Math Problem Statement

Apply the comparison test with the series sum, from, n, equals, 1, to, infinity, of, start fraction, 1, divided by, n, to the power p , end fraction n=1 ∑ ∞ ​

n p

1 ​ where p, equalsp= . If a, start subscript, n, end subscript, equals, start fraction, n, start superscript, start fraction, 3, divided by, 2, end fraction, end superscript, plus, 6, divided by, 6, n, start superscript, start fraction, 5, divided by, 2, end fraction, end superscript, plus, 3, n, minus, 3, end fractiona n ​

6n 2 5 ​

+3n−3 n 2 3 ​

+6 ​ and b, start subscript, n, end subscript, equals, start fraction, 1, divided by, n, start superscript, end superscript, end fractionb n ​

n

1 ​ , then limit, start subscript, n, right arrow, infinity, end subscript, start fraction, a, start subscript, n, end subscript, divided by, b, start subscript, n, end subscript, end fraction, equals n→∞ lim ​

b n ​

a n ​

. Since a, start subscript, n, end subscript, comma, b, start subscript, n, end subscript, is greater than, 0a n ​ ,b n ​

0 and the limit is a finite and positive (non-zero) number, the limit comparison test applies. sum, from, n, equals, 1, to, infinity, of, start fraction, 1, divided by, n, start superscript, end superscript, end fraction n=1 ∑ ∞ ​

n

1 ​

since a p-series will diverge if and only if . Therefore,sum, from, n, equals, 1, to, infinity, of, start fraction, n, start superscript, start fraction, 3, divided by, 2, end fraction, end superscript, plus, 6, divided by, 6, n, start superscript, start fraction, 5, divided by, 2, end fraction, end superscript, plus, 3, n, minus, 3, end fraction n=1 ∑ ∞ ​

6n 2 5 ​

+3n−3 n 2 3 ​

+6 ​

.

Solution

The problem asks to apply the Limit Comparison Test to the series:

n=11np\sum_{n=1}^{\infty} \frac{1}{n^p} where pp is a specific value, and we are comparing it with the given terms for ana_n and bnb_n.


Step 1: Define ana_n and bnb_n

Given:

  • an=6n32+6n52+3n3a_n = 6n^{\frac{3}{2}} + 6n^{\frac{5}{2}} + 3n - 3
  • bn=1nb_n = \frac{1}{n} (which represents a typical harmonic series, i.e., a p-series with p=1p = 1).

Step 2: Apply the Limit Comparison Test

We need to compute the following limit:

L=limnanbnL = \lim_{n \to \infty} \frac{a_n}{b_n}

Substitute the expressions for ana_n and bnb_n:

L=limn6n32+6n52+3n31nL = \lim_{n \to \infty} \frac{6n^{\frac{3}{2}} + 6n^{\frac{5}{2}} + 3n - 3}{\frac{1}{n}}

This can be rewritten as:

L=limn(n(6n32+6n52+3n3))L = \lim_{n \to \infty} \left( n \cdot (6n^{\frac{3}{2}} + 6n^{\frac{5}{2}} + 3n - 3) \right)

Simplifying the expression:

L=limn(6n52+6n72+3n23n)L = \lim_{n \to \infty} \left( 6n^{\frac{5}{2}} + 6n^{\frac{7}{2}} + 3n^2 - 3n \right)

For large nn, the term with the highest power of nn dominates, which is 6n726n^{\frac{7}{2}}. So the limit simplifies to:

L=limn6n72=L = \lim_{n \to \infty} 6n^{\frac{7}{2}} = \infty

Step 3: Conclusion

Since the limit is infinite, the Limit Comparison Test tells us that the series:

n=1(6n32+6n52+3n3)\sum_{n=1}^{\infty} \left( 6n^{\frac{3}{2}} + 6n^{\frac{5}{2}} + 3n - 3 \right)

diverges, because it behaves similarly to the harmonic series, n=11n\sum_{n=1}^{\infty} \frac{1}{n}, which is known to diverge.


Questions for Further Exploration:

  1. How do the powers of nn in the numerator and denominator affect the convergence of a series?
  2. Can the Limit Comparison Test be applied when the limit is zero or infinite? What would those cases imply?
  3. How does the p-series test help determine convergence for specific values of pp?
  4. What would happen if the exponent of nn in bnb_n were changed to a value less than 1?
  5. How can you use the Ratio Test to analyze the convergence of the series?

Tip:

When performing the Limit Comparison Test, always focus on the terms with the highest powers of nn in both the numerator and denominator, as they dominate the behavior of the series as nn tends to infinity.

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Math Problem Analysis

Mathematical Concepts

Series Convergence
Limit Comparison Test
P-Series
Asymptotic Analysis

Formulas

Limit Comparison Test: lim (n → ∞) (a_n / b_n) = L
P-Series: Σ (1/n^p) converges if p > 1, diverges if p ≤ 1

Theorems

Limit Comparison Test
P-Series Test

Suitable Grade Level

Undergraduate Mathematics