The Term Test Away from the Center

Why Check the Terms at a Candidate Point?

A power series is a sum whose terms depend on the chosen value of its variable. At one point, those terms may shrink toward zero; at another, they may stay constant or grow. Before trying a more involved convergence test, it is useful to check this basic requirement: if the series converges, its individual terms must tend to zero.

This gives a direct way to rule out candidate points. If substituting a point into the terms produces a sequence that does not approach zero, the power series cannot converge there. The reverse implication does not hold: terms approaching zero is necessary for convergence, but does not by itself establish convergence.

The Term Test for a Power Series

Consider a power series centered at $c$, written as $\sum_{n=0}^{\infty}a_n(x-c)^n$. For a fixed real number $x$, its $n$th term is $a_n(x-c)^n$. The question here is what convergence at that particular $x$ forces those terms to do.

Theorem: If the power series $\sum_{n=0}^{\infty}a_n(x-c)^n$ converges at a fixed point $x$, then $$ \lim_{n\to\infty}a_n(x-c)^n=0. $$ Consequently, if these terms do not tend to zero, the power series diverges at $x$.

The implication applies at any fixed $x$, including $x=c$. At the center, every term with $n\geq 1$ is zero. The most informative use of the result is away from the center, where $x-c\neq 0$ and the terms depend on both the coefficients and the distance from the center.

Why Convergence Forces the Terms to Vanish

The result follows from the definition of convergence of a series. For a fixed $x$, write $u_n=a_n(x-c)^n$, and let $S_N=\sum_{n=0}^{N}u_n$ be the $N$th partial sum. If the series converges, the partial sums approach a finite limit $S$.

Express a term using partial sums.
For every $n\geq 1$, subtract consecutive partial sums: $u_n=S_n-S_{n-1}$.

Use convergence of the partial sums.
Since $S_n\to S$, the shifted sequence $S_{n-1}$ also tends to $S$.

Take the difference.
Thus $u_n=S_n-S_{n-1}\to S-S=0$, as required.

This proof also explains the limitation of the test. It derives a necessary condition from convergence. It does not show that a sequence of terms tending to zero has partial sums approaching a finite limit.

Using the Condition to Rule Out Points

At a candidate point, substitute its value into the general term and simplify. Then determine whether the resulting sequence tends to zero. If it does not, the theorem rules out convergence immediately. The series need not be summed, and no estimate of its partial sums is needed.

Worked Example: A Candidate Point Where the Terms Stay Constant

Consider the power series $\sum_{n=0}^{\infty}2^n(x-1)^n$. Test $x=\frac{3}{2}$, which is away from the center $c=1$.

The $n$th term at this point is

a_n(x-c)^n =2^n\left(\frac{3}{2}-1\right)^n =2^n\left(\frac{1}{2}\right)^n =1.

Thus the terms have limit $1$, not $0$. The term test says the power series diverges at $x=\frac{3}{2}$.

Check: Substituting $x=\frac{3}{2}$ makes each summand exactly $1$, so the terms do not approach zero. The divergence conclusion follows from the necessary condition.

Worked Example: Terms That Grow at a Candidate Point

Consider $\sum_{n=1}^{\infty}n(x-2)^n$, and test the point $x=3$.

Here $c=2$, so $x-c=3-2=1$. The $n$th term becomes

n(x-2)^n=n(3-2)^n=n\cdot 1^n=n.

As $n\to\infty$, these terms grow without bound; in particular, they do not tend to zero. Therefore the power series diverges at $x=3$.

Check: The substituted term is $n$ for every $n\geq 1$, and $\lim_{n\to\infty}n$ is not zero. The term test rules out convergence.

These calculations show why it is important to test the complete term $a_n(x-c)^n$, not just the coefficient $a_n$. The value of $x-c$ changes the powers in the term and can change whether those terms approach zero.

When the Terms Do Approach Zero

If the terms do approach zero, the term test has passed, but that alone does not tell us whether the series converges. A familiar example makes the distinction clear: the harmonic series has terms that tend to zero but still diverges. This can also occur at a point of a power series.

Worked Example: Passing the Term Test but Still Diverging

Consider the power series $\sum_{n=1}^{\infty}\frac{(x-4)^n}{n}$, and test $x=5$. This point is away from its center $c=4$.

Since $5-4=1$, its terms at $x=5$ are

\frac{(5-4)^n}{n}=\frac{1^n}{n}=\frac{1}{n}.

These terms do tend to zero, because $\lim_{n\to\infty}1/n=0$. So the term test does not rule out convergence. To see that the series nevertheless diverges here, group its terms in blocks. For each integer $k\geq 1$, the block from $n=2^{k-1}+1$ to $n=2^k$ contains $2^{k-1}$ terms. Each term in that block is at least $1/2^k$, so the block sum is at least

2^{k-1}\cdot\frac{1}{2^k}=\frac{1}{2}.

Every such block contributes at least $1/2$. Therefore the partial sums increase by at least $1/2$ for each successive block and cannot approach a finite limit. The series diverges at $x=5$.

Check: The terms are $1/n$, which tend to zero, but the grouped partial sums grow without bound. This verifies that passing the term test is not sufficient for convergence.

A Common Mistake: Reversing the Implication

A frequent error is to treat “the terms tend to zero” as if it meant “the series converges.” The theorem gives only the implication from convergence to terms tending to zero. Its converse is false.

Keep the direction straight: $$ \text{series converges} \quad\Longrightarrow\quad \text{terms tend to }0. $$ The fact that the terms tend to zero alone does not imply that the series converges.

For example, at $x=5$ in the previous worked example, the terms are $1/n$, so they tend to zero. Yet the series diverges. A correct use of the test is therefore to say: “The terms do not tend to zero, so the series diverges.” If the terms do tend to zero, the correct conclusion is only: “The term test gives no decision.”

Another error is to check only the coefficients. Even if $a_n\to 0$, that fact alone does not establish that $a_n(x-c)^n\to 0$ at a chosen point. The term test concerns the full summand after $x$ has been fixed.

A Reliable Check at a Candidate Point

For each candidate point away from the center, use the same sequence of steps. This keeps the calculation focused on the necessary condition and avoids drawing a stronger conclusion than the test supports.

Identify the center and candidate point.
Compute $x-c$, retaining its sign.

Substitute into the general term.
Simplify $a_n(x-c)^n$ as a sequence in $n$.

Check its limit.
If the limit is not zero, conclude that the series diverges at that point.

State the limitation when the limit is zero.
Say that this test alone gives no decision; use an appropriate further argument if convergence must be determined.

Why This Test Matters

The term test is simple, but it is a useful first check whenever a power series is evaluated at a specific point. A single nonzero term limit is enough to rule out convergence, even when the full sum would be difficult to analyze. It also provides a discipline for interpreting convergence tests: a necessary condition can disprove a claim, but satisfying that condition is not automatically proof.

For a power series, the test is point-specific. It asks whether the terms $a_n(x-c)^n$ tend to zero at the chosen $x$. It does not, by itself, describe all points of convergence or determine what happens at other points. Its conclusion should be kept to the candidate point actually tested.

Check Your Understanding

For each question, distinguish what the term test proves from what it leaves undecided.

State the necessary condition on $a_n(x-c)^n$ if a power series converges at $x$.
At $x=2$, the terms of a series are $3^n(x-2)^n$. Do they rule out convergence? Explain.
At a fixed point, the terms of a series approach $1$. What conclusion follows?
Why does $\lim_{n\to\infty}a_n(x-c)^n=0$ not by itself prove convergence?
For $\sum_{n=1}^{\infty}(x-3)^n/n$, evaluate the terms at $x=4$. What does the term test conclude, and what does it not conclude?