Use the Integral Test to determine whether the series is convergent or divergent given #sum 1 / n^5# from n=1 to infinity?

Question

Charles Arocha · Accepted Answer

It converges.

The Integral Test says that if #a_n>0# is a sequence and #f(x)>0# is monotonic decrescent and #a_n=f(n) forall n#,

#int_1^(+infty)f(x)dx# exists finite #<=> sum_(n=1)^(+infty)a_n# converges.

We know #int_1^(+infty)1/x^5dx=(-1/(4x^4))_1^(+infty)=1/4#, which is finite, so the series converges.

Notice that we don't know the sum of the series, we just know it converges!!

Ezekiel Alexander · Answer

undefined To determine whether the series $ \sum_{n=1}^{\infty} \frac{1}{n^5} $ is convergent or divergent using the Integral Test, we compare it to the corresponding improper integral:

$$ \int_{1}^{\infty} \frac{1}{x^5} \, dx $$

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

Let's evaluate the integral:

$$ \int_{1}^{\infty} \frac{1}{x^5} \, dx = \lim_{b \to \infty} \int_{1}^{b} \frac{1}{x^5} \, dx = \lim_{b \to \infty} \left[ -\frac{1}{4x^4} \right]_{1}^{b} $$

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

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