Mathematical Framework and Problem Constraints

Component Analysis of the Geometric Expression

The evaluation of an infinite geometric series requires a rigorous identification of its fundamental parameters. In the given expression ##\sum_{n=1}^{\infty} 5 \left(\frac{2}{3}\right)^{n-1}##, we observe a structure that adheres to the standard geometric form ##\sum a r^{n-1}##. Here, the constant multiplier, represented by the variable a, is identified as 5. This value serves as the initial magnitude of the sequence when the index n is equal to the lower bound of one. The term in the parentheses, ##\frac{2}{3}##, represents the common ratio, denoted by r, which determines the scale of each successive term in the progression.

To technically validate the components, we must ensure the power of the ratio corresponds correctly to the index. Since the index begins at ##n=1## and the exponent is ##n-1##, the first term in the series is calculated as ###a_1 = 5 \left(\frac{2}{3}\right)^{1-1} = 5 \cdot 1 = 5.### This confirms that the series begins with a discrete integer value, establishing the baseline for the summation. The common ratio ##r = \frac{2}{3}## implies that every subsequent term is exactly two-thirds of its predecessor, creating a monotonic decrease in the magnitude of the individual terms.

The convergence of such a series is not guaranteed but depends entirely on the absolute value of the common ratio. In this specific case, we evaluate the condition ###|r| < 1.### Given that ###\left| \frac{2}{3} \right| \approx 0.6667,### the condition for convergence is satisfied. This analytical confirmation is critical because it ensures that the sum of the infinite number of terms will approach a finite limit rather than diverging toward infinity. Without this prerequisite check, the application of summation formulas would be mathematically invalid and theoretically unsound.

Boundary Conditions and Indexing Logic

The indexing logic of the summation operator ##\sum## defines the operational boundaries for the variable n. In our series, n starts at 1 and proceeds toward infinity, which is the standard convention for geometric sequences where the first term is simply the coefficient a. If the index were shifted to start at ##n=0## without adjusting the exponent, the first term would differ, potentially altering the entire sum. Precise attention to the n-1 exponent is necessary to align the first term with the coefficient 5.

Technically, we can describe the terms of the series as a sequence ##\{a_n\}## defined by the function ###f(n) = 5 \left(\frac{2}{3}\right)^{n-1}.### The summation then represents the accumulation of these discrete values across the entire domain of natural numbers. As n increases, the value of the exponent grows, causing the fraction to diminish toward zero. This behavior is the hallmark of a convergent geometric progression, where the infinitesimal nature of higher-order terms allows the total sum to stabilize at a specific numerical constant known as the limit.

We must also consider the implications of the constant coefficient 5 being outside or inside the power term. In this expression, 5 acts as a linear scalar for the entire summation. According to the distributive property of limits and series, we can factor the constant out of the summation operator if necessary: ###5 \sum_{n=1}^{\infty} \left(\frac{2}{3}\right)^{n-1}.### This simplifies the internal calculation of the sum of powers before multiplying by the external factor. Such a structural understanding is vital for complex algebraic manipulations involving multiple series or nested summations in advanced calculus.

Analytical Derivation of the Summation Formula

The Method of Partial Sums

The proof for the sum of an infinite geometric series relies on the behavior of its k-th partial sum, denoted as ##S_k##. A partial sum is the sum of the first k terms of the series, represented as ###S_k = \sum_{n=1}^{k} a r^{n-1} = a + ar + ar^2 + \dots + ar^{k-1}.### To find a closed-form expression for ##S_k##, we utilize an algebraic maneuver where we multiply the entire sum by the ratio r, yielding ###r S_k = ar + ar^2 + ar^3 + \dots + ar^{k}.### This shifted version of the sum allows for a clean subtraction of terms.

By subtracting the shifted sum ##r S_k## from the original partial sum ##S_k##, we observe a telescoping effect where most intermediate terms cancel out: ###S_k – r S_k = (a + ar + \dots + ar^{k-1}) – (ar + ar^2 + \dots + ar^k).### After the cancellation of the interior terms, only the first term of the original sum and the last term of the shifted sum remain. This results in the simplified linear equation ###S_k(1 – r) = a – ar^k.### Solving for ##S_k## provides the standard formula for the sum of a finite geometric sequence.

The resulting formula for the k-th partial sum is ###S_k = \frac{a(1 – r^k)}{1 – r}.### This equation is valid for any value of r except ##r=1##, where the denominator would become zero and the series would behave linearly. In our specific context, this partial sum formula serves as the foundation for determining the infinite limit. By examining how ##r^k## behaves as k approaches infinity, we can transition from the finite world of discrete addition to the continuous realm of infinitesimal analysis and limits.

Limits and the Criterion for Convergence

The transition from a finite partial sum to an infinite sum involves taking the limit as the number of terms k grows without bound. We define the infinite sum S as ###S = \lim_{k \to \infty} S_k = \lim_{k \to \infty} \frac{a(1 – r^k)}{1 – r}.### In this expression, the variables a and r are constants relative to the limit operator. Therefore, the only term that changes as k increases is ##r^k##. The behavior of the entire series depends exclusively on whether this exponential term converges or diverges.

For our series, where ##|r| < 1##, the term ##r^k## represents a fraction that is repeatedly multiplied by itself. Mathematically, the limit of a fraction between -1 and 1 raised to an increasing power is zero: ###\lim_{k \to \infty} r^k = 0 \text{ for } |r| < 1.### Substituting this limit back into the partial sum formula, the numerator simplifies from ##a(1 – r^k)## to simply a. This yields the elegant and powerful formula for the sum of an infinite geometric series, which is ###S = \frac{a}{1 – r}.###

It is important to emphasize that if ##|r| \geq 1##, the limit of ##r^k## would either stay constant, alternate, or grow infinitely large. In such cases, the sequence of partial sums does not approach a single value, and the series is classified as divergent. The technical requirement for convergence is a strict inequality. This rigorous derivation ensures that our calculation for the specific series ##\sum_{n=1}^{\infty} 5 \left(\frac{2}{3}\right)^{n-1}## is grounded in established calculus principles regarding the convergence of sequences and series within the complex or real number planes.

Quantitative Implementation and Arithmetic Proof

Iterative Expansion and Substitution

With the analytical formula ###S = \frac{a}{1 – r}### firmly established, we can proceed with the quantitative evaluation of the specific problem. We begin by substituting the known values into the equation. Given that the coefficient ##a = 5## and the common ratio ##r = \frac{2}{3}##, the expression for the sum becomes ###S = \frac{5}{1 – \frac{2}{3}}.### This setup requires careful arithmetic manipulation of the denominator to ensure the final ratio is calculated correctly without losing precision during the fraction subtraction.

To simplify the denominator, we represent the integer 1 as a fraction with a common denominator of 3, resulting in ##\frac{3}{3}##. The subtraction operation yields ###1 – \frac{2}{3} = \frac{3}{3} – \frac{2}{3} = \frac{1}{3}.### Now, the expression for the sum is reduced to a division problem involving a whole number and a fraction: ###S = \frac{5}{1/3}.### Division by a fraction is equivalent to multiplication by its reciprocal, which in this case is 3. Performing this final operation gives ###S = 5 \times 3 = 15.###

The result of 15 represents the exact limit that the accumulated sum of the terms will approach. To provide further clarity, we can list the first few terms of the series to visualize the progression: ###5 + \frac{10}{3} + \frac{20}{9} + \frac{40}{27} + \dots### Converting these to decimals, we have ###5 + 3.3333 + 2.2222 + 1.4815 + \dots### Each term is smaller than the previous one, and as we add them, the total grows ever closer to the ceiling of 15 but will never exceed it, illustrating the asymptotic nature of the convergence.

Verification through Recursive Relationships

To verify the result using an alternative technical approach, we can employ a recursive definition of the infinite sum S. If we assume the sum exists and is finite, we can write the series expansion as ###S = 5 + 5\left(\frac{2}{3}\right) + 5\left(\frac{2}{3}\right)^2 + 5\left(\frac{2}{3}\right)^3 + \dots### By factoring out the term ##\frac{2}{3}## from every term starting from the second one, we observe that the remainder of the series is actually proportional to the original sum itself. This is a property unique to infinite geometric structures.

Factoring out the ratio, the equation becomes ###S = 5 + \frac{2}{3} \left[ 5 + 5\left(\frac{2}{3}\right) + 5\left(\frac{2}{3}\right)^2 + \dots \right].### The expression inside the brackets is identical to our original definition of S. Therefore, we can set up a simple linear algebraic equation: ###S = 5 + \frac{2}{3}S.### This recursive relationship provides a secondary method for solving the problem without explicitly relying on the limit of partial sums, although the existence of the limit must still be presupposed for the algebra to hold.

Solving this linear equation for S involves isolating the variable on one side. Subtracting ##\frac{2}{3}S## from both sides of the equation yields ###S – \frac{2}{3}S = 5,### which simplifies to ###\frac{1}{3}S = 5.### Multiplying both sides by 3, we once again arrive at the conclusion that ###S = 15.### This internal consistency between the limit-based derivation and the recursive algebraic method provides a robust verification of the solution, confirming that the sum of the infinite geometric series is indeed 15.

Theoretical Deep Dive and Literature Context

Historical Origins and the Resolution of Paradox

The study of infinite geometric series dates back to ancient Greek mathematics, most notably through the work of Zeno of Elea. Zeno’s paradoxes, such as the dichotomy paradox, suggested that motion was impossible because reaching a destination would require completing an infinite number of tasks. For instance, to travel a distance of 1, one must first travel 1/2, then 1/4, then 1/8, and so on. This creates the series ###\sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^n,### which Zeno argued could never be completed in a finite amount of time.

It was not until the development of rigorous calculus and the formalization of limits that these paradoxes were mathematically resolved. The realization that an infinite sum of terms can converge to a finite value allowed mathematicians to bridge the gap between discrete logic and physical reality. Archimedes had early intuitions about this through his method of exhaustion, which he used to find the area under a parabolic curve by summing a geometric series. However, the formal epsilon-delta definitions of the 19th century provided the necessary rigor to handle infinity safely.

In the context of modern mathematical literature, the geometric series is often cited as the simplest example of a power series. It serves as a pedagogical gateway to more complex topics like Taylor series expansions and analytic functions. The formula for the geometric series is essentially the Taylor expansion of the function ##f(x) = \frac{1}{1-x}## centered at zero. Understanding how a simple arithmetic ratio can lead to a closed-form rational function is fundamental for students transitioning from algebra to advanced real and complex analysis.

Practical Significance in Scientific Modeling

Beyond pure theory, infinite geometric series are utilized extensively in various scientific and economic fields to model systems that involve compounding or decay. In physics, the radioactive decay of isotopes can be modeled as a series of discrete events that follow a geometric progression over specific time intervals. Similarly, in the study of optics and wave interference, the total intensity of light reflecting between two parallel mirrors involves summing an infinite number of decreasingly intense reflections, which is modeled precisely by a geometric summation.

In the realm of finance, geometric series are the mathematical backbone of annuity calculations and the determination of present value. When calculating the value of a perpetual stream of payments, such as a consol bond, economists use the infinite series sum to discount future cash flows back to the present day. If a payment is C and the interest rate is i, the present value is ###PV = \sum_{n=1}^{\infty} \frac{C}{(1+i)^n},### which is an infinite geometric series that simplifies to ##C/i##, a standard formula in financial engineering.

Finally, in computer science, geometric series are used in the analysis of algorithm complexity, particularly for recursive functions that process data in progressively smaller chunks. The total work performed in a divide-and-conquer algorithm can often be expressed as a summation of terms that follow a geometric ratio. Understanding the convergence properties of these series allows engineers to prove the efficiency and stability of software systems. Whether in the physical world or digital space, the geometric series remains one of the most versatile and essential tools in the mathematician’s arsenal.