Power Series Techniques: Ratio Test, Endpoints & Interval Analysis

The series is centered at c = -5. Convergence at x = -8 means the distance is |-8 - (-5)| = 3. Divergence at x = 0 means the distance is |0 - (-5)| = 5. The radius of convergence R is at least the distance to a convergent point (3) and at most the distance to a divergent point (5). Therefore, the smallest possible R is 3. It could be any number between 3 and 5.

First, rewrite 1/(2+x) as (½) * [1 / (1 + x/2)] = (½) * [1 / (1 - (-x/2))]. We know 1/(1 - u) = sum of uⁿ for |u|

This is a substitution. We take the known series for ln(1+u) and substitute u = 3x. So, ln(1+3x) = sum from n=1 to infinity of (-1)ⁿ⁺¹ (3x)ⁿ / n. This is valid when |3x|

If the limit L from the Ratio Test is 0 for any x, then the condition L

Term-by-term differentiation preserves the radius of convergence, so the new series will also have R = 3. However, the behavior at the endpoints can change. We only know the new series will converge for |x|

First, substitute t² for x in the eˣ series: e^(t²) = Σ from n=0 to infinity of (t²)ⁿ / n! = sum of t^(2n)/n!. Then, integrate this series term-by-term from 0 to x: integral of t^(2n) dt = t^(2n+1)/(2n+1). Evaluating from 0 to x gives x^(2n+1)/(2n+1). Multiply by 1/n!, so the integrated series is Σ from n=0 to infinity of x^(2n+1) / (n! (2n+1)).

The condition from the Ratio Test is L

The center is c = 4 and the radius is R = 5. The endpoints of the interval of convergence are at c - R and c + R, which are 4 - 5 = -1 and 4 + 5 = 9. We must substitute x = -1 and x = 9 into the original series and test for convergence or divergence using tests appropriate for the resulting numerical series.

Multiplying a power series by x² is a simple operation. We multiply every term in the series Σ aₙ xⁿ by x². This gives Σ aₙ xⁿ * x² = Σ aₙ x^(n+2). The radius of convergence remains R, as multiplication by a power of x does not affect the radius (though it could affect behavior at x=0, but 0 is within the interval).

Differentiating term-by-term: d/dx [xⁿ / n!] = n xⁿ⁻¹ / n!. Since n! = n*(n-1)!, this simplifies to xⁿ⁻¹ / (n-1)!. The n=0 term (which is 1/0! = 1) has derivative 0, so we can start the sum from n=1. Therefore, f'(x) = sum from n=1 to infinity of xⁿ⁻¹ / (n-1)!. If we let k = n-1, this becomes sum from k=0 to infinity of x^k / k!, which is the original series, as expected for eˣ.

This is a consequence of the identity theorem for power series. If two power series centered at the same point converge to the same function on some interval containing the center, then their coefficients must be identical. This follows from the uniqueness of Taylor coefficients, as each coefficient is determined by the derivatives of the function at the center point via aₙ = f⁽ⁿ⁾(0)/n! and bₙ = f⁽ⁿ⁾(0)/n!, forcing aₙ = bₙ for all n.

First, note 3/(1 - x²) = 3 * [1/(1 - x²)]. We know 1/(1 - u) = sum of uⁿ. Here, u = x². So, 1/(1 - x²) = Σ from n=0 to infinity of (x²)ⁿ = sum of x^(2n). Multiplying by 3 gives the series: Σ from n=0 to infinity of 3x^(2n). This is valid when |x²|

Term-by-term integration of a power series yields a new power series with the same radius of convergence, R. The interval of convergence might differ at the endpoints (one or both endpoints may become included after integration), but the radius R remains unchanged. So, the new series for g(x) will converge at least for |x|

Substitute √x for x in the cos(x) series. cos(√x) = Σ from n=0 to infinity of (-1)ⁿ (√x)^(2n) / (2n)! = sum of (-1)ⁿ xⁿ / (2n)!. Now write the first three terms (n=0,1,2): n=0: (-1)⁰x^0/(0!) = 1/1 = 1. n=1: (-1)^1 x^1/(2!) = -x/2. n=2: (-1)² x²/(4!) = x²/24. So the series starts 1 - x/2 + x²/24.

We know f(x) = sum xⁿ. Then f'(x) = sum n*x^(n-1). The series in question is sum n*(1/2)ⁿ. We can rewrite the term n*(1/2)ⁿ as n*(1/2)^(n-1) * (1/2). This matches the form (1/2) * f'(1/2).