Proving Divergence From a Nonzero Term Limit

When the Terms Settle Away from Zero

A series is built by adding terms, one after another. For its sum to settle toward a finite value, the additions must eventually become arbitrarily small. If the terms instead approach a fixed, nonzero number, they do not become small: eventually, they continue adding an amount bounded away from zero, always with the same sign. The partial sums therefore cannot settle at a finite value.

This gives a direct way to prove divergence. The key is to distinguish the limit of the terms from the sum of the series: one describes the behavior of individual additions, while the other describes the behavior of their accumulated totals.

Central idea: If the terms of a series approach a finite, nonzero limit, then the series diverges. The terms eventually have a fixed sign and have magnitude bounded below by a positive number.

Terms and Partial Sums

For a sequence of terms \(a_1,a_2,a_3,\ldots\), the series \(\sum_{n=1}^{\infty}a_n\) is defined through its sequence of partial sums. The \(N\)th partial sum is the total of the first \(N\) terms:

S_N=\sum_{n=1}^{N}a_n=a_1+a_2+\cdots+a_N.

The infinite series converges to a finite sum if the sequence of partial sums \(S_N\) has a finite limit as \(N\) tends to infinity. If the partial sums do not approach a finite limit, the series diverges.

Keep the two limits separate. The limit \(\lim_{n\to\infty}a_n\) concerns the individual terms. The limit \(\lim_{N\to\infty}S_N\), if it exists as a finite number, is the sum of the series.

Theorem (nonzero-term-limit divergence test): Suppose the terms of a series satisfy \(\lim_{n\to\infty}a_n=L\), where \(L\) is finite and \(L\ne 0\). Then the series \(\sum_{n=1}^{\infty}a_n\) diverges.

This result is one direction of the basic term test: convergence of a series requires its terms to approach zero. Here we use the contrapositive of that requirement for the specific case in which the terms approach a finite, nonzero limit. The conclusion is divergence, not that the series has sum \(L\).

Why a Nonzero Limit Forces Divergence

We can prove the theorem directly by showing that the partial sums grow without bound in one direction. The argument uses the definition of a limit to control all sufficiently late terms at once.

Choose a tolerance.
Since \(a_n\to L\ne 0\), use \(\varepsilon=|L|/2\). By the definition of a sequence limit, there is an index \(N\) such that \(|a_n-L|<|L|/2\) whenever \(n\ge N\).

Determine the eventual sign and size.
If \(L>0\), then \(a_n>L-|L|/2=L/2>0\). If \(L<0\), then \(a_n<L+|L|/2=L/2<0\). Thus, after \(N\), all terms have the sign of \(L\) and their magnitudes exceed \(|L|/2\).

Apply the bound to partial sums.
If \(L>0\), each additional term from index \(N\) onward is greater than \(L/2\), so the partial sums increase without bound. If \(L<0\), each such term is less than \(L/2<0\), so the partial sums decrease without bound.

For completeness, write those bounds explicitly. When \(L>0\), for \(M\ge N\),

S_M = S_{N-1}+\sum_{n=N}^{M}a_n > S_{N-1}+(M-N+1)\frac{L}{2}.

The right-hand side tends to positive infinity as \(M\) increases. When \(L<0\), the corresponding bound is

S_M = S_{N-1}+\sum_{n=N}^{M}a_n < S_{N-1}+(M-N+1)\frac{L}{2}.

Because \(L/2<0\), this upper bound tends to negative infinity. In either case, the partial sums cannot approach a finite limit. This proves the theorem.

The proof also explains why the conclusion holds: finitely many early terms can shift the partial sums, but they cannot counteract the infinitely many later additions, each of which has the same sign and at least a fixed positive magnitude.

Worked Applications

Worked Example: A Positive Rational Term Limit

Determine whether the series \(\sum_{n=1}^{\infty}\frac{2n}{n+3}\) converges.

First find the limit of its terms by dividing the numerator and denominator by \(n\):

a_n=\frac{2n}{n+3} =\frac{2}{1+3/n} \quad\Longrightarrow\quad \lim_{n\to\infty}a_n = \frac{2}{1+0} =2.

The limit is finite and nonzero, so the theorem already proves divergence. We can also see the growth in the partial sums explicitly. For \(n\ge4\),

\frac{2n}{n+3}>1 \quad\Longleftrightarrow\quad 2n>n+3 \quad\Longleftrightarrow\quad n>3.

The first three terms are \(1/2\), \(4/5\), and \(1\), giving \(S_3=1/2+4/5+1=23/10\). Therefore, for \(M\ge4\),

S_M=S_3+\sum_{n=4}^{M}\frac{2n}{n+3} > \frac{23}{10}+(M-3).

This lower bound grows without bound, so the series diverges to positive infinity. Check: the terms do approach \(2\), as calculated, and the explicit bound shows that the partial sums cannot have a finite limit.

Worked Example: A Negative Rational Term Limit

Determine whether the series \(\sum_{n=1}^{\infty}-\frac{5n}{n+2}\) converges.

Its term limit is

a_n=-\frac{5n}{n+2} =-\frac{5}{1+2/n} \quad\Longrightarrow\quad \lim_{n\to\infty}a_n=-5.

Since \(-5\ne0\), the nonzero-term-limit test proves divergence. To verify the direction of divergence, for \(n\ge4\) we have

\frac{5n}{n+2}>3 \quad\Longleftrightarrow\quad 5n>3n+6 \quad\Longleftrightarrow\quad n>3,

so \(a_n<-3\) for every \(n\ge4\). The first three terms are \(-5/3\), \(-5/2\), and \(-3\), hence \(S_3=-10/6-15/6-18/6=-43/6\). For \(M\ge4\),

S_M=S_3+\sum_{n=4}^{M}a_n < -\frac{43}{6}-3(M-3).

The right-hand side tends to negative infinity, so the partial sums decrease without bound. Check: the terms approach \(-5\), and the bound confirms that their continuing negative contributions prevent a finite sum.

Worked Example: A Constant Multiple of a Familiar Sequence

Determine whether the series \(\sum_{n=1}^{\infty}6\frac{n}{n+1}\) converges.

The sequence \(n/(n+1)\) approaches \(1\). Multiplying by the constant \(6\) gives

a_n=6\frac{n}{n+1} =\frac{6}{1+1/n} \quad\Longrightarrow\quad \lim_{n\to\infty}a_n =\frac{6}{1+0} =6.

This limit is finite and nonzero, so the series diverges. For a direct bound, whenever \(n\ge2\),

6\frac{n}{n+1}>3 \quad\Longleftrightarrow\quad 6n>3n+3 \quad\Longleftrightarrow\quad n>1.

The first term is \(6(1/2)=3\), so \(S_1=3\). Every term from \(n=2\) onward exceeds \(3\); therefore, for \(M\ge2\),

S_M = 3+\sum_{n=2}^{M}6\frac{n}{n+1} > 3+3(M-1) =3M.

Since \(3M\) grows without bound, so do the partial sums. Check: the constant multiple has term limit \(6\), consistent with the divergence conclusion.

Common Mistake: Confusing the Term Limit with the Sum

A nonzero limit of the terms is not the value of the series. For example, the calculation \(\lim_{n\to\infty}2n/(n+3)=2\) does not mean that \(\sum_{n=1}^{\infty}2n/(n+3)=2\). It means the individual terms approach \(2\). The series diverges because its partial sums keep receiving positive additions bounded away from zero.

In particular, do not use the term test as an estimate for a sum. Its conclusion in this situation is simply that no finite sum exists. The positive and negative examples above show the two possible directions of unbounded partial sums.

Takeaway: A finite, nonzero term limit is decisive: the series diverges. A positive limit forces the partial sums upward without bound; a negative limit forces them downward without bound.

Why the Test Is Useful

This test is often the first check to make when considering a series. It can settle divergence immediately, without finding a formula for the partial sums or comparing the series with another one. For rational expressions and constant multiples, simplifying the term and taking its limit may be enough.

The condition matters. This result applies when the term limit exists and is finite and nonzero. If the limit is zero, this particular test gives no conclusion; if a limit does not exist or the terms are unbounded, those situations require separate treatment. For this test, the reliable sequence is: calculate the term limit, check that it is finite and not zero, and then conclude divergence.

Check Your Understanding

For each question, focus on the limit of the terms and what it implies about the partial sums.

State the condition on \(\lim_{n\to\infty}a_n\) that allows the nonzero-term-limit divergence test to be applied.
If \(a_n\to 4\), explain why the terms eventually have a fixed sign and stay bounded away from zero.
If \(a_n\to -3\), in which direction do the partial sums diverge? Give the reason.
For \(a_n=8n/(n+4)\), find the term limit and state whether \(\sum_{n=1}^{\infty}a_n\) converges.
Why does a term limit of \(L\ne0\) not mean the series has sum \(L\)?