The integral test and positive decreasing functions

Understand the idea

The integral test connects a positive series with the accumulated area of a related continuous function. When the function is positive, continuous, and decreasing beyond some starting point, the series and the corresponding improper integral either both converge or both diverge.

Check the conditions on the function used for the comparison, not just on isolated values at integer inputs. Eventual satisfaction is enough: changing finitely many initial terms cannot change convergence. The integral's finite value is generally not the sum of the series.

For a convergent positive decreasing series, neighboring improper integrals also bound the tail after the $\displaystyle N$ th term. This gives a numerical error estimate, but it still depends on the same positivity and monotonicity conditions.

\int_{N+1}^{\infty}f(x)\,dx\le R_N=\sum_{n=N+1}^{\infty}f(n)\le\int_N^{\infty}f(x)\,dx

A method you can apply

Choose a continuous function $\displaystyle f$ with $\displaystyle f(n)=a_n$ .
Verify positivity, continuity, and decreasing behavior on a suitable tail.
Evaluate the improper integral using a limit, then transfer its convergence conclusion to the series.

Worked example

Use the integral test for $\displaystyle \sum_{n=1}^{\infty}1/n^2$ and bound its tail after ten terms.

The function $\displaystyle f(x)=1/x^2$ is positive, continuous, and decreasing for $\displaystyle x\ge1$ .
The improper integral $\displaystyle \int_1^\infty x^{-2}\,dx=1$ is finite, so the series converges.
The tail bounds are $\displaystyle \int_{11}^\infty x^{-2}\,dx=1/11$ and $\displaystyle \int_{10}^\infty x^{-2}\,dx=1/10$ .

The series converges, and its tail satisfies

\displaystyle 1/11\le R_{10}\le1/10

. The integral value

\displaystyle 1

is not the series sum.

Try it yourself

Determine the convergence of $\displaystyle \sum_{n=2}^{\infty}1/(n\ln n)$ using an integral.

Show solution and reasoning

The function is positive, continuous, and decreasing for $\displaystyle x\ge2$ . The integral is $\displaystyle \lim_{b\to\infty}[\ln(\ln x)]_2^b$ , which diverges. Therefore the series diverges.