Teste de integrais

TODO: fix formatting, add questions/pauses; will fix later

One of the ideas behind the integral test:

We start with the harmonic series $$\sum_{n=1}^\infty\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\cdots$$ (Note that it passes the Divergence Test since the terms go to zero.) This is a pretty famous series which diverges, as in, $$\sum_{n=1}^\infty\frac{1}{n}=+\infty.$$ We will show this fact as follows.

We draw a picture with the graph of $y=\dfrac{1}{x}$.

Now we look at the rectangles drawn. They are left hand rectangles, all with width $1$. Looking at the picture, each rectangle overestimates the area under the curve (since $y=\frac{1}{x}$ is a decreasing function). Notice also that the areas of the rectangles, if we add them, correspond to the harmonic series since each rectangle has area $\text{width}\cdot\text{height}=1\cdot\frac{1}{n}$.

TODO picture

Since the sum of the areas of the rectangles overestimate the area under the curve, we have $$\sum_{k=1}^n\frac{1}{k}=\sum\text{areas of the 1st }n\text{ rect.}\ge\int_1^{n+1}\frac{1}{x}\,dx=\log(n+1)$$

Why is there an $n+1$ in the upper limit for the integral? Take a look at the graph. For example, the first 3 rectangles ($n=3$) correspond to the interval $[1,4]$.

Since $\displaystyle\lim_{n\to\infty}\log(n+1)=+\infty$, and $\displaystyle\sum_{k=1}^n\frac{1}{k}\ge\log(n+1)$ for all $n=0,1,2,\ldots$, we conclude $\displaystyle\sum_{k=1}^n\frac{1}{k}=+\infty$.

This generalizes into the Integral Test, which is a test for series convergence. (It is a test by comparison to an improper integral.)

Integral Test

If $f$ is positive, continuous, and decreasing for all $x\ge1$ and $a_n=f(n)$ for all $n=1,2,3,\ldots$

Then $\displaystyle\sum_{n=1}^\infty a_n$ and $\displaystyle\int_1^\infty f(x)\,dx$ either both converge or both diverge.

Remark 1: We do not actually need the conditions to hold for all $n$, just to hold ``eventually,'' since we can throw out the first few terms that don't satisfy the conditions without affecting convergence of the series.

Remark 2: The Integral Test is about convergence/divergence, not equality. The series and the improper integral need not converge to the same value.

Our first application for the Integral Test will be for $p$-series.

Definition: A $p$-series is a series $\displaystyle\sum_{n=1}^\infty\frac{1}{n^p}$, where $p>0$.

(If $p\le0$, then it's not as interesting because the series will be something like $\sum_{n=1}^\infty n^2$, which diverges by the Divergence Test.)

We want to find out for what values of $p>0$ does the $p$-series $\sum_{n=1}^\infty\frac{1}{n^p}$ converge. We have already seen that the harmonic series ($p=1$), $\sum_{n=1}^\infty\frac{1}{n}$, diverges.

We will use the Integral Test to determine convergence of $\sum_{n=1}^\infty\frac{1}{n^p}$.

First we check the requirements of the Integral Test.

• The function $f(x)=\frac{1}{x^p}$ is positive and continuous for $x\ge1$.

• Since $p>0$, then $x^p$ is increasing, so $\frac{1}{x^p}$ is decreasing for $x\ge1$.

Thus we can conclude by the Integral Test that $\sum_{n=1}^\infty\frac{1}{n^p}$ converges if and only if $\int_1^\infty\frac{1}{x^p}\,dx$ converges. We integrate: $$\int_1^\infty\frac{1}{x^p}\,dx=\lim_{t\to\infty}\left.\frac{x^{-p+1}}{-p+1}\right|_1^t \qquad \text{for }p\ne1$$ $$=\lim_{t\to\infty}\left(\frac{1}{-p+1}\cdot{\color{blue}{\frac{1}{t^{p-1}}}}-\frac{1}{-p+1}\right)$$ The only part that depends on $t$ is the blue part, $\dfrac{1}{t^{p-1}}$.

We have $\displaystyle\lim_{t\to\infty}\frac{1}{t^{p-1}}=0$ iff $p-1>0$, or $p>1$.

And $\displaystyle\lim_{t\to\infty}\frac{1}{t^{p-1}}=\infty$ if $p-1<0$, or $p<1$.

We also know that the series and integral will diverge for $p=1$, the harmonic series.

So $\int_1^\infty\frac{1}{x^p}\,dx$ converges iff $p>1$.

To summarize: The series $\displaystyle\sum_{n=1}^\infty\frac{1}{n^p}$ converges if $p>1$, and diverges if $p\le1$.

The above fact is worth remembering, as we will be using it the rest of the chapter on series. The $p$-series and geometric series will be very useful in other tests by comparison.

Example 1: The series $\sum_{n=1}^{\infty}\frac{1}{n^{\sqrt{2}}}$ converges because it is a $p-$series with $p=\sqrt{2}>1$.

Example 2: The series $\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}$ diverges because it is a $p-$series with $p=1/2\le1$.

TODO solution

Remark 1: We should always check if the 3 conditions of the Integral Test are met before applying it. To show it is decreasing, we can take the derivative and show that it is negative, or, if the function is of the form $\frac{1}{\text{stuff}}$, we can show that the denominator is increasing, and so its reciprocal $\frac{1}{\text{stuff}}$ is decreasing.

Remark 2: We don't want to have to work hard to integrate to evaluate the improper integral. That is, we don't want to spend a lot of time using integration techniques and figuring out how to find an antiderivative; there are other, much faster series tests. In particular, rational functions and functions with only powers of $n$ are better handled by the Limit Comparison Test. You don't want to have to try Partial Fractions and Trig Substitution and Power Substitution just to integrate a rational function so you can use the Integral Test; the Limit Comparison Test is much faster. Also, when you have a term that grows geometrically fast or faster, the series is often better handled by the Direct Comparison Test, Ratio Test, or Root Test. For example, $\sum ne^{-n}$ could be handled by the Integral Test, but it involves an Integration by Parts. It is faster to just use the Ratio or Root test to determine that it converges.

Remark 3: From Remark 2, you might ask, ``What good is the Integral Test?'' As seen earlier, it can be used to determine when the $p-$series converges ($p>1$), which is very important for other comparison tests. It is also good when you see an easy to spot antiderivative; one where it doesn't take much work to integrate it. The Integral Test is also especially good for series that consist of only powers of $n$ combined with logarithms.

Remark 4: There is an alternative to the Integral Test, called the Cauchy Condensation Test, that generally works well in the same situations as the Integral Test.

TODO: CCT

Our last example will be determining convergence of the Bertrand series.

Example: Bertrand series...TODO

The terms are positive and continuous. Since the denominator is increasing, the terms (reciprocals) are decreasing. So we may use the Integral Test. $$\int_2^{\infty}\frac{1}{x(\log x)^p}\,dx=\int_{\log2}^{\infty}\frac{1}{u^p}\,du,$$ by substitution $u=\log x$, $du=\frac{1}{x}\,dx$. (This substitution is common when there are $n$s and logarithms. Often it is written backwards as $x=e^u$.)

The integral on the right converges iff $\int_2^{\infty}\frac{1}{u^p}\,du$ converges, since $$\int_{\log2}^{\infty}\frac{1}{u^p}\,du=\int_{\log2}^2\frac{1}{u^p}\,du+\int_2^\infty\frac{1}{u^p}\,du,$$ and the left integral is a convergent integral of a bounded continuous function over a bounded interval. By the Integral Test again, the integral $\int_2^{\infty}\frac{1}{u^p}\,du$ converges iff the series $\sum_{n=2}^\infty \frac{1}{n^p}$ converges. This is a $p-$series, which converges for $p>1$ (the first term when $n=1$ is neglected, but that does not affect convergence or divergence). Thus the original series converges for $p>1$, and diverges for $p\le 1$.

Alternatively, you can stop at $\int_{\log2}^{\infty}\frac{1}{u^p}\,du$ and evaluate it directly to find that it converges.

TODO examples here using ask

Integral test for series convergence

We can compare series to not just (simpler) series, but also (simpler) integrals. The most common version of the integral test says that, if $f$ is a non-negative decreasing function such that $f(n) = a_n$ for all positive integers $n$, then the series $\sum_{n=1}^{\infty} a_n := \lim_{t\to\infty} \sum_{n=1}^{t}a_n$ converges if and only if the integral $\int_1^\infty f(x)\,dx := \lim_{t\to\infty} \int_1^t f(x)\,dx$ (an improper integral with endpoint $\infty$) converges.