Divergence and the Integral Test

Start with the fastest necessary condition

Every convergent series has terms that approach zero. Therefore, if

$$\lim_{n\to\infty}a_n\ne0$$

or if the limit does not exist, then $\sum a_n$ diverges. This divergence test is efficient because it inspects the terms before attempting a more elaborate comparison. For the series $\sum(3n+1)/(n+2)$, the terms approach $3$, not zero, so the series diverges immediately. No partial-sum formula is needed.

The test has only one direction. When $a_n\to0$, the series might converge or diverge. The harmonic series $\sum1/n$ is the standard warning: its terms vanish, but its partial sums increase without bound. A zero term limit means that later additions are small; it does not say that their total remaining effect is finite. Always state the logical result accurately: nonzero term limit proves divergence, while zero term limit is inconclusive.

Link a decreasing function to a series

The integral test applies when $a_n=f(n)$ for a function $f$ that is positive, continuous, and decreasing for sufficiently large $x$. Under those conditions,

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

either both converge or both diverge. The test does not say that the sum equals the integral. It says that the rectangular series contributions and the curved area have matching long-run finiteness behavior because decreasing rectangles trap the area above and below in controlled ways.

The hypotheses matter. Positivity makes the geometric area comparison meaningful, decreasing behavior puts the rectangle heights in the needed order, and continuity lets the improper integral represent the curve cleanly. A function can fail one of these conditions near the beginning without ruining the conclusion if the conditions hold after some index. Adding or removing finitely many terms changes a partial sum by a finite amount, so it cannot turn convergence into divergence or vice versa.

This “eventually” principle is more useful than it first appears. A series may have a few unusual first terms, a denominator that is zero at a small index, or a function that becomes decreasing only after a threshold. Record the threshold and start the comparison after it; then handle the finite beginning separately if it needs to be defined at all. What matters for convergence is the behavior of the endless tail. Treating a finite prefix as though it controls an infinite process is a common category error.

The p-series is a reference family

Apply the integral test to $f(x)=1/x^p$ for $x\ge1$. For $p\ne1$,

$$\int_1^B\frac{1}{x^p}\,dx =\left[\frac{x^{1-p}}{1-p}\right]_1^B.$$

As $B\to\infty$, this improper integral converges when $p>1$ and diverges when $p\le1$. Thus

$$\sum_{n=1}^{\infty}\frac{1}{n^p} \begin{cases} \text{converges},&p>1,\\ \text{diverges},&p\le1. \end{cases}$$

The boundary $p=1$ is worth remembering because it is the harmonic series. A larger exponent makes terms shrink fast enough to have a finite total; an exponent at or below one does not. The statement does not depend on a vague impression that “powers in the denominator are good.” It depends on precisely how the power affects the infinite-area limit.

For example, $\sum1/n^{3/2}$ converges because $3/2>1$, while $\sum1/\sqrt{n}$ diverges because $1/2\le1$. Neither conclusion requires calculating a sum. The p-series test classifies the behavior, which is often the goal. When later tests compare a new series to one of these reference forms, the p-series threshold supplies the known benchmark.

Be careful with notation such as $1/\sqrt[3]{n}$. Rewrite it as $n^{-1/3}$ or $1/n^{1/3}$ before identifying $p$. The exponent belongs to the whole denominator factor, not merely to the symbol $n$ in an informal visual sense. Rewriting makes the comparison to $1/n^p$ explicit and avoids accidentally treating a root as a power greater than one. Clear algebra at this stage keeps the convergence argument from becoming a pattern-matching guess.

Integral remainders estimate what is left

For a positive decreasing series that converges, the integral test also bounds the tail after $N$ terms:

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

where $R_N=\sum_{n=N+1}^{\infty}a_n$ is the remainder. These inequalities give a way to decide how many terms are needed for a desired accuracy. They also clarify why a partial sum is an approximation: the omitted tail is positive and can be bounded by a familiar area.

Suppose $a_n=1/n^2$. Since $\int_N^\infty x^{-2}dx=1/N$, the error after $N$ terms is at most $1/N$. To guarantee an error below $0.01$, it is enough to use $N>100$ terms. The estimate is conservative; it guarantees accuracy without claiming it is the smallest possible number of terms. State whether an error bound is an exact error or a guaranteed upper bound—those are different claims.

Retrieval check

Classify each series with a reason: $\sum_{n=1}^{\infty}(n+1)/(2n)$, $\sum_{n=1}^{\infty}1/n^{5/4}$, and $\sum_{n=1}^{\infty}1/\sqrt[3]{n}$. For the second, write the integral-test comparison function and determine a simple number of terms sufficient to make the remainder less than $0.02$. Explain why that term count is a guarantee rather than an exact necessity. The next lesson compares positive-term series directly to known reference series when an integral is not the most natural tool.