Convergence Tests for Series of Positive Terms
Decide convergence with the integral, comparison, limit comparison, ratio and root tests, and choose the test that matches the shape of the terms.
What a learner can do afterwards
- Choose and apply p-series, integral, ratio and root tests from the form of the terms
- Use direct or limit comparison with a known series and justify the conclusion
- Explain why a ratio limit of 1 is inconclusive and choose a follow-up argument
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You have defined a series sum using partial sums. Here, sum a_n means the infinite sum of the terms a_n; n starts at 1 unless stated otherwise. The p-series sum 1/n^p converges exactly when p > 1 and diverges when p <= 1. Combine denominator powers: n^2 sqrt(n)=n^(5/2). Changing finitely many starting terms does not change convergence.
Direct comparison requires nonnegative terms. If 0 <= a_n <= b_n eventually and sum b_n converges, sum a_n converges. If a_n >= b_n >= 0 eventually and sum b_n diverges, sum a_n diverges. For example, 0 < 1/(n^2+7) < 1/n^2 proves convergence. A convergent lower bound or a divergent upper bound gives no conclusion by itself.
For limit comparison, take positive a_n and b_n. If L=lim(a_n/b_n) is finite and strictly positive, both series converge or both diverge. Use dominant powers to choose a benchmark, then calculate the limit. With a_n=(2n+1)/(n^2+4) and b_n=1/n, the ratio is (2n^2+n)/(n^2+4). Divide by n^2 to find L=2. The harmonic benchmark diverges, so sum a_n diverges.
For positive terms, calculate the ratio-test limit L=lim(a_(n+1)/a_n), when it exists. L < 1 proves convergence; L > 1, including infinity, proves divergence. L=1 is inconclusive. For a p-series the ratio is (n/(n+1))^p, which tends to 1. Yet p=1 diverges and p=2 converges. Both outcomes are possible when the ratio limit is 1.
For a_n=3^n/n!, use (n+1)!=(n+1)n! to simplify the consecutive ratio to 3/(n+1). Its limit 0 is below 1, proving convergence. When n^n occurs, the limit (1+1/n)^n=e can help; reciprocals tend to 1/e. Cancel within terms too: n!/(n+1)!=1/(n+1). Comparing its terms with 1/n gives limit 1 and proves divergence, although its consecutive ratio also tends to 1.
For a_n=n^k/c^n, with fixed real k and c > 0, the consecutive ratio is (1+1/n)^k/c. Its limit is 1/c. Thus c > 1 proves convergence by the ratio test. For example, n/5^n gives ratio (n+1)/(5n), tending to 1/5. An exponential in a formula is a reason to try the test, not a conclusion without calculation.
For nonnegative terms, the root test uses R=lim(a_n^(1/n)), when it exists. R < 1 proves convergence; R > 1, including infinity, proves divergence; R=1 is inconclusive. When a_n=(b_n)^n with b_n positive, taking the nth root leaves b_n. For [(n+2)/(4n+1)]^n, the root tends to 1/4, so the series converges.
The integral test needs f positive, continuous and decreasing on [N,infinity), with f(n)=a_n. The series and the improper integral from N to infinity both converge or both diverge; their values need not agree. For f(x)=1/(x+2)^2 on x >= 1, f'(x)=-2/(x+2)^3 < 0. Its antiderivative is -1/(x+2). The integral from 1 to t is 1/3-1/(t+2), tending to 1/3. Thus the series converges; its sum is not determined.
Check each test's conditions and calculate its limit. Ratio and root limits of 1 are inconclusive. Comparison needs a justified benchmark; an integral test does not give the series sum.
The lesson and practice material were written for this dataset using standard mathematical definitions and convergence tests.
The readings, questions and flashcards were written for this dataset. The linked videos are from Khan Academy and remain their publishers' work.
More to read
Which way must a comparison go?
Start with nonnegative terms and a series whose convergence is known. A convergent upper benchmark proves convergence of the smaller series. A divergent lower benchmark proves divergence of the larger series. The comparison need only hold from some fixed index onward.
The other two directions do not decide the question. A series below a divergent benchmark might converge or diverge. A series above a convergent benchmark might also do either. Check the direction before using the benchmark's conclusion.
Limit comparison replaces the inequality check with a ratio calculation. For positive terms a_n and b_n, find the limit of a_n/b_n. A finite, strictly positive limit proves that the two series have the same convergence outcome.
Dominant powers can suggest a benchmark. They do not finish the proof. Write the ratio, reduce it and take its limit. A limit of zero or infinity does not meet the finite-positive version of the test stated here.
Two different limits to calculate
The ratio test divides the next term by the current term. Replace every n in the formula with n+1 before dividing. Then cancel factors. For factorials, the identity (n+1)!=(n+1)n! often removes most of the expression.
The root test takes the nth root of the nth term. If a positive term is an expression raised to the nth power, the root removes that outer power. The remaining expression may have a much easier limit.
For positive terms, either test proves convergence when its limit exists and is below 1. A limit above 1, including infinity, proves divergence. A limit equal to 1 gives no conclusion. The harmonic series and the convergent p-series with p=2 both have ratio limit 1.
For signed terms, these tests use absolute values. A limit below 1 then proves absolute convergence, meaning that the series of absolute values converges.
Also reduce the original term before choosing a test. Factorials can cancel into a rational expression. Its ratio limit may be 1, while comparison with a p-series still settles convergence.
What an integral test tells you
Choose a function f whose values at the integer inputs equal the series terms. To apply the integral test, f must be positive, continuous and decreasing throughout an interval [N,infinity). Checking only the integer inputs is not enough to establish these conditions.
Integrate from N to a finite upper bound t. Then let t grow without bound. If the resulting improper integral converges, the series converges. If the integral diverges, the series diverges.
For example, take f(x)=1/(x+4)^2 for x >= 1. It is positive and continuous there. Its derivative is -2/(x+4)^3, which is negative, so it is decreasing. An antiderivative is -1/(x+4). The integral from 1 to t equals 1/5-1/(t+4), with limit 1/5. The corresponding series therefore converges.
The integral's value is not a formula for the series sum. The test decides whether both are finite; it does not say their finite values agree. Starting at a later fixed index is allowed because removing finitely many terms does not change convergence.
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