Applying Integral and Comparison Tests (BC)

Why These Tests Are Useful

A positive-term series can be difficult to classify by looking at its terms alone. The integral test connects a series to an improper integral, while comparison tests relate its terms to a series whose behavior is already known. In each case, the key is to verify the hypotheses and compare the tails correctly.

These methods do not require a formula for the sum. They answer whether the partial sums approach a finite limit or grow without bound. When using them, distinguish carefully between comparing individual terms and comparing sums, and remember that an inequality may hold only after some starting index.

For positive terms, the partial sums are increasing. Thus, proving that they are bounded above proves convergence; showing that they grow at least as fast as the partial sums of a divergent positive-term series proves divergence. The tests below make those comparisons precise.

The Integral Test

The integral test is useful when the terms \(a_n\) are values of a positive, decreasing function. Decreasing means that \(f(x_2)\leq f(x_1)\) whenever \(x_2>x_1\) in the interval under consideration.

The integral is improper: by definition, \(\int_N^\infty f(x)\,dx=\lim_{B\to\infty}\int_N^B f(x)\,dx\), if that limit exists as a finite number. The test compares areas of rectangles with the area under the graph. Since \(f\) is decreasing, on each interval \([k,k+1]\), the function lies between the endpoint heights \(f(k+1)\) and \(f(k)\).

Why the Integral Test Works

For an integer \(M\geq N\), decreasingness gives the inequalities below on each unit interval. Adding them compares a sum with an integral. The bounds concern tails starting at \(N\), so they do not assume anything about terms before \(N\).

For the left inequality, add \(\int_k^{k+1}f(x)\,dx\geq f(k+1)\) for \(k=N,\ldots,M\). For the right inequality, separate \(f(N)\), then use \(f(k)\leq\int_{k-1}^{k}f(x)\,dx\) for \(k=N+1,\ldots,M\). These inequalities show both directions: a finite improper integral bounds the series’ partial sums, while an unbounded integral forces the series’ partial sums to be unbounded. This proves the test.

Direct Comparison

Direct comparison uses term-by-term inequalities. The benchmark series must have known behavior, and both the terms being compared must be positive. The comparison is needed only from some index onward: the finitely many earlier terms cannot change convergence or divergence.

To see why, compare tail partial sums \(\sum_{n=N}^{M}a_n\) and \(\sum_{n=N}^{M}b_n\), not necessarily the partial sums starting at 1. In the first case, every tail sum of \(a_n\) is bounded above by the corresponding tail sum of \(b_n\). In the second case, every tail sum of \(a_n\) is at least as large as the corresponding tail sum of \(b_n\). The first comparison bounds the increasing partial sums of \(a_n\); the second forces them to be unbounded. The omitted initial terms contribute only a fixed finite amount.

Limit Comparison

Sometimes a useful term-by-term inequality is awkward to find, but two terms have the same long-term scale. Limit comparison formalizes this idea using their ratio.

Because the ratio approaches a finite positive number, it is eventually bounded above and below by positive constants. Specifically, for sufficiently large \(n\), \(L/2<a_n/b_n<3L/2\). Multiplying by \(b_n>0\) gives \((L/2)b_n<a_n<(3L/2)b_n\). Thus each series’ tail terms are bounded by constant multiples of the other’s. Direct comparison proves that convergence or divergence is shared. As before, the conclusion is about tails; any finite initial segment has no effect.

Common Pitfall: Comparing Full Partial Sums Too Soon

If an inequality holds only for \(n\geq N\), it does not necessarily hold for partial sums that include earlier terms. For example, let \(a_1=5\), \(b_1=1\), and set \(a_n=b_n=1/n^2\) for \(n\geq2\). Then \(a_n\leq b_n\) for every \(n\geq2\), and \(\sum b_n\) converges. But at \(M=1\), the full partial sums satisfy \(S_1^{(a)}=5>1=S_1^{(b)}\). It would be wrong to claim \(S_M^{(a)}\leq S_M^{(b)}\) for every \(M\) from the eventual term inequality.

The correction is to compare only tail sums beginning at \(N=2\). For every \(M\geq2\), \(\sum_{n=2}^{M}a_n\leq\sum_{n=2}^{M}b_n\). These tail sums are bounded because the comparison series converges. Adding the single initial term \(a_1=5\) adds a fixed amount, so the full series still converges. The same tail-first reasoning is essential when the comparison is used to prove divergence.

Choosing a Test and Interpreting Its Result

Start by confirming that the terms are positive from some index onward. Then choose a test that fits the form of the terms. Use the integral test when the terms come from a continuous, positive, decreasing function. Use direct comparison when a useful inequality with a known series is apparent. Use limit comparison when the ratio with a known positive benchmark has a finite, nonzero limit.

Direction matters. A series with terms smaller than those of a convergent series converges; a series with terms larger than those of a divergent positive-term series diverges. The other two pairings do not establish a conclusion: being smaller than a divergent series, or larger than a convergent one, is inconclusive.

For limit comparison, a limit \(L\) strictly between zero and infinity gives the two-sided bounds needed to transfer behavior. A limit of zero or infinity does not meet this theorem’s stated condition. Sometimes such a limit can support a one-sided direct comparison, but the appropriate inequality and benchmark behavior must still be established separately.