How do you use the integral test to determine the convergence or divergence of #Sigma 1/n^2# from #[1,oo)#?

Question

Cameron Alexander · Accepted Answer

The series #sum_(n=1)^oo 1/(n^2)# is convergent.

The integral test states that, given: #sum_(n=1)^oo a_n# if we have a function #f(x)# such that: (i).... #f(x)# is positive (ii)... #f(x)# is decreasing (iii) #lim_(x->+oo) f(x) = 0# (iv)...#f(n) = a_n# then a necessary and sufficient condition for the series above to converge is that the integral: #int_1^(+oo) f(x)dx# converges as well. It's easy to see that for the series: #sum_(n=1)^oo 1/(n^2)# the function #f(x) = 1/(x^2)# satisfies all the conditions above, and we have that: #int_1^(+oo) dx/(x^2)= [-1/x]_1^( +oo) = 1# so that the series must be convergent as well.

Ezekiel Alexander · Answer

undefined To use the integral test to determine the convergence or divergence of $ \sum_{n=1}^{\infty} \frac{1}{n^2} $ from $ n = 1 $ to $ n = \infty $, we first integrate the function $ f(x) = \frac{1}{x^2} $ from $ x = 1 $ to $ x = \infty $. If the integral converges, then the series $ \sum_{n=1}^{\infty} \frac{1}{n^2} $ converges. If the integral diverges, then the series $ \sum_{n=1}^{\infty} \frac{1}{n^2} $ also diverges.

The integral of $ f(x) = \frac{1}{x^2} $ from $ x = 1 $ to $ x = \infty $ is given by:

$$ \int_{1}^{\infty} \frac{1}{x^2} \, dx $$

We evaluate this integral to determine its convergence or divergence. If the integral converges, then the series $ \sum_{n=1}^{\infty} \frac{1}{n^2} $ converges. If the integral diverges, then the series $ \sum_{n=1}^{\infty} \frac{1}{n^2} $ also diverges.

Isaiah Allen · Answer

undefined To use the integral test to determine the convergence or divergence of the series Σ 1/n^2 from 1 to infinity:

1. Consider the function f(x) = 1/x^2.
2. Integrate f(x) from 1 to infinity. If the integral converges, then the series converges. If the integral diverges, then the series also diverges.
3. Evaluate the integral ∫(1/x^2) dx from 1 to infinity.
4. The integral ∫(1/x^2) dx = [-1/x] from 1 to infinity = [-1/infinity] - [-1/1] = 0 - (-1) = 1.
5. Since the integral equals 1, it converges.
6. Therefore, by the integral test, the series Σ 1/n^2 from 1 to infinity also converges.