Tuesday 9 February 2021

Integral Comparison Tests


Understanding the behaviour of continuous functions is often easier than discrete functions. We can gain insights into discrete sums like $\sum\frac{1}{x}$ by exploring the related continuous integral $\int\frac{1}{x}dx$.

This is a simple but powerful technique used a lot in number theory, and worth becoming familiar with.


Video: [youtube]

Slides: [pdf]


Lower & Upper Bounds For The Growth Of $\sum1/n$


The picture below shows a graph of $y=\frac{1}{x}$, together with rectangles representing the fractions $\frac{1}{n}$.


Lower Bound

If we consider the range $1\leq x\leq4$ we can see the area of the three taller rectangles $1+\frac{1}{2}+\frac{1}{3}$ is greater than the area under the curve $\int_{1}^{4}\frac{1}{x}dx$. By extending the range to $n$ we can make a general observation.

$$\sum_{1}^{n}\frac{1}{x}>\int_{1}^{n+1}\frac{1}{x}dx$$

The integral has an upper limit of $n+1$ because the width of the last rectangle extends from $x=n$ to $x=n+1$. We can perform the integral to simplify the expression.

$$\boxed{\sum_{1}^{n}\frac{1}{x}>\ln(n+1)}$$

This is a rather nice lower bound on the growth of the harmonic series.


Upper Bound

Let's now look at the shorter rectangles. In the range $1\leq x\leq4$ we can see the area of the three shorter rectangles $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}$ is less than the area under the curve $\int_{1}^{4}\frac{1}{x}dx$. Again, by extending the range to n we can make a general observation.

$$\sum_{2}^{n}\frac{1}{x}<\int_{1}^{n}\frac{1}{x}dx$$

The harmonic sum starts at 2 because this time we're looking at rectangles extending to the left of a given $x$. We can adjust the limit of the sum using $\sum_{1}^{n}\frac{1}{x}=1+\sum_{2}^{n}\frac{1}{x}$.

$$\sum_{1}^{n}\frac{1}{x}-1<\int_{1}^{n}\frac{1}{x}dx$$

Again, we can perform the integral.

$$\boxed{\sum_{1}^{n}\frac{1}{x}<\ln(n)+1}$$

This is a nice upper bound to the growth of the harmonic series.


Convergence Of $\zeta(s)=\sum1/n^{s}$


The picture below shows a graph of $y=\frac{1}{x^{s}}$, together with rectangles representing the fractions $\frac{1}{x^{s}}$. The shape of the graph assumes $s>0$. If $s$ was $\leq0$ then it is easy to see $\sum1/n^{s}$ would diverge because each term would be $\geq1$.


If we consider the range $1\leq x\leq4$ we can see the area of the three shorter pink rectangles is less than the area under the curve $\int_{1}^{4}\frac{1}{x^{s}}dx$. By extending the range to $k$ we can make a general observation.

$$\sum_{2}^{k}\frac{1}{x^{s}}<\int_{1}^{k}\frac{1}{x^{s}}dx$$

The sum starts at 2 because we're looking at rectangles extending to the left of a given x. We can adjust the limit of the sum using $\sum_{1}^{k}\frac{1}{x^{s}}=1+\sum_{2}^{k}\frac{1}{x^{s}}$.

$$\sum_{1}^{k}\frac{1}{x^{s}}-1<\int_{1}^{k}\frac{1}{x^{s}}dx$$

The integral is easily evaluated.

$$\sum_{1}^{k}\frac{1}{x^{s}}<\frac{k^{1-s}-1}{1-s}+1$$

As $k\rightarrow\infty$, the right hand side only converges when $s>1$. Because it is less than the right hand side, the sum $\sum1/x^{s}$ also converges when $s>1$. We haven't yet ruled out the possibility the sum might also converge for some $s \leq 1$.

If we now consider the three taller rectangles $1+\frac{1}{2^{s}}+\frac{1}{3^{s}}$ in the range $1\leq x\leq4$, we can see their area is greater than the area under the curve $\int_{1}^{4}\frac{1}{x^{s}}dx$. By extending the range to $k$, we can make a general observation.

$$\sum_{1}^{k}\frac{1}{x^{s}}>\int_{1}^{k+1}\frac{1}{x^{s}}dx$$

The integral has an upper limit of $k+1$ because the width of the last rectangle extends from $x=k$ to $x=k+1$. We can perform the integral to simplify the expression.

$$\sum_{1}^{k}\frac{1}{x^{s}}>\frac{(k+1)^{1-s}-1}{1-s}$$

As $k\rightarrow\infty$, the right hand side diverges when $s\leq1$. Because it is greater than the right hand side, the sum $\sum1/x^{s}$ also diverges when $s\leq1$. We have now ruled out the possibility the sum might converge for some $s \leq 1$.

$$\boxed{\zeta(s)=\sum1/n^{s}\text{ only converges for }s>1}$$

We can go further. The two inequalities together provide a lower and upper bound for the zeta function.

$$\boxed{\frac{1}{s-1}<\zeta(s)<\frac{1}{s-1}+1}$$

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