Using the integral test, how do you show whether #sum (n + 2) / (n + 1)# diverges or converges from n=1 to infinity?

Question

Cameron Alexander · Accepted Answer

The series diverges.

Without the integral test: The terms do not go to #0# as #nrarroo# (they go to #1#), so the series diverges. If we must use the integral test let's do so as quickly as possible. #f(x) = (x+2)/(x+1) = (x+1+1)/(x+1) = 1+1/(x+1)# #f'(x) = -1/(x+1)^2# which is negative, so #f# is a decreasing function. #int_1^oo f(x) dx = lim_(brarroo) int_1^b (1+1/(x+1))dx# # = lim_(brarroo) (x+ln abs(x+1))]_1^b# # = lim_(brarroo)[ (b+ln(b+1)) - (1+ln2)]# # = oo + oo -1-ln2 = oo# The series diverges.

Paisley Johnson · Answer

undefined To determine whether the series $ \sum_{n=1}^{\infty} \frac{n + 2}{n + 1} $ converges or diverges, we can use the integral test.

1. Let $ f(x) = \frac{x + 2}{x + 1} $. 
2. Check if $ f(x) $ is continuous, positive, and decreasing for $ x \geq 1 $.
3. If $ f(x) $ meets these conditions, integrate $ f(x) $ from 1 to infinity.
4. If the integral converges, then the series converges; if the integral diverges, then the series diverges.

Let's analyze:

1. $ f(x) = \frac{x + 2}{x + 1} $
2. $ f(x) $ is continuous, positive, and decreasing for $ x \geq 1 $.
3. Integrate $ f(x) $ from 1 to infinity:

$$ \int_{1}^{\infty} \frac{x + 2}{x + 1} \, dx = \lim_{b \to \infty} \int_{1}^{b} \frac{x + 2}{x + 1} \, dx $$

4. After integrating, evaluate the limit. If the limit exists and is finite, the series converges; if it's infinite or doesn't exist, the series diverges.

By evaluating the limit of the integral, we can determine whether the series converges or diverges.