Math Problem Statement

Select ALL the series which are convergent. (Hint: Use Ratio Test, Root Test, or Integral Test) 1. \(\sum_{n=0}^{\infty} \frac{n^{40} \cdot 40^n}{n!}\) 2. \(\sum_{n=0}^{\infty} n \cdot 4^{-n}\) 3. \(\sum_{n=2}^{\infty} \left(\frac{4n}{\ln n}\right)^n\) 4. \(\sum_{n=0}^{\infty} \left(\frac{9n + 7}{7n + 9}\right)^n\)

Solution

To analyze the convergence of each series, let's examine each individually using appropriate convergence tests as suggested: the Ratio Test, Root Test, or Integral Test.

  1. First Series: n=0n4040nn!\sum_{n=0}^{\infty} \frac{n^{40} \cdot 40^n}{n!}

    • Ratio Test is useful here because we have terms involving factorials and exponential growth.
    • Applying the Ratio Test: limn(n+1)4040n+1(n+1)!n!n4040n=limn(n+1)4040(n+1)n40=0\lim_{n \to \infty} \left| \frac{(n+1)^{40} \cdot 40^{n+1}}{(n+1)!} \cdot \frac{n!}{n^{40} \cdot 40^n} \right| = \lim_{n \to \infty} \frac{(n+1)^{40} \cdot 40}{(n+1) \cdot n^{40}} = 0
    • Since the limit is 0, this series converges by the Ratio Test.
  2. Second Series: n=0n4n\sum_{n=0}^{\infty} n \cdot 4^{-n}

    • Here, each term is n4n=n4nn \cdot 4^{-n} = \frac{n}{4^n}.
    • Ratio Test or Root Test is effective for terms with exponential decay.
    • Applying the Ratio Test: limn(n+1)4(n+1)n4n=limnn+14n=0\lim_{n \to \infty} \left| \frac{(n+1) \cdot 4^{-(n+1)}}{n \cdot 4^{-n}} \right| = \lim_{n \to \infty} \frac{n+1}{4n} = 0
    • The limit is 0, so this series converges by the Ratio Test.
  3. Third Series: n=2(4nlnn)n\sum_{n=2}^{\infty} \left(\frac{4n}{\ln n}\right)^n

    • This series has the form (4nlnn)n\left(\frac{4n}{\ln n}\right)^n, where the base grows with nn.
    • Root Test is particularly suitable here because each term is raised to the nn-th power.
    • Applying the Root Test: limn(4nlnn)nn=limn4nlnn=\lim_{n \to \infty} \sqrt[n]{\left(\frac{4n}{\ln n}\right)^n} = \lim_{n \to \infty} \frac{4n}{\ln n} = \infty
    • Since the limit is infinite, this series diverges by the Root Test.
  4. Fourth Series: n=0(9n+77n+9)n\sum_{n=0}^{\infty} \left(\frac{9n + 7}{7n + 9}\right)^n

    • Similar to the third series, the terms have the form (9n+77n+9)n\left(\frac{9n + 7}{7n + 9}\right)^n.
    • Using the Root Test again: limn(9n+77n+9)nn=limn9n+77n+9=97>1\lim_{n \to \infty} \sqrt[n]{\left(\frac{9n + 7}{7n + 9}\right)^n} = \lim_{n \to \infty} \frac{9n + 7}{7n + 9} = \frac{9}{7} > 1
    • Since the limit is greater than 1, this series diverges by the Root Test.

Conclusion

  • Convergent Series: Series 1 and Series 2.
  • Divergent Series: Series 3 and Series 4.

Would you like more details on these tests, or have any questions about convergence?

Here are some related questions:

  1. How does the Ratio Test determine convergence for factorials?
  2. Why is the Root Test effective for exponential terms?
  3. Can the Integral Test be applied to series involving factorials?
  4. What happens if the Ratio Test yields a limit of 1?
  5. Are there alternative methods to test for convergence here?

Tip: Use the Ratio Test when you see factorials or powers involving n!n!.

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Math Problem Analysis

Mathematical Concepts

Series Convergence
Ratio Test
Root Test
Integral Test

Formulas

Ratio Test: \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\)
Root Test: \(\lim_{n \to \infty} \sqrt[n]{|a_n|}\)

Theorems

Ratio Test for Convergence
Root Test for Convergence
Integral Test for Convergence

Suitable Grade Level

College level or advanced high school calculus