Evaluating ∑(n=0 To ∞) [(n+2)/(n+1)]xⁿ Two Calculation Methods

Introduction

In the realm of calculus and sequences and series, evaluating infinite sums is a fundamental task with applications spanning various branches of mathematics, physics, and engineering. This article delves into the evaluation of the infinite series ∑(n=0 to ∞) [(n+2)/(n+1)]xⁿ, where x is a real number within the interval of convergence (|x| < 1). We will explore two distinct methods to arrive at the solution, highlighting the techniques and underlying principles involved. Our journey begins with a crucial prerequisite: the well-known geometric series formula, which serves as a cornerstone for our analysis.

Understanding the Foundation: The Geometric Series

The geometric series formula states that for any real number x such that |x| < 1, the infinite sum ∑(n=0 to ∞) xⁿ converges to 1/(1-x). This result is derived from the limit of a finite geometric series as the number of terms approaches infinity. The geometric series is a fundamental concept in calculus and serves as a building block for evaluating more complex series. Its elegant simplicity and wide applicability make it an indispensable tool in mathematical analysis. The convergence condition |x| < 1 ensures that the terms of the series diminish rapidly enough for the sum to approach a finite value. When |x| ≥ 1, the series either diverges or oscillates, and the sum does not have a well-defined value.

Method 1: Decomposition and Series Manipulation

Our first approach involves decomposing the fraction (n+2)/(n+1) into simpler terms and then manipulating the series to leverage the geometric series formula. This method showcases the power of algebraic manipulation in simplifying complex expressions and revealing underlying structures. The initial step is to rewrite the fraction (n+2)/(n+1) as 1 + 1/(n+1). This decomposition allows us to separate the original series into two distinct sums: ∑(n=0 to ∞) xⁿ and ∑(n=0 to ∞) [xⁿ/(n+1)]. The first sum is simply a geometric series, which we know converges to 1/(1-x) for |x| < 1. The second sum requires further manipulation. To tackle the second sum, we introduce a clever trick: integrating the series term by term. Let's consider the power series representation of the natural logarithm function: -log(1-x) = ∑(n=1 to ∞) xⁿ/n for |x| < 1. Dividing both sides by x and adjusting the index of summation, we obtain ∑(n=0 to ∞) [xⁿ/(n+1)] = (-1/x)log(1-x). Combining these results, we find that ∑(n=0 to ∞) [(n+2)/(n+1)]xⁿ = 1/(1-x) - (1/x)log(1-x) for |x| < 1. This method demonstrates the versatility of series manipulation and the interplay between algebraic and analytic techniques.

Method 2: Differentiation and Series Recognition

Our second approach utilizes differentiation to transform the series into a more recognizable form. This method highlights the connection between differentiation and series manipulation, providing a different perspective on evaluating infinite sums. We start by recognizing that the derivative of xⁿ⁺¹ is (n+1)xⁿ. This observation suggests that differentiating a related series might lead us to the desired sum. Let's consider the series ∑(n=0 to ∞) xⁿ⁺¹/(n+1). Differentiating this series term by term with respect to x, we obtain ∑(n=0 to ∞) xⁿ, which is again the geometric series. Therefore, the derivative of ∑(n=0 to ∞) xⁿ⁺¹/(n+1) is 1/(1-x). Now, we integrate both sides with respect to x to recover the original series. Integrating 1/(1-x) gives us -log(1-x) + C, where C is the constant of integration. Evaluating the series at x=0, we find that C=0. Thus, ∑(n=0 to ∞) xⁿ⁺¹/(n+1) = -log(1-x). To relate this to our target series, we multiply both sides by x⁻¹ and adjust the index of summation to get ∑(n=0 to ∞) [xⁿ/(n+1)] = (-1/x)log(1-x). Finally, we combine this with the geometric series result, as in Method 1, to obtain ∑(n=0 to ∞) [(n+2)/(n+1)]xⁿ = 1/(1-x) - (1/x)log(1-x) for |x| < 1. This method elegantly demonstrates the power of differentiation in simplifying series and connecting them to known functions.

Detailed Explanation of Method 1: Decomposition and Series Manipulation

Decomposition and Separation of Series

In our quest to evaluate the infinite series ∑(n=0 to ∞) [(n+2)/(n+1)]xⁿ, the initial step is a crucial algebraic maneuver. We aim to simplify the fraction (n+2)/(n+1) to make it more amenable to series manipulations. The key insight here is to decompose this fraction into a sum of simpler terms. By performing polynomial long division or using algebraic manipulation, we rewrite (n+2)/(n+1) as 1 + 1/(n+1). This seemingly small change has a profound impact on the subsequent steps. It allows us to separate the original series into two distinct sums: ∑(n=0 to ∞) xⁿ and ∑(n=0 to ∞) [xⁿ/(n+1)]. This separation is a powerful technique because the first sum is a familiar friend: the geometric series. The second sum, while not immediately recognizable, is now in a form that we can tackle using other tools.

The geometric series, ∑(n=0 to ∞) xⁿ, has a well-known closed-form expression: 1/(1-x), valid for |x| < 1. This result stems from the fundamental properties of geometric progressions and is a cornerstone of series analysis. The convergence condition |x| < 1 is crucial; it ensures that the terms of the series diminish rapidly enough for the sum to approach a finite value. The geometric series serves as a foundation upon which we can build solutions for more complex series. By isolating this term, we have effectively handled a significant portion of the original problem. The remaining sum, ∑(n=0 to ∞) [xⁿ/(n+1)], presents a new challenge, but its structure hints at a possible connection to logarithmic functions.

Integrating the Series Term by Term

To evaluate the second sum, ∑(n=0 to ∞) [xⁿ/(n+1)], we employ a clever technique: integrating the series term by term. This approach is justified by the uniform convergence of the power series within its interval of convergence. Term-by-term integration is a powerful tool in series manipulation, allowing us to transform sums into more manageable forms. The key idea is to recognize that the integral of xⁿ is xⁿ⁺¹/(n+1), which closely resembles the terms in our second sum. To proceed, we consider the power series representation of the natural logarithm function: -log(1-x) = ∑(n=1 to ∞) xⁿ/n for |x| < 1. This series is a classic result in calculus and is often derived using Taylor series expansions. The natural logarithm function plays a crucial role in many areas of mathematics and physics, and its power series representation provides a powerful connection between continuous and discrete mathematics.

To make the connection more explicit, we divide both sides of the logarithmic series by x, resulting in -log(1-x)/x = ∑(n=1 to ∞) xⁿ⁻¹/n. Now, we need to adjust the index of summation to match the form of our second sum. By shifting the index, we obtain ∑(n=0 to ∞) [xⁿ/(n+1)] = (-1/x)log(1-x). This crucial step reveals the closed-form expression for the second sum. We have successfully transformed the infinite sum into an expression involving the natural logarithm function. The beauty of this approach lies in its ability to leverage the properties of well-known functions to evaluate seemingly complex series. By integrating term by term and manipulating the index of summation, we have unlocked the solution for the second sum.

Combining the Results

Having evaluated both parts of the separated series, we can now combine the results to obtain the final solution. We found that ∑(n=0 to ∞) xⁿ = 1/(1-x) and ∑(n=0 to ∞) [xⁿ/(n+1)] = (-1/x)log(1-x) for |x| < 1. Therefore, the original series can be expressed as ∑(n=0 to ∞) [(n+2)/(n+1)]xⁿ = 1/(1-x) - (1/x)log(1-x) for |x| < 1. This is the closed-form expression for the given infinite series. The solution involves a combination of algebraic manipulation, geometric series, and the power series representation of the natural logarithm function. The final result highlights the interplay between different mathematical concepts and techniques in solving complex problems. The expression 1/(1-x) represents the contribution from the constant term in the decomposition, while -(1/x)log(1-x) captures the contribution from the fraction 1/(n+1). The domain of convergence remains |x| < 1, as dictated by the geometric series and the logarithmic series.

Detailed Explanation of Method 2: Differentiation and Series Recognition

Leveraging Differentiation to Uncover Series Relationships

In our second approach to evaluating the infinite series ∑(n=0 to ∞) [(n+2)/(n+1)]xⁿ, we employ a different strategy: differentiation. This method showcases the power of calculus in manipulating and simplifying series expressions. The central idea is to recognize that differentiation can transform a series into a more recognizable form. The key observation here is that the derivative of xⁿ⁺¹ is (n+1)xⁿ. This suggests that if we can find a series whose terms involve xⁿ⁺¹/(n+1), differentiating it might lead us to a simpler series related to the geometric series.

To implement this strategy, we consider the series ∑(n=0 to ∞) xⁿ⁺¹/(n+1). This series is closely related to the second term we encountered in Method 1, which involved 1/(n+1). The presence of xⁿ⁺¹ in the numerator and (n+1) in the denominator suggests a possible connection to integration, but we will explore the differentiation route in this method. The critical step now is to differentiate this series term by term with respect to x. Term-by-term differentiation is a valid operation within the interval of convergence of a power series. Differentiating each term, we obtain ∑(n=0 to ∞) d/dx [xⁿ⁺¹/(n+1)] = ∑(n=0 to ∞) xⁿ. This is a remarkable result! The derivative of our chosen series is precisely the geometric series, which we know converges to 1/(1-x) for |x| < 1. By differentiating, we have transformed a somewhat complex series into a fundamental one. The geometric series serves as a bridge, connecting our target series to a well-understood function.

Integrating Back to Find the Series

Having found that the derivative of ∑(n=0 to ∞) xⁿ⁺¹/(n+1) is 1/(1-x), we now need to reverse the process to recover the original series. This involves integrating both sides of the equation with respect to x. Integrating 1/(1-x) gives us -log(1-x) + C, where C is the constant of integration. The constant of integration is a crucial element in indefinite integration, and we need to determine its value to obtain the correct closed-form expression. To find the value of C, we evaluate the series ∑(n=0 to ∞) xⁿ⁺¹/(n+1) at x=0. When x=0, all terms in the series vanish, so the sum is 0. Therefore, we have 0 = -log(1-0) + C, which implies that C = 0. With the constant of integration determined, we have ∑(n=0 to ∞) xⁿ⁺¹/(n+1) = -log(1-x). This equation provides a closed-form expression for the series ∑(n=0 to ∞) xⁿ⁺¹/(n+1). However, our target series involves xⁿ/(n+1), so we need to make a slight adjustment.

To relate this result to our target series, we divide both sides of the equation by x, giving us ∑(n=0 to ∞) xⁿ/(n+1) = (-1/x)log(1-x). This manipulation reveals the closed-form expression for the series that we encountered in Method 1. We have successfully expressed this series in terms of the natural logarithm function. Now, we can use the same approach as in Method 1 to combine this result with the geometric series. Recall that the original series was ∑(n=0 to ∞) [(n+2)/(n+1)]xⁿ, which we decomposed into ∑(n=0 to ∞) xⁿ + ∑(n=0 to ∞) [xⁿ/(n+1)]. We know that ∑(n=0 to ∞) xⁿ = 1/(1-x) and ∑(n=0 to ∞) [xⁿ/(n+1)] = (-1/x)log(1-x). Therefore, combining these results, we obtain ∑(n=0 to ∞) [(n+2)/(n+1)]xⁿ = 1/(1-x) - (1/x)log(1-x) for |x| < 1. This is the same solution we obtained using Method 1, demonstrating the consistency of the two approaches.

Conclusion

In this article, we have successfully evaluated the infinite series ∑(n=0 to ∞) [(n+2)/(n+1)]xⁿ using two distinct methods. The first method, decomposition and series manipulation, involved breaking the fraction (n+2)/(n+1) into simpler terms and utilizing the geometric series formula along with term-by-term integration. The second method, differentiation and series recognition, employed differentiation to transform the series into a more recognizable form and then integrated back to obtain the original series. Both methods yielded the same result: ∑(n=0 to ∞) [(n+2)/(n+1)]xⁿ = 1/(1-x) - (1/x)log(1-x) for |x| < 1. This result showcases the power and versatility of calculus techniques in evaluating infinite sums. The two methods provide different perspectives on the problem, highlighting the interconnectedness of various mathematical concepts.

The decomposition and series manipulation method is particularly useful when dealing with rational functions in series. By breaking down the fraction into simpler terms, we can often express the series as a combination of known series, such as the geometric series and the power series representation of the natural logarithm. This method relies on algebraic manipulation and the ability to recognize common series patterns. The differentiation and series recognition method, on the other hand, leverages the relationship between differentiation and series. By differentiating a related series, we can sometimes transform it into a simpler form that we can easily evaluate. This method highlights the connection between calculus and series manipulation and provides a powerful tool for evaluating more complex series. Both methods demonstrate the importance of having a diverse toolkit of mathematical techniques and the ability to choose the most appropriate method for a given problem. The evaluation of infinite series is a fundamental topic in calculus with applications in various fields, and these two methods provide valuable insights into the techniques involved.

The result we obtained, ∑(n=0 to ∞) [(n+2)/(n+1)]xⁿ = 1/(1-x) - (1/x)log(1-x) for |x| < 1, is a closed-form expression for the infinite series. This expression allows us to compute the sum of the series for any value of x within the interval of convergence. The closed-form expression is a valuable tool in applications where we need to evaluate the series repeatedly or understand its behavior as a function of x. The series converges for |x| < 1, which means that the terms of the series must diminish rapidly enough for the sum to approach a finite value. The convergence condition is determined by the behavior of the terms as n approaches infinity. In this case, the terms (n+2)/(n+1) approach 1 as n approaches infinity, while xⁿ diminishes rapidly for |x| < 1. The combination of these factors ensures the convergence of the series. The divergence of the series for |x| ≥ 1 can be understood by considering the behavior of the terms. When |x| ≥ 1, the terms do not approach 0 as n approaches infinity, which violates a necessary condition for convergence. Therefore, the series diverges for |x| ≥ 1. The analysis of convergence and divergence is a crucial aspect of working with infinite series, and it provides insights into the behavior of the series and the validity of the closed-form expression.

In conclusion, the evaluation of the series ∑(n=0 to ∞) [(n+2)/(n+1)]xⁿ through two distinct methodologies underscores the richness and interconnectedness of mathematical techniques within calculus and series analysis. Both the decomposition method and the differentiation method provide unique insights into manipulating and simplifying complex infinite sums. The final closed-form expression, 1/(1-x) - (1/x)log(1-x), not only offers a practical means of computation but also serves as a testament to the power of analytical approaches in solving mathematical problems. This exploration encourages a deeper understanding of series, convergence, and the elegance of mathematical problem-solving.