Advanced Power Series Methods: Integration, Differentiation & Substitution

A power series is defined in the standard form Σ cₙ(x - a)ⁿ, where 'a' represents the center of the series. In the given expression, the term raised to the power of n is (x + 5). We can rewrite (x + 5) as (x - (-5)). By comparing this to the standard form (x - a), we can identify that a = -5. Therefore, the series is centered at x = -5.

The Ratio Test requires us to evaluate the lim_{n 🠒∞}  the absolute value of the ratio of the (n+1)-th term to the n-th term, denoted as |a_{n+1} / aₙ|. Here, aₙ = (x - 2)ⁿ / n. The (n+1)-th term is a_{n+1} = (x - 2)ⁿ⁺¹ / (n + 1). When we divide a_{n+1} by aₙ, we multiply a_{n+1} by the reciprocal of aₙ. This gives us [ (x - 2)ⁿ⁺¹ / (n + 1) ] * [ n / (x - 2)ⁿ ]. Simplifying the powers of (x - 2) leaves a single factor of (x - 2) in the numerator. The n terms group as n / (n + 1). Thus, the expression inside the limit is | (n(x - 2)) / (n + 1) |.

We can identify this series as a geometric series with the common ratio r = 3x. A geometric series converges if and only if the absolute value of the ratio is less than 1. Therefore, we set |3x| < 1. Solving for x, we divide both sides by 3 to get |x| < 1/3. This inequality implies that the distance from the center (which is 0) must be less than 1/3. Consequently, the radius of convergence R is 1/3. Alternatively, using the Ratio Test, we find the limit of |3x| as n approaches infinity, which must be less than 1, leading to the same result.

First, we apply the Ratio Test to find the radius of convergence. The limit of |a_{n+1}/aₙ| simplifies to |x - 1|. Setting this less than 1 gives |x - 1| < 1, so the radius is 1 and the open interval is (0, 2). Next, we must test the endpoints. At x = 2, the series becomes Σ (1)ⁿ / n = Σ 1/n, which is the harmonic series and diverges. At x = 0, the series becomes Σ (-1)ⁿ / n, which is the alternating harmonic series. By the Alternating Series Test, this converges because the terms 1/n decrease to 0. Therefore, x = 0 is included, but x = 2 is excluded. The interval is [0, 2).

A fundamental property of power series is that term-by-term differentiation does not change the radius of convergence. While the interval of convergence might change regarding the inclusion of endpoints, the radius R remains identical to that of the original series. Since the original series has a radius of R = 4, the differentiated series also has a radius of convergence equal to 4.

We start with the standard geometric series formula 1 / (1 - u) = Σ uⁿ, which converges for |u| < 1. To obtain the series for 1 / (1 + x²), we can rewrite the function as 1 / (1 - (-x²)). We then substitute u = -x² into the standard geometric series. This yields Σ (-x²)ⁿ. Applying the power of n to both -1 and x² results in Σ (-1)ⁿ x²ⁿ. This series is valid where |-x²| < 1, or |x| < 1.

We recognize this series relates to the integral of the geometric series. We know sum xⁿ = 1/(1-x). Integrating term-by-term from 0 to x gives sum x^(n+1)/(n+1) (or adjusted indices sum xⁿ/n) = -ln(1-x). Here, x = 1/2. So the sum is -ln(1 - 1/2) = -ln(1/2) = -ln(2⁻¹) = ln(2).

The general formula for the coefficients of a Taylor series centered at a is aₙ = f⁽ⁿ⁾(a) / n!. Here, the center is a = 0 and we are looking for the coefficient a₃ where n = 3. Plugging these values into the formula gives a₃ = f'''(0) / 3!. Since 3! (3 factorial) equals 3 × 2 × 1 = 6, the coefficient is f'''(0) / 6.

We apply the Ratio Test. We examine the lim_{n 🠒∞}  |a_{n+1} / aₙ|. Here, aₙ = xⁿ / n!. The ratio is | (xⁿ⁺¹ / (n+1)!) * (n! / xⁿ) |. Simplifying the x terms leaves |x| in the numerator. Simplifying the factorials leaves (n+1) in the denominator, because (n+1)! = (n+1)n!. The limit expression becomes lim (n→∞) |x| / (n+1). For any fixed real number x, as n goes to infinity, this limit is 0. Since 0 is always less than 1, the series converges for all real numbers x. Therefore, the radius of convergence is infinite.

The interval of convergence is given as [1, 3). The notation using a parenthesis ")" after the 3 indicates that the value x = 3 is excluded from the set of values for which the series converges. Therefore, the series does not converge at x = 3. If it converged (either conditionally or absolutely), the interval would use a square bracket "]" at that end.

To find the derivative of a power series, we differentiate term-by-term with respect to x. The general term of the function is x²ⁿ. Using the power rule for differentiation, d/dx(xᵏ) = kxᵏ⁻¹, we treat 2n as the exponent k. The derivative of x²ⁿ is 2n * x²ⁿ⁻¹. Therefore, the series for the derivative is the sum of these differentiated terms: Σ 2n x²ⁿ⁻¹. Note that if the sum starts at n=0, the first term is constant (x⁰=1), so its derivative is 0, and the new sum effectively has non-zero terms starting from n=1, though writing it from n=0 is valid as the n=0 term is just 0.

An even function satisfies f(-x) = f(x). In a polynomial or power series, this symmetry is achieved only if the variable x appears only with even powers (0, 2, 4...). Therefore, all coefficients of odd powers (a₁, a₃, etc.) must be zero.

We must recognize the structure of this series. The Maclaurin series for cos(x) is Σ (-1)ⁿ x²ⁿ / (2n)!. If we substitute x = π into this formula, we get exactly the series given in the problem: Σ (-1)ⁿ (π)²ⁿ / (2n)!. Therefore, the sum of the series is the value of cos(π). Since cos(π) = -1, the series sums to -1, but the question asks for the function value representation, which is cos(π).

A power series must be of the form Σ aₙ (x-c)ⁿ, where the coefficients aₙ are constants that do not depend on x. In option C, the coefficient of xⁿ is sin(nx). Since sin(nx) depends on x, this is not a constant coefficient relative to x. It is a series of functions, but it does not fit the strict definition of a power series. Options A, B, and D all fit the form where the coefficients depend only on n, not x.