How do you use the Integral test on the infinite series #sum_(n=1)^oo1/n^5# ?

Question

How do you use the Integral test on the infinite series  #sum_(n=1)^oo1/n^5# ?

Aurora Alvarado · Accepted Answer

By Integral Test,

#sum_{n=1}^infty 1/n^5# converges.

Let us look at some details.

Let us evaluate the corresponding improper integral.

#int_1^infty 1/x^5 dx#

#=lim_{t to infty}int_1^tx^{-5} dx#

#=lim_{t to infty}[x^{-4}/-4]_1^t#

#=-1/4lim_{t to infty}[1/x^4]_1^t#

#=-1/4 lim_{t to infty}[1/t^4-1]#

#=-1/4(0-1)=1/4#

Since the integral

#int_1^infty 1/x^5 dx#

converges to #1/4#,

#sum_{n=1}^infty 1/n^5#

also converges by Integral Test.

Christopher Allers · Answer

undefined To use the Integral Test on the infinite series $ \sum_{n=1}^\infty \frac{1}{n^5} $, you compare it to the integral of the corresponding continuous function. The integral test states that if the integral of the function $ f(x) $ over the interval [1, ∞) converges or diverges, then the series $ \sum_{n=1}^\infty f(n) $ converges or diverges accordingly.

For this series, the corresponding function is $ f(x) = \frac{1}{x^5} $. Now, integrate $ f(x) $ over the interval [1, ∞):

$$ \int_{1}^{\infty} \frac{1}{x^5} dx = \lim_{b \to \infty} \int_{1}^{b} \frac{1}{x^5} dx $$

$$ = \lim_{b \to \infty} \left[ -\frac{1}{4x^4} \right]_{1}^{b} $$

$$ = \lim_{b \to \infty} \left( -\frac{1}{4b^4} + \frac{1}{4} \right) $$

Since the limit converges to a finite value (1/4), the integral converges. Therefore, by the Integral Test, the series $ \sum_{n=1}^\infty \frac{1}{n^5} $ also converges.