Integral
Test: If $f$ is a continuous,positive
and decreasing function where $f(n)=a_n$ on the
interval $[1,\infty)$, then
the improper integral $\displaystyle\int_1^\infty f(x)\,
dx$ and the infinite series
$\displaystyle\sum_{n=1}^\infty a_n$
either both converge or both diverge.

Picture infinitely many rectangles of width 1 and height $a_n$, so
the area of the $n^{th}$ rectangle is $a_n$. Then the series
$\displaystyle\sum_{n=1}^\infty a_n$ is equal to the sum of the
areas of these infinitely many rectangles. See the graphic
examples below.

Consider this graph. We see that the value of the
series from $a_2$ on is less than the area under the curve
$f$ from 1 to infinity; i.e.
$\displaystyle\sum_{n=2}^\infty a_n<\int_1^\infty
f(x)\,dx$. If this
integral converges to some finite value
$C$, then by using this inequality and doing a little
work, we get $\displaystyle\sum_{n=1}^\infty a_n=
a_1+\sum_{n=2}^\infty a_n<a_1+\int_1^\infty
f(x)\,dx=a_1+C<\infty$, so the series is finite, and thus the series converges.

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Now consider this graph. We see that the sum of
the same series beginning with $a_1$ is larger
than the area under the same curve $f$ from 1 to
infinity; i.e.
$\displaystyle\int_1^\infty f(x)\,dx<\sum_{n=1}^\infty
a_n$. If this
integral diverges, then because of our
constraints on $f$ it diverges to infinity. Since
the area under $f$ is infinite, then the sum of the areas
of the rectangles must also be infinite, i.e.
$\displaystyle\sum_{n=1}^\infty a_n$ is infinite, and thus the series diverges.
We see that if the integral diverges, so does the series.

archive.cnx.org

Summary: either both the integral and
the series converge, or both diverge.

From our work with improper integrals, you may have seen that the
improper integral

$\displaystyle\int_1^\infty\frac{1}{x^p}\,dx$
converges if $p>1$, and diverges if $p\le 1$.

By using the integral test, we therefore get our
$p$-series test, which is
extremely useful, especially when used to find comparable series
for the comparison tests.

$\displaystyle{\sum_{n=1}^\infty \frac{1}{n^p}}$ converges
if $p>1$ and diverges if $p \le 1$.

Explanation and examples of the integral test, as well as
determining the above integral of $\frac{1}{x^p}$ and the $p$-series
test are included on the first video. The second video
includes detail of the graphical information above.