Dominant Terms in Sums of Positive Sequences

Why the Leading Term Can Determine a Series

A positive term in a series may be made from several simpler pieces. If one piece is much larger than all the others for large $n$, then the whole term can have the same asymptotic behavior as that dominant piece. This can simplify a convergence question: instead of studying a complicated sum, we can compare it with a familiar positive sequence.

The key requirements are that the sum contains a fixed finite number of positive terms and that every term other than the proposed dominant one becomes negligible relative to it. Under those conditions, their ratios add to a quantity that tends to zero.

Definition: For positive sequences $a_n$ and $b_n$, we say that $a_n$ is asymptotic to $b_n$, and write $a_n\sim b_n$, if $$ \lim_{n\to\infty}\frac{a_n}{b_n}=1. $$

In particular, $a_n\sim b_n$ means that their ratio tends to one; it does not mean that their values are equal for each $n$. For positive series, asymptotic equivalence is useful because the limit comparison test says that if $a_n/b_n$ tends to a finite positive number, then $\sum a_n$ and $\sum b_n$ either both converge or both diverge.

The Finite-Sum Result

Suppose that $u_{1,n},u_{2,n},\ldots,u_{m,n}$ are positive sequences, where $m$ is fixed, and suppose $u_{1,n}$ is proposed as the dominant term. The condition to check is that each of the other terms is negligible relative to it:

\lim_{n\to\infty}\frac{u_{j,n}}{u_{1,n}}=0 \qquad (j=2,\ldots,m).

The conclusion follows by dividing the whole sum by $u_{1,n}$. The proof also shows why the number of terms must remain fixed: it uses a finite sum of quantities that each tend to zero.

Theorem: Let $m$ be a fixed positive integer, and let $u_{j,n}>0$ for $j=1,\ldots,m$. If $$ \lim_{n\to\infty}\frac{u_{j,n}}{u_{1,n}}=0 \qquad\text{for every }j=2,\ldots,m, $$ then $$ u_{1,n}+u_{2,n}+\cdots+u_{m,n}\sim u_{1,n}. $$

Here is the complete derivation. Because $u_{1,n}>0$, division by it is valid. The ratio of the sum to the proposed dominant term is

\frac{u_{1,n}+u_{2,n}+\cdots+u_{m,n}}{u_{1,n}} = 1+\sum_{j=2}^{m}\frac{u_{j,n}}{u_{1,n}}.

There are only finitely many ratios in the sum on the right, and each tends to zero. The limit of a finite sum is the sum of its limits, so

\lim_{n\to\infty} \left(1+\sum_{j=2}^{m}\frac{u_{j,n}}{u_{1,n}}\right) = 1+\sum_{j=2}^{m}0 = 1.

Thus the ratio of the original sum to $u_{1,n}$ tends to one, which is exactly the definition of asymptotic equivalence. This proves the theorem.

From the Equivalence to a Series Test

Write $S_n=u_{1,n}+\cdots+u_{m,n}$. If the theorem gives $S_n\sim u_{1,n}$, then $S_n/u_{1,n}\to1$. In particular, for all sufficiently large $n$, this ratio lies between $1/2$ and $3/2$. Since $u_{1,n}>0$, multiplying by $u_{1,n}$ gives

\frac12u_{1,n}\le S_n\le\frac32u_{1,n}.

These inequalities show directly that $\sum S_n$ converges if and only if $\sum u_{1,n}$ converges: each series is bounded above by a constant multiple of the other, apart from finitely many initial terms. Equivalently, the limit comparison test applies because $S_n/u_{1,n}\to1$. The finitely many initial terms do not affect convergence.

Worked Example: A Convergent Sum with a Power as Its Dominant Term

For $n\ge1$, let $$ S_n=\frac{1}{n^2}+\frac{3}{n^4}. $$ Both summands are positive. Choose $u_{1,n}=1/n^2$. The ratio of the second term to the first is

\frac{3/n^4}{1/n^2} = \frac{3}{n^4}\cdot n^2 = \frac{3}{n^2} \longrightarrow 0.

Therefore,

\frac{S_n}{1/n^2} = \frac{1/n^2+3/n^4}{1/n^2} = 1+\frac{3}{n^2} \longrightarrow 1, \qquad\text{so}\qquad S_n\sim\frac{1}{n^2}.

The series $\sum_{n=1}^{\infty}1/n^2$ converges by the $p$-series test, since $p=2>1$. Limit comparison therefore shows that $\sum_{n=1}^{\infty}S_n$ converges as well. The check is the computed ratio: it tends to the finite positive value $1$, as required.

Worked Example: A Divergent Sum with a Harmonic Dominant Term

For $n\ge1$, consider $$ T_n=\frac{1}{n}+\frac{1}{n^2}. $$ Choose $d_n=1/n$. The other term divided by $d_n$ is

\frac{1/n^2}{1/n} = \frac{1}{n^2}\cdot n = \frac{1}{n} \longrightarrow 0.

Consequently, the ratio of the full sum to the selected term is

\frac{T_n}{1/n} = \frac{1/n+1/n^2}{1/n} = 1+\frac{1}{n} \longrightarrow 1, \qquad\text{so}\qquad T_n\sim\frac{1}{n}.

The harmonic series $\sum_{n=1}^{\infty}1/n$ diverges. The limit comparison test, with $T_n/(1/n)\to1$, shows that $\sum_{n=1}^{\infty}T_n$ diverges. This confirms that adding the smaller positive term does not change the convergence classification.

Worked Example: A Logarithmic Term That Is Negligible Beside a Power

For $n\ge2$, let $$ R_n=\frac{1}{n^{3/2}}+\frac{\ln n}{n^2}. $$ Select $d_n=1/n^{3/2}$. Dividing the logarithmic term by this power gives

\frac{(\ln n)/n^2}{1/n^{3/2}} = \frac{\ln n}{n^2}\cdot n^{3/2} = \frac{\ln n}{\sqrt n} \longrightarrow 0.

The limit is zero because a positive power of $n$ eventually dominates $\ln n$, as established in the earlier material on why powers dominate logarithms. It follows that

\frac{R_n}{1/n^{3/2}} = 1+\frac{\ln n}{\sqrt n} \longrightarrow 1, \qquad\text{so}\qquad R_n\sim\frac{1}{n^{3/2}}.

The comparison series $\sum_{n=2}^{\infty}1/n^{3/2}$ converges by the $p$-series test, since $3/2>1$. Hence the limit comparison test proves that $\sum_{n=2}^{\infty}R_n$ converges. As a check, the ratio of $R_n$ to its proposed dominant term has been explicitly shown to tend to $1$.

When There Is No Single Dominant Term

The theorem requires every other term divided by the chosen dominant term to tend to zero. Being positive and no larger at each index is not enough. If another term remains comparable in size, it contributes to the leading-order sum and cannot be discarded.

Worked Example: Two Terms of the Same Order

Let $$ Q_n=\frac{1}{n}+\frac{2}{n}=\frac{3}{n}. $$ If we choose $1/n$ as the sole dominant term, the ratio of the other term to it is

\frac{2/n}{1/n} = \frac{2}{n}\cdot n = 2,

which does not tend to zero. In fact, the full ratio is

\frac{Q_n}{1/n} = \frac{3/n}{1/n} = 3,

not $1$. Thus $Q_n\not\sim1/n$. The correct equivalence is $Q_n\sim3/n$, since $Q_n/(3/n)=1$ for every $n$. The series $\sum Q_n$ diverges because it is three times the harmonic series.

A Common Mistake

A tempting but incorrect shortcut is to call one term “largest” and immediately replace the whole sum by that term, without checking the ratios. For the concrete sum $1/n+2/n$, that shortcut would assert

\frac{1/n+2/n}{1/n}=1.

But calculating the left side gives

\frac{1/n+2/n}{1/n} = \frac{3/n}{1/n} = 3\ne1.

The correction is to test each remaining term relative to the proposed dominant term. Here the second-to-first ratio is $2$, not a quantity tending to zero. The two terms have the same order, so their contributions must both be retained.

Why Positivity and Finiteness Matter

Positivity makes the comparisons meaningful and allows the limit comparison test to transfer convergence behavior. The proof also depends on having a fixed finite number of summands: finitely many ratios tending to zero have a sum tending to zero. If the number of summands changes with $n$, this argument alone does not establish that their total contribution is negligible.

The practical method is therefore precise: identify a candidate term, divide every other term by it, and check that each resulting ratio tends to zero. Only after that check may the sum be replaced asymptotically by the candidate term. Once the equivalence is established, compare its series with a known convergence benchmark.

Check Your Understanding

For each question, verify the relevant ratio before making a conclusion about the whole sum or its series.

For $A_n=1/n^3+5/n^5$, which term is a candidate dominant term, and what is the ratio of the other term to it?
If $B_n=1/n+1/n^2$, what is the limit of $B_n/(1/n)$, and what does it imply about the series $\sum B_n$?
Why does a fixed number of negligible positive terms have a negligible total contribution in the theorem?
Does $1/n+4/n$ have $1/n$ as an asymptotic equivalent? Check the ratio and give a correct equivalent.
If $C_n\sim1/n^{5/4}$, what convergence conclusion follows for $\sum C_n$ by limit comparison?