Math Problem Statement

Select all series which are convergent.

Solution

To determine which of the series are convergent, let's analyze each series individually using various convergence tests:

1. Series 1:

n=0(9)n+172n\sum_{n=0}^{\infty} \frac{(-9)^{n+1}}{7 \cdot 2^n}

This series resembles a geometric series because it has terms in the form of a constant raised to a power of nn. We can rewrite it as: n=0(92)n17\sum_{n=0}^{\infty} \left( -\frac{9}{2} \right)^n \frac{1}{7}

Since 92>1\left| \frac{-9}{2} \right| > 1, this series diverges because the terms do not approach zero, violating the necessary condition for convergence of a geometric series.

2. Series 2:

n=0sin(4n+19n+4)\sum_{n=0}^{\infty} \sin \left( \frac{4n + 1}{9n + 4} \right)

For convergence, the terms of the series must approach zero as nn \to \infty. Examining the argument of the sine function, we have: limn4n+19n+4=49\lim_{n \to \infty} \frac{4n + 1}{9n + 4} = \frac{4}{9} Thus, sin(4n+19n+4)\sin \left( \frac{4n + 1}{9n + 4} \right) does not tend to zero as nn \to \infty, so this series diverges.

3. Series 3:

n=1(1)nn94\sum_{n=1}^{\infty} (-1)^n n^{-\frac{9}{4}}

This is an alternating series with terms an=n94a_n = n^{-\frac{9}{4}}. To test for convergence, we can apply the Alternating Series Test (Leibniz's Test). The requirements are:

  1. ana_n must be positive, decreasing, and approach zero as nn \to \infty.

Since n94n^{-\frac{9}{4}} is positive, decreases for n1n \geq 1, and limnn94=0\lim_{n \to \infty} n^{-\frac{9}{4}} = 0, this series converges by the Alternating Series Test.

4. Series 4:

n=052n23n\sum_{n=0}^{\infty} \frac{5^{2n}}{2^{3n}}

This series can be simplified as: n=0(5223)n=n=0(258)n\sum_{n=0}^{\infty} \left( \frac{5^2}{2^3} \right)^n = \sum_{n=0}^{\infty} \left( \frac{25}{8} \right)^n

Since 258>1\frac{25}{8} > 1, this is a divergent geometric series because the ratio exceeds 1.

Conclusion:

Only Series 3 is convergent.

Would you like more details on any particular series or convergence tests?

Related Questions

  1. How does the Alternating Series Test determine convergence?
  2. What other tests could we use for convergence if the Alternating Series Test fails?
  3. Can all geometric series be tested for convergence simply by checking the ratio?
  4. How do we determine convergence for trigonometric series like Series 2?
  5. What happens to a series if its terms do not approach zero?

Tip:

For a series to converge, especially a geometric one, ensure that the common ratio rr satisfies r<1|r| < 1.

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Math Problem Analysis

Mathematical Concepts

Series Convergence
Alternating Series Test
Geometric Series
Limit Comparison

Formulas

Alternating Series Test: A series ∑(-1)^n * a_n converges if a_n is positive, decreasing, and lim(n→∞) a_n = 0
Geometric Series: A series ∑ar^n converges if |r| < 1
Convergence criterion for terms approaching zero: lim(n→∞) a_n = 0

Theorems

Alternating Series Test
Geometric Series Convergence Test

Suitable Grade Level

College Level