How do you do these two?

For the power series on the left ∑ 5^nx^n/n^5, the radius of convergence is 1/5, and the interval of convergence is [-1/5,1/5].

For the power series on the right, given ∑ cn x^n has a convergence of R, I believe that the radius of convergence for ∑ c_n x^8n the eighth root of R

Step-by-step explanation:

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

30 units^2

step-by-step explanation:

area is l × w

the length is 3

and the width is 10

3 × 10= 30 units squared

42.4

step-by-step explanation:

convert to decimal