How the Integral Test Determines Series Convergence

The test connects a positive series to an improper integral

Suppose $a_n=f(n)$ for a function $f$ that is positive, continuous, and decreasing for all $x\ge N$. The integral test states that

$$\sum_{n=N}^{\infty}a_n \quad\text{and}\quad \int_N^{\infty}f(x)\,dx$$

either both converge or both diverge. The test compares convergence behavior. It does not say that the sum of the series equals the value of the integral.

Why decreasing positive terms matter

For a decreasing positive function, unit-width rectangles with heights $f(n)$ sit in a controlled relationship with the area under the curve. Those upper and lower area comparisons keep the tail of the series and the tail of the integral bounded together.

Positivity prevents cancellation from hiding large terms. Continuity makes the improper integral well-defined on the interval, and decreasing behavior supports the rectangle comparison. If one of these conditions fails, a different convergence test may be needed.

A reliable workflow

1. Choose a continuous extension $f(x)$ with $f(n)=a_n$.
2. Verify that $f$ is positive, continuous, and decreasing from some $N$ onward.
3. Evaluate $\int_N^b f(x)\,dx$.
4. Take the limit as $b\to\infty$.
5. Transfer only the convergence or divergence conclusion back to the series.

Changing or removing finitely many starting terms does not change whether an infinite series converges, so the conditions only need to hold eventually.

Example with a different series

Consider

$$\sum_{n=1}^{\infty}\frac{1}{(n+1)^2}.$$

Take $f(x)=1/(x+1)^2$. It is positive, continuous, and decreasing for $x\ge1$, and

$$\int_1^{\infty}\frac{1}{(x+1)^2}\,dx =\left[-\frac{1}{x+1}\right]_1^{\infty} =\frac12.$$

The improper integral converges, so the series converges. Its sum is not being claimed to equal $1/2$; only its convergence classification transfers.