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10.5 Harmonic Series and p-Series

1 min read•june 7, 2020

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    Welcome back to Calc BC! As you keep chugging on Unit 10, you’ll find that there’s more to number series (and the course as a whole) than the geometric series we defined in 10.2 Working with Geometric Series:

    🧠 P-Series in Calculus

    Before we define them, let’s take a look at the following series of numbers:

    1+18+127+164+...1+\frac{1}{8}+\frac{1}{27}+\frac{1}{64}+...

    What do the terms have in common? If you look at the denominator, you’ll notice that it is the cube (multiply a number by itself three times).

    1+18+127+164+...=113+123+133+143+...1+\frac{1}{8}+\frac{1}{27}+\frac{1}{64}+...=\frac{1}{1^3}+\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+...

    Another way to describe this patterns is by using the summation notation:

    ∑n=1∞1n3\sum_{n=1}^{\infty} \frac{1}{n^{3}}

    What about this following series?

    1+14+19+116+125+...1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+...

    This time, the pattern is…

    112+122+132+142+152+...=∑n=1∞1n2\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...=\sum_{n=1}^{\infty} \frac{1}{n^{2}}

    What do these two patterns have in common? Let’s put them side by side.

    ∑n=1∞1n3∑n=1∞1n2\sum_{n=1}^{\infty} \frac{1}{n^{3}}\\\sum_{n=1}^{\infty} \frac{1}{n^{2}}

    You’ll notice that the only differentiating factor is their exponents! We can represent this as a p-series via the following notation.

    ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^{p}}

    Pretty straightforward compared to, say, the Integral Test or the nth-term Test, eh?

    ✏️ P-Series Practice Problems

    Let’s take a look at a couple examples! In the following set of 6 series, determine whether each series converges or diverges.

    🔢 P-Series Example 1

    ∑n=1∞1n13\sum_{n=1}^{\infty} \frac{1}{{\sqrt{n}}^{13}}

    We can rewrite the following summation as

    ∑n=1∞1(n1/2)13=∑n=1∞1n13/2\sum_{n=1}^{\infty} \frac{1}{{({n}^{1/2})}^{13}}=\sum_{n=1}^{\infty} \frac{1}{{{n}^{13/2}}}

    Since 13/2 = 6.5 > 1, our series converges! ✅

    🔢 P-Series Example 2

    ∑n=1∞1n4\sum_{n=1}^{\infty} \frac{1}{{n}^{4}}

    This one’s pretty easy. Our p in this case is 4, and thus our series converges because 4 > 1. ✅

    🔢 P-Series Example 3

    ∑n=1∞1n=1+12+13+14+15+...\sum_{n=1}^{\infty} \frac{1}{{n}}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...

    Series #3 actually has a unique name: the harmonic series! Per our definition of the p-series test, p = 1 is not greater than 1 (they’re equal)… so the harmonic series diverges. ✅

    🔢 P-Series Example 4

    ∑n=1∞n−3\sum_{n=1}^{\infty} n^{-3}

    Don’t be fooled by the -3! Just because the number says -3 and -3 < 1 doesn’t mean that our series diverges. We can rewrite the following equation as:

    ∑n=1∞n−3=∑n=1∞1n3\sum_{n=1}^{\infty} n^{-3}=\sum_{n=1}^{\infty} \frac{1}{n^3}

    And just like that, our series actually converges as p = 3 > 1! ✅

    🔢 P-Series Example 5

    ∑n=1∞n5n6\sum_{n=1}^{\infty} \frac{n^5}{{n^6}}

    Again, do not ignore the numerator and jump straight of the numerator and conclude that the series converges because 6 > 1. Simplify the fraction within the summation first! In this case, we get:

    ∑n=1∞n5n6=∑n=1∞1n\sum_{n=1}^{\infty} \frac{n^5}{{n^6}}=\sum_{n=1}^{\infty} \frac{1}{{n}}

    This looks like an old friend from (3), the harmonic series! Recall that the harmonic series diverges per the definition of the p-series test.

    🔢 P-Series Example 6

    ∑n=1∞1n0.75n\sum_{n=1}^{\infty} \frac{1}{{n^{0.75}\sqrt{n}}}

    Last but not least: like (1), let’s simplify the denominator and see what we get:

    ∑n=1∞1n0.75n=∑n=1∞1n0.75∗n0.5=∑n=1∞1n1.25\sum_{n=1}^{\infty} \frac{1}{{n^{0.75}\sqrt{n}}}=\sum_{n=1}^{\infty} \frac{1}{{n^{0.75}*n^{0.5}}}=\sum_{n=1}^{\infty} \frac{1}{{n^{1.25}}}

    Since p = 1.25 > 1, our series converges.

    🪐 Closing

    Out of all the funky and eclectic tests in Unit 10 of AP Calc BC, this test is the friendliest and least complicated. However, you should still be careful in making sure that you simplify the fraction within the summation term and transform it into it’s 1/np1/n^p form before deciding—using the definition—whether the series converges or diverges.

    Good luck! ⭐

    Key Terms to Review (1)

    P-Series Test

    : The P-Series Test is a convergence test used to determine if an infinite series converges or diverges based on the value of the exponent in the denominator of each term. If the exponent is greater than 1, the series converges; if it is less than or equal to 1, the series diverges.

    10.5 Harmonic Series and p-Series

    1 min read•june 7, 2020

    Attend a live cram event

    Review all units live with expert teachers & students

      Welcome back to Calc BC! As you keep chugging on Unit 10, you’ll find that there’s more to number series (and the course as a whole) than the geometric series we defined in 10.2 Working with Geometric Series:

      🧠 P-Series in Calculus

      Before we define them, let’s take a look at the following series of numbers:

      1+18+127+164+...1+\frac{1}{8}+\frac{1}{27}+\frac{1}{64}+...

      What do the terms have in common? If you look at the denominator, you’ll notice that it is the cube (multiply a number by itself three times).

      1+18+127+164+...=113+123+133+143+...1+\frac{1}{8}+\frac{1}{27}+\frac{1}{64}+...=\frac{1}{1^3}+\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+...

      Another way to describe this patterns is by using the summation notation:

      ∑n=1∞1n3\sum_{n=1}^{\infty} \frac{1}{n^{3}}

      What about this following series?

      1+14+19+116+125+...1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+...

      This time, the pattern is…

      112+122+132+142+152+...=∑n=1∞1n2\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...=\sum_{n=1}^{\infty} \frac{1}{n^{2}}

      What do these two patterns have in common? Let’s put them side by side.

      ∑n=1∞1n3∑n=1∞1n2\sum_{n=1}^{\infty} \frac{1}{n^{3}}\\\sum_{n=1}^{\infty} \frac{1}{n^{2}}

      You’ll notice that the only differentiating factor is their exponents! We can represent this as a p-series via the following notation.

      ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^{p}}

      Pretty straightforward compared to, say, the Integral Test or the nth-term Test, eh?

      ✏️ P-Series Practice Problems

      Let’s take a look at a couple examples! In the following set of 6 series, determine whether each series converges or diverges.

      🔢 P-Series Example 1

      ∑n=1∞1n13\sum_{n=1}^{\infty} \frac{1}{{\sqrt{n}}^{13}}

      We can rewrite the following summation as

      ∑n=1∞1(n1/2)13=∑n=1∞1n13/2\sum_{n=1}^{\infty} \frac{1}{{({n}^{1/2})}^{13}}=\sum_{n=1}^{\infty} \frac{1}{{{n}^{13/2}}}

      Since 13/2 = 6.5 > 1, our series converges! ✅

      🔢 P-Series Example 2

      ∑n=1∞1n4\sum_{n=1}^{\infty} \frac{1}{{n}^{4}}

      This one’s pretty easy. Our p in this case is 4, and thus our series converges because 4 > 1. ✅

      🔢 P-Series Example 3

      ∑n=1∞1n=1+12+13+14+15+...\sum_{n=1}^{\infty} \frac{1}{{n}}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...

      Series #3 actually has a unique name: the harmonic series! Per our definition of the p-series test, p = 1 is not greater than 1 (they’re equal)… so the harmonic series diverges. ✅

      🔢 P-Series Example 4

      ∑n=1∞n−3\sum_{n=1}^{\infty} n^{-3}

      Don’t be fooled by the -3! Just because the number says -3 and -3 < 1 doesn’t mean that our series diverges. We can rewrite the following equation as:

      ∑n=1∞n−3=∑n=1∞1n3\sum_{n=1}^{\infty} n^{-3}=\sum_{n=1}^{\infty} \frac{1}{n^3}

      And just like that, our series actually converges as p = 3 > 1! ✅

      🔢 P-Series Example 5

      ∑n=1∞n5n6\sum_{n=1}^{\infty} \frac{n^5}{{n^6}}

      Again, do not ignore the numerator and jump straight of the numerator and conclude that the series converges because 6 > 1. Simplify the fraction within the summation first! In this case, we get:

      ∑n=1∞n5n6=∑n=1∞1n\sum_{n=1}^{\infty} \frac{n^5}{{n^6}}=\sum_{n=1}^{\infty} \frac{1}{{n}}

      This looks like an old friend from (3), the harmonic series! Recall that the harmonic series diverges per the definition of the p-series test.

      🔢 P-Series Example 6

      ∑n=1∞1n0.75n\sum_{n=1}^{\infty} \frac{1}{{n^{0.75}\sqrt{n}}}

      Last but not least: like (1), let’s simplify the denominator and see what we get:

      ∑n=1∞1n0.75n=∑n=1∞1n0.75∗n0.5=∑n=1∞1n1.25\sum_{n=1}^{\infty} \frac{1}{{n^{0.75}\sqrt{n}}}=\sum_{n=1}^{\infty} \frac{1}{{n^{0.75}*n^{0.5}}}=\sum_{n=1}^{\infty} \frac{1}{{n^{1.25}}}

      Since p = 1.25 > 1, our series converges.

      🪐 Closing

      Out of all the funky and eclectic tests in Unit 10 of AP Calc BC, this test is the friendliest and least complicated. However, you should still be careful in making sure that you simplify the fraction within the summation term and transform it into it’s 1/np1/n^p form before deciding—using the definition—whether the series converges or diverges.

      Good luck! ⭐

      Key Terms to Review (1)

      P-Series Test

      : The P-Series Test is a convergence test used to determine if an infinite series converges or diverges based on the value of the exponent in the denominator of each term. If the exponent is greater than 1, the series converges; if it is less than or equal to 1, the series diverges.
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      About Us
      About FiveableBlogCareersCode of ConductTerms of UsePrivacy PolicyCCPA Privacy Policy
      Resources
      Cram ModeAP Score CalculatorsStudy GuidesPractice QuizzesGlossaryCram EventsMerch ShopCrisis Text LineHelp Center

      © 2024 Fiveable Inc. All rights reserved.

      AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.

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