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Explanation:
f(x) is continuous when x >= 1. The only discontinuity for f(x) is when x = 0, but 0 is not part of this interval.
f(x) is positive for any valid x value in the domain since x^6 is always positive. In general, x^n is positive for all x when n is any even number.
f(x) is decreasing. You can see this through a table of values or through a graph. For anything in the form 1/(x^k), it will be a decreasing function because x^k gets larger, so 1/(x^k) gets smaller, when x goes to infinity.
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The conditions to use the integral test have been met. So we have to see if [tex]\displaystyle \int_1^{\infty}f(x)dx[/tex] converges or not.
Let's integrate and find out
[tex]\displaystyle \int \frac{1}{x^6} dx = \int x^{-6} dx\\\\\\ \displaystyle \int \frac{1}{x^6} dx = \frac{1}{1+(-6)}x^{-6+1}+C\\\\\\ \displaystyle \int \frac{1}{x^6} dx = \frac{1}{-5}x^{-5}+C\\\\\\ \displaystyle \int \frac{1}{x^6} dx = -\frac{1}{5}*\frac{1}{x^5}+C\\\\\\ \displaystyle \int \frac{1}{x^6} dx = -\frac{1}{5x^5}+C\\\\[/tex]
So we have
[tex]\displaystyle g(x) = \int f(x) dx\\\\\\\displaystyle g(x) = \int \frac{1}{x^6} dx\\\\\\\displaystyle g(x) = -\frac{1}{5x^5}+C\\\\\\[/tex]
Meaning that,
[tex]\displaystyle \int_{a}^{b} f(x) dx = g(b)-g(a)\\\\\\\displaystyle \int_{a}^{b} \frac{1}{x^6} dx = \left(-\frac{1}{5b^5}+C\right)-\left(-\frac{1}{5a^5}+C\right)\\\\\\\displaystyle \int_{a}^{b} \frac{1}{x^6} dx = -\frac{1}{5b^5}+\frac{1}{5a^5}\\\\\\[/tex]
If we plug in a = 1 and apply the limit as b approaches positive infinity, then the expression [tex]-\frac{1}{5b^5}+\frac{1}{5a^5}[/tex] will turn into [tex]\frac{1}{5}[/tex]
Therefore,
[tex]\displaystyle \int_{1}^{\infty} \frac{1}{x^6} dx = \frac{1}{5}\\\\\\[/tex]
Because this integral converges, this means the series [tex]\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^6}[/tex] also converges as well by the integral test.
