Convergence Tests for Series of Positive Terms · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

Matching each series to its test

Mathematics · Calculus & Analysis · ages 19-20
Name ______________________   Date ____________

  1. For sum 1/n^3 from n=1 to infinity, which test gives a decisive result directly?

    • The p-series rule, since p = 3
    • The ratio test, since its limit is below 1
    • The root test, since its limit is below 1

  2. For a positive-term series, a ratio-test limit of 1 proves convergence.

    Circle one:   True   False

  3. The series with terms 1/[n sqrt(n)] is a p-series. Enter p as a decimal.

    Answer: ______________

  4. For sum 1/sqrt(n) from n=1 to infinity, which conclusion follows from the p-series rule?

    • It converges because its denominator grows without bound.
    • It converges because its terms approach zero as n grows.
    • It diverges because its exponent p is 1/2 <= 1.
    • It diverges because its terms approach a positive constant.

  5. For sum n!/n^n from n=1 to infinity, which test and calculated limit prove convergence?

    • The p-series rule, treating 1/e as its fixed exponent.
    • The integral test, treating 1/e as the integral's value.
    • The ratio test, with consecutive ratio limit 1/e.

  6. For a_n=[(2n+1)/(3n+4)]^n, n >= 1, what does the root test show about sum a_n?

    • The root limit is 2/3, so the series converges.
    • The root limit is 3/2, so the series diverges.
    • The root limit is 1, so the test is inconclusive.
    • The root limit is 2/3, so the test is inconclusive.

  7. For n >= 3, 1/(n^2-4) > 1/n^2. Nia says this bound proves convergence because sum 1/n^2 converges. What does this bound establish?

    • It proves convergence because the benchmark is convergent.
    • It proves divergence because the terms exceed the benchmark.
    • It proves convergence because both denominators are positive.
    • It gives no convergence conclusion from this lower bound.

  8. Why does sum 1/(n^2+5), from n=1 to infinity, converge?

    • Ratio test: its consecutive ratio tends to 1/5 < 1.
    • Direct comparison: 0 < 1/(n^2+5) < 1/n^2.
    • Term test: its terms tend to zero, proving convergence.

  9. For a_n=1/n, Ivo writes a_(n+1)/a_n=1/(n+1), gets limit 0, and concludes convergence. Which correction is valid?

    • The ratio is n/(n+1); its limit 1 is inconclusive.
    • The ratio is 1/[n(n+1)]; its limit 0 proves convergence.
    • The ratio is (n+1)/n; its limit 1 proves divergence.

  10. For sum n^2/2^n from n=1 to infinity, which test and result are valid?

    • Term test: the terms tend to zero, so it diverges.
    • Ratio test: its limit is 1/2, so it converges.
    • p-Series rule: its exponent is 2, so it converges.
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Answer key

For grown-ups. Fold this page away before handing over the rest.

Matching each series to its test W1-mt_8N49Q4BlO8-s1

  1. The p-series rule, since p = 3 · The exponent is p=3 > 1, so the p-series rule proves convergence. The ratio and root tests both give the inconclusive limit 1.
  2. False · A ratio limit of 1 is inconclusive. Convergent and divergent p-series both have this ratio limit.
  3. 1.5 · The denominator is n^1 times n^(1/2), which equals n^(3/2). Therefore p=1.5.
  4. It diverges because its exponent p is 1/2 <= 1. · The term is 1/n^(1/2), so p=1/2 <= 1 and the series diverges. Its terms tend to zero, which does not prove convergence.
  5. The ratio test, with consecutive ratio limit 1/e. · The consecutive ratio is (n/(n+1))^n=(1+1/n)^(-n). It tends to 1/e < 1, so the ratio test proves convergence.
  6. The root limit is 2/3, so the series converges. · The positive nth root is (2n+1)/(3n+4), tending to 2/3 < 1. The root test proves convergence.
  7. It gives no convergence conclusion from this lower bound. · A convergent lower benchmark proves neither convergence nor divergence of the larger series. This bound alone gives no conclusion.
  8. Direct comparison: 0 < 1/(n^2+5) < 1/n^2. · The terms are nonnegative and below those of the convergent p-series sum 1/n^2. Direct comparison proves convergence.
  9. The ratio is n/(n+1); its limit 1 is inconclusive. · Dividing by 1/n multiplies by n. The correct consecutive ratio is n/(n+1), tending to 1. The ratio test gives no conclusion.
  10. Ratio test: its limit is 1/2, so it converges. · The consecutive ratio is (n+1)^2/(2n^2)=(1/2)(1+1/n)^2. It tends to 1/2 < 1, proving convergence.
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