Question

Find the first 10 partial sums of the series. (Round your answers to five decimal places.)

[infinity]
∑ 1/ n(n²)
n=2

A) 0.08333
B) 0.18056
C) 0.29167
D) 0.41667

Answer 1

The first 10 partial sums of the series are approximately:

A) 0.125

B) 0.291

C) 0.422

D) 0.547

E) 0.667

F) 0.782

G) 0.893

H) 1.000

I) 1.104

To find the first 10 partial sums of the series
`(\sum_(n=2)^(\infty) (1)/(n(n^2)))`, we can manually calculate the sum for each value of (n) from 2 to 11. The formula for the nth partial sum
`(S_n)` of a series is given by:

`[S_n = \sum_(i=2)^(n) (1)/(i(i^2))]`

Using this formula, we can calculate the first 10 partial sums as follows:

`[S_2 = (1)/(2(2^2)) = (1)/(8) \approx 0.125]`

`[S_3 = S_2 + (1)/(3(3^2)) = (1)/(8) + (1)/(27) \approx 0.291]`

`[S_4 = S_3 + (1)/(4(4^2)) = (1)/(8) + (1)/(27) + (1)/(64) \approx 0.422]`

`[S_5 = S_4 + (1)/(5(5^2)) = (1)/(8) + (1)/(27) + (1)/(64) + (1)/(125) \approx 0.547]`

`[S_6 = S_5 + (1)/(6(6^2)) = (1)/(8) + (1)/(27) + (1)/(64) + (1)/(125) + (1)/(216) \approx 0.667]`

`[S_7 = S_6 + (1)/(7(7^2)) = (1)/(8) + (1)/(27) + (1)/(64) + (1)/(125) + (1)/(216) + (1)/(343) \approx 0.782]`

`[S_8 = S_7 + (1)/(8(8^2)) = (1)/(8) + (1)/(27) + (1)/(64) + (1)/(125) + (1)/(216) + (1)/(343) + (1)/(512) \approx 0.893]`

`[S_9 = S_8 + (1)/(9(9^2)) = (1)/(8) + (1)/(27) + (1)/(64) + (1)/(125) + (1)/(216) + (1)/(343) + (1)/(512) + (1)/(729) \approx 1.000]`

`[S_(10) = S_9 + (1)/(10(10^2)) = (1)/(8) + (1)/(27) + (1)/(64) + (1)/(125) + (1)/(216) + (1)/(343) + (1)/(512) + (1)/(729) + (1)/(1000) \approx 1.104]`

Therefore, the first 10 partial sums of the series are approximately:

A) 0.125

B) 0.291

C) 0.422

D) 0.547

E) 0.667

F) 0.782

G) 0.893

H) 1.000

I) 1.104