Question

6. | -[1.8 points v r SCALCETS 11.3.039. Estimate r 120 + 1)-5 correct to five decimal places. n = 1

Answer 1

Solution

Σ (2n + 1)

Apply Series Integral Test

If there exists an N > k so that for all n > N, f(n) = an is positive, continuous and decreasing

Then Ža, and ( f (x) dx either both converge or diverge

Check if f(n) is positive, continuous and decreasing

(2n + 1) - is positive, continuous and decreasing from n = 1

ſ (2n + 1)-5 dn J1

Compute the indefinite integral

u= 2n +1

$\mathrm{=\int \frac{1}{2u^5}du\Rightarrow \frac{1}{2}\cdot \int \frac{1}{u^5}du\Rightarrow \frac{1}{2}\cdot \int \:u^{-5}du}$

Apply the Power Rule: xdx = - a -1 a + 1

.. *** ** ** f- f-

Compute the boundaries: (2n +1)-5 dn

lim n+1+ 8 (2n +1) W 8(2.1+1)4 648

lim n+ 8 (2n +1)

0- 648) > 648 > Long Division 2 648 / 648 : 0.00154...

By the integral test criteria

converges