Answer:
R = ⅕
I = [-⅕, ⅕]
R^⅛
Step-by-step explanation:
Use ratio test:
lim(n→∞)│aₙ₊₁ / aₙ│
lim(n→∞)│[5ⁿ⁺¹ xⁿ⁺¹ / (n+1)⁵] / (5ⁿ xⁿ / n⁵)│
lim(n→∞)│[5ⁿ⁺¹ xⁿ⁺¹ / (n+1)⁵] × n⁵ / (5ⁿ xⁿ)│
lim(n→∞)│[5x n⁵ / (n+1)⁵│
│5x│
5│x│
The series converges if the limit is less than 1.
5│x│< 1
│x│< ⅕
The radius of convergence is ⅕.
Solving for x:
-⅕ < x < ⅕
Check the endpoints:
If x = -⅕:
∑ 5ⁿ (-⅕)ⁿ / n⁵ = ∑ (-1)ⁿ / n⁵, which converges
If x = ⅕:
∑ 5ⁿ (⅕)ⁿ / n⁵ = ∑ 1 / n⁵, which converges
The interval of convergence is [-⅕, ⅕].
∑ cₙ xⁿ has a radius of convergence of R. Using ratio test:
lim(n→∞)│aₙ₊₁ / aₙ│< 1
lim(n→∞)│(cₙ₊₁ xⁿ⁺¹) / (cₙ xⁿ)│< 1
lim(n→∞)│x (cₙ₊₁ / cₙ)│< 1
│x│lim(n→∞)│cₙ₊₁ / cₙ│< 1
│x│< lim(n→∞)│cₙ / cₙ₊₁│
R = lim(n→∞)│cₙ / cₙ₊₁│
Now using ratio test on ∑ cₙ x⁸ⁿ:
lim(n→∞)│aₙ₊₁ / aₙ│< 1
lim(n→∞)│(cₙ₊₁ x⁸⁽ⁿ⁺¹⁾) / (cₙ x⁸ⁿ)│< 1
lim(n→∞)│x⁸ (cₙ₊₁ / cₙ)│< 1
│x⁸│lim(n→∞)│cₙ₊₁ / cₙ│< 1
│x⁸│< lim(n→∞)│cₙ / cₙ₊₁│
│x⁸│< R
│x│< R^⅛
