the first term `$a$` and the common ratio `$r$`. \n\n**Solution:**\n1. **Formulate equations from the given information.**\n The sum of the first two terms is `$a + ar = 9$`. Factoring gives: `$a(1+r) = 9$` (Equation 1).\n The sum to infinity is `$\\frac{a}{1-r} = 25$` (Equation 2).\n The condition of convergence, `$|r|<1$`, must be satisfied.\n\n2. **Solve the system of equations.** A good strategy is to express `$a$` in terms of `$r$` using one equation and substitute it into the other.\n From Equation 2, we can write `$a = 25(1-r)$`.\n\n3. **Substitute this expression for `$a$` into Equation 1.**\n `$25(1-r)(1+r) = 9$`\n Using the difference of two squares, `$(1-r)(1+r) = 1-r^2$`.\n `$25(1-r^2) = 9$`\n\n4. **Solve for `$r$`**.\n `$1-r^2 = \\frac{9}{25}$`\n `$r^2 = 1 - \\frac{9}{25} = \\frac{16}{25}$`\n `$r = \\pm\\sqrt{\\frac{16}{25}} = \\pm\\frac{4}{5}$`\n Both values satisfy the convergence condition `$|r|<1$`, so we have two possible solutions.\n\n5. **Find the corresponding values of `$a$` for each `$r$`**.\n Case 1: If `$r = 4/5$`\n `$a = 25(1 - 4/5) = 25(1/5) = 5$`\n Case 2: If `$r = -4/5$`\n `$a = 25(1 - (-4/5)) = 25(1 + 4/5) = 25(9/5) = 45$`\n\n Thus, the two possible pairs of `$(a, r)$` are `$(5, 4/5)$` and `$(45, -4/5)`.", "tips": "Before calculating a sum to infinity, always state and check the condition `$|r|<1$`. In exam questions, this check can be worth a mark and ensures you are applying the correct concept.", "cautions": "Be careful not to confuse the formula for the `$n^{th}$` term, `$u_n = ar^{n-1}$`, with the formula for the sum of the first `$n$` terms, `$S_n = \\frac{a(1-r^n)}{1-r}$`. This is a frequent source of error under pressure.", "insights": "The concept of a convergent series is a cornerstone of calculus and analysis. It allows us to define functions like `$e^x$` and `$\\sin(x)$` as infinite polynomials (power series), which is crucial for both theoretical mathematics and practical approximations." }