Math Problem Statement

Evaluate the convergence of two series and calculate the limits of two sequences based on the expressions provided.

Solution

The image contains handwritten mathematical problems involving series and limits. Here’s an interpretation and solution of the problems based on the content:

Problem 1: Series Convergence

  1. (a) Evaluate the series:

    k=11k2+k\sum_{k=1}^{\infty} \frac{1}{k^2 + k}

    Solution:

    We rewrite the term inside the summation:

    1k2+k=1k(k+1)\frac{1}{k^2 + k} = \frac{1}{k(k+1)}

    Use the limit comparison test:

    limk1k(k+1)=limk1k2=0\lim_{k \to \infty} \frac{1}{k(k+1)} = \lim_{k \to \infty} \frac{1}{k^2} = 0

    Since this approaches 0 as kk \to \infty, and the series 1k2\sum \frac{1}{k^2} converges (p-series with p > 1), the given series also converges.

    Conclusion: Convergent (eli suppenee).

  2. (b) Evaluate the series:

    k=11+k2+k\sum_{k=1}^{\infty} \frac{1 + k}{2 + k}

    Solution:

    Find the limit of the terms:

    limk1+k2+k=limkk(1+1k)k(1+2k)=limk1+1k1+2k=11=1\lim_{k \to \infty} \frac{1+k}{2+k} = \lim_{k \to \infty} \frac{k(1 + \frac{1}{k})}{k(1 + \frac{2}{k})} = \lim_{k \to \infty} \frac{1 + \frac{1}{k}}{1 + \frac{2}{k}} = \frac{1}{1} = 1

    Since the limit of the terms does not go to 0, by the divergence test, the series diverges.

    Conclusion: Divergent (eli tämä hajoaa).

Problem 2: Sequence Limits

  1. (a) Evaluate the limit:

    ak=kk+2,limkaka_k = \frac{k}{k + 2}, \quad \lim_{k \to \infty} a_k

    Solution:

    Simplify:

    limkkk+2=limkk(1+2k)k=limk(1+2k)=1+0=1\lim_{k \to \infty} \frac{k}{k+2} = \lim_{k \to \infty} \frac{k \left(1 + \frac{2}{k}\right)}{k} = \lim_{k \to \infty} \left(1 + \frac{2}{k}\right) = 1 + 0 = 1

    Conclusion: The limit is 1.

  2. (b) Evaluate the limit:

    bk=7sin2(kcos2(π2))k5/2,limkbkb_k = \frac{7 \sin^2(k \cos^2(\frac{\pi}{2}))}{k^{5/2}}, \quad \lim_{k \to \infty} b_k

    Solution:

    Note that:

    sin2(kcos2(π2))1\sin^2(k \cos^2(\frac{\pi}{2})) \leq 1

    Thus,

    bk=7sin2(kcos2(π2))k5/2b_k = \frac{7 \sin^2(k \cos^2(\frac{\pi}{2}))}{k^{5/2}}

    As kk \to \infty, since the numerator is bounded by 7 and the denominator grows without bound, bk0b_k \to 0.

    Conclusion: The limit is 0.

Summary:

  1. Series Convergence:
    • (a) Convergent.
    • (b) Divergent.
  2. Sequence Limits:
    • (a) Limit is 1.
    • (b) Limit is 0.

Would you like more details or have any questions about these solutions?

Here are 5 related questions you might find helpful:

  1. What are common convergence tests for infinite series?
  2. How does the limit comparison test work for different series?
  3. What is the importance of the divergence test in series analysis?
  4. Can you explain how to determine the behavior of sequences as they approach infinity?
  5. What are some examples of divergent series and their properties?

Tip: When dealing with infinite series, always check if the terms go to zero. If not, the series diverges immediately!

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Math Problem Analysis

Mathematical Concepts

Series
Limits
Convergence
Divergence

Formulas

lim k→∞ (1/k(k+1))
lim k→∞ (1+k)/(2+k)
lim k→∞ (k/(k+2))
lim k→∞ (7sin²(kcos²(π/2))/k^(5/2))

Theorems

Limit Comparison Test
Divergence Test

Suitable Grade Level

University level (Calculus II or Analysis)