How do you use the integral test to determine the convergence or divergence of #1+1/(2sqrt2)+1/(3sqrt3)+1/(4sqrt4)+1/(5sqrt5)+...#?

Question

Samantha Adkison · Accepted Answer

The series:

#sum_(n=1)^oo 1/(nsqrt(n))#

is convergent.

We have the series: #sum_(n=1)^oo 1/(nsqrt(n))# we can apply the integral test using as test function: #f(x) = 1/(xsqrt(x)) = x^(-3/2)# since the function is positive and decreasing in the interval #(1.+oo)# and we have: #lim_(x->oo) x^(-3/2) = 0# Calculating the improper integral: #int_1^oo x^(-3/2)dx = [-2x^(-1/2)]_1^oo# #int_1^oo x^(-3/2)dx = 2-lim_(x->oo) -2/sqrt(x) = 2# As the integral is convergent then the series is proven to be convergent.

Luke Alvarez · Answer

undefined To use the integral test to determine the convergence or divergence of the series $ 1 + \frac{1}{2\sqrt{2}} + \frac{1}{3\sqrt{3}} + \frac{1}{4\sqrt{4}} + \frac{1}{5\sqrt{5}} + \ldots $:

1. Define the function $ f(x) $ corresponding to the terms of the series. In this case, $ f(x) = \frac{1}{x\sqrt{x}} $.

2. Check if the function $ f(x) $ is continuous, positive, and decreasing for all $ x \geq 1 $.
   
3. Evaluate the integral $ \int_{1}^{\infty} f(x) \, dx $.

4. If the integral converges, then the series converges. If the integral diverges, then the series diverges.