MAT187 L10 - Integral Test for Series

#MAT187 PCE
The integral test compares an infinite sum of positive numbers to an improper integral to prove either divergence or convergence. Only works if:

$f$ is continuous
$f$ is decreasing

For an infinite series $\sum_{k = 1}^{n} f (k)$ , the integral proves that:

the series converges if $\sum_{k = 1}^{n} f (k) < \int_{1}^{k + 1} f (x) d x$ and $\int_{1}^{k + 1} f (x) d x$ converges.
the series diverges if $\sum_{k = 1}^{n} f (k) > \int_{1}^{k + 1} f (x) d x$ and $\int_{1}^{k + 1} f (x) d x$ diverges.
The integral we use follows the form:
$\int_{1}^{k + 1} f (x) d x$
...because we treat the infinite series as a Left-endpoint Riemann sum with interval width 1, so to get the full area under the curve, we need to consider $k + 1$ .

Example: $\sum_{k = 1}^{n} \frac{1}{k}$

We can use integral test by examining the behaviour of $\int_{1}^{k + 1} \frac{1}{x} d x$ . As a graph, this looks like:

Each rectangle represents a term in the infinite series. It is analogous to a left endpoint Riemann sum with interval length 1. Since $\frac{1}{x}$ is a decreasing function as $x \to \infty$ , this summation is an overestimate of the actual area beneath $\frac{1}{x}$ , and thus we derive the following inequality:

\sum_{k = 1}^{n} \frac{1}{k} > \int_{1}^{k + 1} \frac{1}{x} d x $ $ S o, w e c a n i n t e g r a t e $ \frac{1}{x} $ t o e x a m i n e t h e b e h a v i o u r o f t h e f u n c t i o n .

\begin{align}
\int_{1}^{k+1} \frac{1}{x}dx &= (\ln \lvert x \rvert)\big|_{1}^{k+1} \
&= \ln(k+1)-\ln(1) \
&=\ln(k+1)
\end

S i n c e $ \ln (k + 1) $ d i v e r g e s, a n d $ \sum_{k = 1}^{n} \frac{1}{k} > \int_{1}^{k + 1} \frac{1}{x} d x $, b y i n t e g r a l t e s t, w e k n o w t h a t $ \sum_{k = 0}^{n} \frac{1}{k} $ i s a d i v e r g i n g s u m .