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.

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Lower & Upper Bounds For The Growth Of $\sum 1/n$

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 $1\le x\le 4$ 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.

$$\boxed{{\displaystyle \sum _1^n\frac{1}{x}>\ln (n+1)}}$$

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 $1\le x\le 4$ 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.

$$\boxed{{\displaystyle \sum _1^n\frac{1}{x}<\ln (n)+1}}$$

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

Convergence Of $\zeta (s)=\sum 1/n^s$

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 $\le 0$ then it is easy to see $\sum 1/n^s$ would diverge because each term would be $\ge 1$.

If we consider the range $1\le x\le 4$ 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 $k\to \mathrm{\infty }$, the right hand side only converges when $s>1$. Because it is less than the right hand side, the sum $\sum 1/x^s$ also converges when $s>1$. We haven't yet ruled out the possibility the sum might also converge for some $s\le 1$.

If we now consider the three taller rectangles $1+\frac{1}{2^s}+\frac{1}{3^s}$ in the range $1\le x\le 4$, 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 $k\to \mathrm{\infty }$, the right hand side diverges when $s\le 1$. Because it is greater than the right hand side, the sum $\sum 1/x^s$ also diverges when $s\le 1$. We have now ruled out the possibility the sum might converge for some $s\le 1$.

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

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

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