How do you do these two questions?

a(i) 0

a(ii) π

b) [0, 4)

Step-by-step explanation:

a(i) ∑ₙ₌₁°° aₙ = π

The series converges, which means lim(n→∞) aₙ = 0.

a(ii) sₙ is the partial sum, so lim(n→∞) sₙ = π.

b) Use ratio test:

lim(n→∞)│aₙ₊₁ / aₙ│< 1

lim(n→∞)│[(3x−6)ⁿ⁺¹ / ((n+1)6ⁿ⁺¹)] / [(3x−6)ⁿ / (n6ⁿ)]│< 1

lim(n→∞)│[(3x−6)ⁿ⁺¹ / ((n+1)6ⁿ⁺¹)] × [(n6ⁿ) / (3x−6)ⁿ]│< 1

lim(n→∞)│(3x−6) n / (6(n+1))│< 1

│(3x−6) / 6│< 1

│3x−6│< 6

-6 < 3x − 6 < 6

0 < 3x < 12

0 < x < 4

Check the endpoints.

If x = 0, ∑ₙ₌₁°° (3(0)−6)ⁿ / (n6ⁿ) = ∑ₙ₌₁°° (−1)ⁿ / n, which converges.

If x = 4, ∑ₙ₌₁°° (3(4)−6)ⁿ / (n6ⁿ) = ∑ₙ₌₁°° 1 / n, which diverges.

So the interval of convergence is [0, 4).

cos2400=cos(360−240)0=cos1200=−cos(180−120)0=−cos600sin(−1100)=−sin1100=−sin(180−110)0=−sin700