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Wize University Calculus 2 Textbook > Sequences and Series

Root Test

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Root Test

Consider a series ∑n=1∞an\sum\limits^\infty_{n=1}a_n.
  • If lim⁡n→∞∣an∣n<1\lim\limits_{n\rightarrow\infty}\sqrt[n]{|a_n|}<1, then ∑n=1∞an\sum\limits^\infty_{n=1}a_n converges (in fact it converges absolutely)
  • If lim⁡n→∞∣an∣n>1\lim\limits_{n\rightarrow\infty}\sqrt[n]{|a_n|}>1, then ∑n=1∞an\sum\limits^\infty_{n=1}a_n diverges
  • If lim⁡n→∞∣an∣n=1\lim\limits_{n\rightarrow\infty}\sqrt[n]{|a_n|}=1, then the test fails and we don't know if the series converges or not

Identifying Clues

  • The terms involve powers which are cancelled out by an nth root (...n\sqrt[n]{...})

Practice Question

Determine whether ∑n=1∞32n−1nn\sum\limits^\infty_{n=1}\frac{3^{2n-1}}{n^n} converges or diverges.

Practice Question

Detemine whether the series ∑n=2∞(−1)n(ln⁡n)n{\displaystyle \sum_{n=2}^\infty}\frac{(-1)^n}{(\ln n)^n} converges absolutely, converges conditionally, or diverges.

Extra Practice
Practice: Root Test
Detemine whether the series ∑n=2∞(−1)n(ln⁡n)n{\displaystyle \sum_{n=2}^\infty}\frac{(-1)^n}{(\ln n)^n} converges absolutely, converges conditionally, or diverges.
Practice: Root Test
Detemine whether the series ∑n=2∞(−1)n(ln⁡n)n{\displaystyle \sum_{n=2}^\infty}\frac{(-1)^n}{(\ln n)^n} converges absolutely, converges conditionally, or diverges.
Practice: Root Test
Practice: Root Test
Detemine whether the series ∑n=2∞(−1)n(ln⁡n)n{\displaystyle \sum_{n=2}^\infty}\frac{(-1)^n}{(\ln n)^n} converges absolutely, converges conditionally, or diverges.
Practice: Root Test
Practice: Root Test
Detemine whether the series ∑n=2∞(−1)n(ln⁡n)n{\displaystyle \sum_{n=2}^\infty}\frac{(-1)^n}{(\ln n)^n} converges absolutely, converges conditionally, or diverges.
Practice: Root Test
Practice: Root Test
Detemine whether the series ∑n=2∞(−1)n(ln⁡n)n{\displaystyle \sum_{n=2}^\infty}\frac{(-1)^n}{(\ln n)^n} converges absolutely, converges conditionally, or diverges.
Practice Question
Practice Question
Determine whether ∑n=1∞32n−1nn\sum\limits^\infty_{n=1}\frac{3^{2n-1}}{n^n} converges or diverges.
Practice Question
Practice Question
Determine whether ∑n=1∞32n−1nn\sum\limits^\infty_{n=1}\frac{3^{2n-1}}{n^n} converges or diverges.
Practice: Root Test
Practice: Root Test
Detemine whether the series ∑n=2∞(−1)n(ln⁡n)n{\displaystyle \sum_{n=2}^\infty}\frac{(-1)^n}{(\ln n)^n} converges absolutely, converges conditionally, or diverges.