Find a formula for the nth term of the sequence(Question 1 - 2). Determine whether the series converges. If it does, find its sum(Question 3 - 8).

Question

Find a formula for the nth term of the sequence(Question 1 - 2).

1 - a_{n =} n^1/n

2 - a_{n =} n - √(n²-n) { (n²-n) is included in the square root }

Determine whether the series converges. If it does, find its sum(Question 3 - 8).

∞

3 - ∑ n = 2 (5 ⁄ 2ⁿ - 1 ⁄ 3ⁿ)

∞

4 - ∑ n = 1 (2ⁿ + 3ⁿ) ⁄ 4ⁿ

∞

5 - ∑ n = 1 cos (1 ⁄ n)

∞

6 - ∑ n = 1 (6) ⁄ (2n - 1) (2n + 1) {Hint: Use partial fractions}

∞

7 - ∑ n = 1 (2ⁿ + 4ⁿ) ⁄ (3ⁿ + 4ⁿ)

∞

8 - FInd the values of x for which the geometric series ∑ n = 1 (-1)ⁿ (x+1)ⁿ converges.

I hope these equations are easy to understand. I really struggled with these problems, So please I need help with these questions as soon as possible!!!

Answer 1

I WILL SOLVE FOR YOU Q6

∑ (6) ⁄ (2n - 1) (2n + 1)

6 ⁄ (2n - 1) (2n + 1) = A / 2n-1 + B / 2n+ 1

6 = A (2n +1) + B ( 2n -1)

for n=1/2

6 = A ( 1+ 1) + 0

A = 3

for n= -1/2

6 = 0 + B ( -1 -1 )

B = - 3

∑ (6) ⁄ (2n - 1) (2n + 1) = ∑ 3 / 2n-1 - 3 / 2n+ 1

= 3 ∑ [ 1 / 2n-1 - 1 / 2n+ 1 ]

= 3 [ (1/1 - 1/3) + ( 1/3 -1/5 ) + ( 1/5 - 1/ 7) + ......+ ( 1/ 2n-3 - 1/2n - 1) + ( 1/2n -1 - 1/ 2n+1) ]

Hint: you can see each term cancel the other except (1) and ( -1/2n+1 )

^∞

∑ (6) ⁄ (2n - 1) (2n + 1) = 3 [ 1 - lim _n→∞ 1 /2n+1 ]

ⁿ⁼¹

= 3 [ 1 - 0 ]

= 3