Show that the series
$$
\sum_{n=1}^\infty \frac{1}{\sqrt{n}}.
$$
is divergent by comparing it with the integral
$$
\int_1^\infty \frac{1}{\sqrt{x}}\, dx.
$$
The idea illustrated in the motivating examples above extend to the following theorem. Pay careful attention to the assumptions on $f$.
Suppose $f$ is a continuous, non-negative, decreasing function on an interval $[b,\infty)$ for some number $b \geq 0$, and suppose $\left\{ a_n\right\}$ is a sequence whose terms are given by $a_n = f(n)$ for all $n \in \mathbb{N}$. The series $\sum a_n$ converges if and only if the following improper integral converges.
$$
\int_b^\infty f(x)\, dx
$$
The proof of this theorem is in our textbook. Reading it is a (strongly!) recommended exercise.
The Integral Test is a mathematical theorem. We can only trust that its conclusion is valid if we are sure that all of its hypotheses have been met. When using the Integral Test it is often tempting to jump straight to the computation of the integral, but it is important that we remember to check all of the hypotheses.
However, the hypotheses can be weakened slightly from our statement of the theorem above. This is done by strategically inserting the word "eventually" into each of the statements of the assumptions.
$f(n)$ must eventually equal the terms of the sequence $a_n$.
$f$ must eventually be non-negative.
$f$ must eventually be decreasing.
We will interpret each of these statements as meaning that there exists a number $N > 0$ such that the desired property of $f$ holds for all $x > N$.
Now that we believe the integral test for series, we can apply everything that we know about improper integrals. In particular, the $p$-test for integrals has a direct analog for series.
$$
\sum_{n=1}^\infty \frac{1}{n^p}
$$
is convergent if and only if $p > 1$.
Recommended Exercise: Prove it!
We now work through a number of examples. If you're feeling comfortable with the ideas in these notes, you should treat these as exercises and try them yourself before watching the video solutions.
As you watch a video solution, be a participant! Pause it frequently to make sure you are following the arguments. Do not take any claims for granted.
Determine whether the series converges or diverges.
$$
\frac{1}{5} + \frac{1}{7} + \frac{1}{9} + \frac{1}{11} + \frac{1}{13} + \cdots
$$
Determine whether the series is convergent or divergent.
$$
\sum_{n=3}^\infty \frac{3n-4}{n^2 - 2n}
$$
Determine whether the series is convergent or divergent.
$$
\sum_{k= 1}^\infty k e^{-k^2}
$$
Determine the values of $p$ for which the series is convergent.
$$
\sum_{n = 3}^\infty\ \frac{1}{n \ln n \big( \ln(\ln n)\big)^p}
$$
In practice applied mathematicians, scientists, and engineers care a lot about infinite series, but we don't always need to know the exact sum. Many times, we will want to know an approximation of the sum correct to a certain tolerance.
Consider a series $\sum a_n$ that converges by the Integral Test, and consider a partial sum
$$
\sum_{n=1}^k a_n = S_k.
$$
The $k^\text{th}$ remainder is $R_k = S - S_k$. The $k^\text{th}$ remainder is bounded by
$$
\int_{k+1}^\infty f(x)\, dx \leq R_k \leq \int_k^\infty f(x)\, dx.
$$
Recommended Exercise. Provide a geometric argument to justify these bounds.
Consider the series
$$
\sum_{n=2}^\infty \frac{1}{n (\ln n)^2}.
$$
Apply the integral test to show that this series converges. Then determine the number of terms of the series that would be required to guarantee that the partial sum is within $\varepsilon = 0.01$ of the exact sum.
Questions? You can ask them below.
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