Proof of the Limit Comparison Test

From a Limit to a Comparison

The Limit Comparison Test replaces a difficult term-by-term inequality with a limit. Its proof rests on a simple fact: if the ratio of two positive sequences approaches a positive finite number, then sufficiently far out the ratio stays between two positive constants. Multiplying those bounds by the comparison terms produces the inequalities needed for the Comparison Test.

The Limit Comparison Test was stated earlier in this course. Here we give its proof from that tail-bound mechanism, and then prove a stronger conclusion about remainders when the series converge. Throughout, “eventually positive” means positive for every index from some integer onward. A finite number of initial terms can be set aside by the earlier Theorem (Finite Changes Preserve Series Convergence).

The Tail-Sandwich Lemma

The key step is a direct use of the definition of a limit. The constants need not be optimal; they only need to be positive, finite, and valid for every sufficiently large index.

Proof. Fix \(\varepsilon\) with \(0<\varepsilon<L\). By the definition of the limit, there is an integer \(N_1\) such that for every \(n\geq N_1\),

This absolute-value inequality is equivalent to

Choose \(N\) large enough that both the ratio is defined and \(b_n>0\) for every \(n\geq N\). Since \(b_n\) is positive, multiplying the last inequality by \(b_n\) preserves its direction and gives

for every \(n\geq N\). The restriction \(\varepsilon<L\) ensures the lower constant \(L-\varepsilon\) is positive. \(\square\)

For instance, taking \(\varepsilon=L/2\) gives the particularly convenient bounds

for all sufficiently large \(n\). The constants \(L/2\) and \(3L/2\) are not special. Their role is to be positive and finite while enclosing the ratio \(a_n/b_n\) eventually.

Proof of the Limit Comparison Test

Proof. Apply the Tail Sandwich Lemma with \(\varepsilon=L/2\). There is an integer \(N\) such that, for every \(n\geq N\),

All terms in these inequalities are positive. Suppose first that \(\sum b_n\) converges. The upper bound gives \(0\leq a_n\leq (3L/2)b_n\) for every \(n\geq N\). The Comparison Test for Nonnegative Series therefore shows that the tail \(\sum_{n=N}^{\infty}a_n\) converges. Adding the finite initial segment does not change convergence, so \(\sum a_n\) converges.

Conversely, suppose that \(\sum a_n\) converges. Rearranging the lower bound gives

for every \(n\geq N\), because \(L>0\). The Comparison Test now shows that \(\sum_{n=N}^{\infty}b_n\) converges. Again, the finite initial segment does not affect convergence, so \(\sum b_n\) converges. We have proved each implication, and hence the two series converge or diverge together. \(\square\)

The proof does not require \(a_n\) and \(b_n\) to be positive at every index. It requires positivity on a tail, where the comparison inequalities are used. Any finite initial segment contributes only a finite sum. The proof also shows why the hypothesis \(L>0\) matters: it supplies a positive lower bound, which is essential for transferring convergence in the reverse direction.

Worked Applications of the Proof

What the Same Bounds Say About Remainders

When both series converge, the same inequalities can be summed over a tail. This gives more than agreement about whether the series converge: their remainders have an asymptotic ratio determined by the original term ratio. For a convergent series \(\sum a_n\), write

These are the sums of the terms after the \(N\)th partial sum. With eventually positive terms, each remainder is positive once \(N\) is sufficiently large.

Proof. By the Limit Comparison Test, both series converge. Fix any \(\varepsilon\) with \(0<\varepsilon<L\). By the Tail Sandwich Lemma, there is an integer \(N_0\) such that for every \(n>N_0\),

For any \(N\geq N_0\), these inequalities hold for every term with index \(n>N\). Summing first over \(N<n\leq M\), for any integer \(M>N\), gives

As \(M\) tends to infinity, the finite partial sums of each convergent tail tend to their respective remainders. Passing to those limits yields

For sufficiently large \(N\), \(R_N^{(b)}>0\), since it is a sum of positive terms. Divide by this remainder to obtain

Given any desired tolerance \(\delta>0\), choose \(\varepsilon\) with \(0<\varepsilon<\min\{L,\delta\}\). The last inequality then places the remainder ratio within \(\delta\) of \(L\) for every sufficiently large \(N\). This is exactly the definition of convergence of that ratio to \(L\). \(\square\)

Why the Proof Is Useful

A ratio limit is not itself a comparison inequality valid at every index. The proof turns it into an inequality only on a tail, where the limit definition guarantees control. This distinction prevents a common mistake: trying to prove a global bound when a tail bound is all the convergence tests require.

The choice of constants can be adapted to the task. To prove convergence in one direction, any finite upper constant is useful; to prove the reverse direction, a positive lower constant is essential. Taking \(\varepsilon=L/2\) guarantees both at once. The remainder theorem uses the same idea with arbitrarily small \(\varepsilon\), which is why it yields a limiting ratio rather than just rough constant-factor bounds.