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Limit theorems for sequences are essential tools in mathematical analysis. They help us understand how sequences behave as they approach infinity, allowing us to manipulate and evaluate complex limits with ease.

These theorems, including the sum, product, and quotient rules, form the foundation for more advanced concepts in calculus. By mastering these rules, we can tackle intricate problems involving sequences and series, paving the way for deeper mathematical understanding.

Algebraic Limit Theorems for Sequences

Fundamental Algebraic Limit Theorems

Top images from around the web for Fundamental Algebraic Limit Theorems
Top images from around the web for Fundamental Algebraic Limit Theorems
  • The Sum Rule states that if lim(an)=L\lim(a_n) = L and lim(bn)=M\lim(b_n) = M, then lim(an+bn)=L+M\lim(a_n + b_n) = L + M
    • Example: If lim(an)=3\lim(a_n) = 3 and lim(bn)=5\lim(b_n) = 5, then lim(an+bn)=3+5=8\lim(a_n + b_n) = 3 + 5 = 8
  • The Difference Rule states that if lim(an)=L\lim(a_n) = L and lim(bn)=M\lim(b_n) = M, then lim(anbn)=LM\lim(a_n - b_n) = L - M
    • Example: If lim(an)=7\lim(a_n) = 7 and lim(bn)=2\lim(b_n) = 2, then lim(anbn)=72=5\lim(a_n - b_n) = 7 - 2 = 5
  • The Constant Multiple Rule states that if lim(an)=L\lim(a_n) = L and cc is a constant, then lim(can)=cL\lim(c * a_n) = c * L
    • Example: If lim(an)=4\lim(a_n) = 4 and c=3c = 3, then lim(3an)=34=12\lim(3 * a_n) = 3 * 4 = 12
  • The Product Rule states that if lim(an)=L\lim(a_n) = L and lim(bn)=M\lim(b_n) = M, then lim(anbn)=LM\lim(a_n * b_n) = L * M
    • Example: If lim(an)=2\lim(a_n) = 2 and lim(bn)=6\lim(b_n) = 6, then lim(anbn)=26=12\lim(a_n * b_n) = 2 * 6 = 12

Advanced Algebraic Limit Theorems

  • The Quotient Rule states that if lim(an)=L\lim(a_n) = L and lim(bn)=M\lim(b_n) = M, where M0M \neq 0, then lim(an/bn)=L/M\lim(a_n / b_n) = L / M
    • Example: If lim(an)=10\lim(a_n) = 10 and lim(bn)=2\lim(b_n) = 2, then lim(an/bn)=10/2=5\lim(a_n / b_n) = 10 / 2 = 5
  • The Quotient Rule requires that the limit of the denominator sequence is non-zero to avoid division by zero
    • If lim(bn)=0\lim(b_n) = 0, the limit of the quotient sequence may not exist or may require further investigation using other techniques (L'Hôpital's Rule)
  • The Power Rule states that if lim(an)=L\lim(a_n) = L and pp is a real number, then lim(anp)=Lp\lim(a_n^p) = L^p, provided that L>0L > 0 if pp is not a rational number with an odd denominator
    • Example: If lim(an)=4\lim(a_n) = 4 and p=1/2p = 1/2, then lim(an1/2)=41/2=2\lim(a_n^{1/2}) = 4^{1/2} = 2

Applying Limit Theorems to Sequences

Evaluating Limits Using Algebraic Limit Theorems

  • To evaluate the using the Algebraic Limit Theorems, first identify the individual limits of the component sequences
    • Example: To find lim(3an2bn)\lim(3a_n - 2b_n), identify lim(an)\lim(a_n) and lim(bn)\lim(b_n)
  • Apply the appropriate Algebraic Limit Theorem based on the operations involved in the sequence (addition, subtraction, multiplication, division, or constant multiplication)
    • Example: If lim(an)=5\lim(a_n) = 5 and lim(bn)=3\lim(b_n) = 3, then lim(3an2bn)=3lim(an)2lim(bn)=3523=9\lim(3a_n - 2b_n) = 3 \cdot \lim(a_n) - 2 \cdot \lim(b_n) = 3 \cdot 5 - 2 \cdot 3 = 9
  • Substitute the individual limits into the theorem to calculate the overall limit of the sequence

Limitations and Special Cases

  • If the limit of a component sequence does not exist or violates the conditions of the theorem (division by zero), the limit of the entire sequence may not exist or may require further investigation
    • Example: If lim(an)=4\lim(a_n) = 4 and lim(bn)=0\lim(b_n) = 0, then lim(an/bn)\lim(a_n / b_n) is undefined due to division by zero
  • In some cases, the limit of a sequence may exist even if the limits of its component sequences do not exist individually
    • Example: If an=(1)na_n = (-1)^n and bn=(1)n+1b_n = (-1)^{n+1}, then lim(an)\lim(a_n) and lim(bn)\lim(b_n) do not exist, but lim(an+bn)=0\lim(a_n + b_n) = 0
  • Algebraic Limit Theorems can be combined to evaluate the limits of more complex sequences involving multiple operations
    • Example: To find lim(2an+3bncndn)\lim(\frac{2a_n + 3b_n}{c_n - d_n}), apply the Sum, Constant Multiple, and Quotient Rules

Monotone Convergence Theorem

Monotone Sequences and Boundedness

  • The Monotone Convergence Theorem states that if a sequence is monotone (either non-decreasing or non-increasing) and bounded, then the sequence converges
  • A sequence (an)(a_n) is non-decreasing if anan+1a_n \leq a_{n+1} for all nNn \in \mathbb{N}, and non-increasing if anan+1a_n \geq a_{n+1} for all nNn \in \mathbb{N}
    • Example: The sequence (11n)(1 - \frac{1}{n}) is non-decreasing because 11n11n+11 - \frac{1}{n} \leq 1 - \frac{1}{n+1} for all nNn \in \mathbb{N}
  • A sequence is bounded if there exist real numbers mm and MM such that manMm \leq a_n \leq M for all nNn \in \mathbb{N}
    • Example: The sequence (1n)(\frac{1}{n}) is bounded because 01n10 \leq \frac{1}{n} \leq 1 for all nNn \in \mathbb{N}

Applying the Monotone Convergence Theorem

  • To apply the Monotone Convergence Theorem, first determine if the sequence is monotone by comparing consecutive terms
    • Example: To show that (nn+1)(\frac{n}{n+1}) converges, note that nn+1n+1n+2\frac{n}{n+1} \leq \frac{n+1}{n+2} for all nNn \in \mathbb{N}, so the sequence is non-decreasing
  • If the sequence is monotone, find the lower and upper bounds of the sequence
    • Example: For (nn+1)(\frac{n}{n+1}), we have 0nn+110 \leq \frac{n}{n+1} \leq 1 for all nNn \in \mathbb{N}
  • If both conditions are satisfied, the sequence converges. The limit of the sequence is equal to the supremum (for non-decreasing sequences) or the infimum (for non-increasing sequences) of the set of terms
    • Example: Since (nn+1)(\frac{n}{n+1}) is non-decreasing and bounded, it converges to its supremum, which is 1

Sequence Convergence Tests

Comparison Test

  • The Comparison Test is used to determine the convergence or divergence of a sequence by comparing it to another sequence with known convergence properties
  • If 0anbn0 \leq a_n \leq b_n for all nNn \geq N (some NNN \in \mathbb{N}) and lim(bn)=0\lim(b_n) = 0, then lim(an)=0\lim(a_n) = 0 ()
    • Example: To show that lim(sinnn)=0\lim(\frac{\sin n}{n}) = 0, note that 0sinnn1n0 \leq |\frac{\sin n}{n}| \leq \frac{1}{n} for all n1n \geq 1 and lim(1n)=0\lim(\frac{1}{n}) = 0
  • If anbna_n \leq b_n for all nNn \geq N and bn\sum b_n converges, then an\sum a_n converges
    • Example: To show that 1n2\sum \frac{1}{n^2} converges, compare it to 1n(n1)\sum \frac{1}{n(n-1)}, which converges by the p-series test
  • If anbn0a_n \geq b_n \geq 0 for all nNn \geq N and bn\sum b_n diverges, then an\sum a_n diverges
    • Example: To show that 1n\sum \frac{1}{\sqrt{n}} diverges, compare it to 1n\sum \frac{1}{n}, which diverges by the p-series test

Ratio Test

  • The Ratio Test is used to determine the convergence or divergence of a series an\sum a_n by examining the limit of the ratio of consecutive terms, lim(an+1an)\lim(|\frac{a_{n+1}}{a_n}|)
    • If lim(an+1an)<1\lim(|\frac{a_{n+1}}{a_n}|) < 1, the series converges absolutely
      • Example: For 2nn!\sum \frac{2^n}{n!}, lim(2n+1/(n+1)!2n/n!)=lim(2n+1)=0<1\lim(|\frac{2^{n+1}/(n+1)!}{2^n/n!}|) = \lim(\frac{2}{n+1}) = 0 < 1, so the series converges absolutely
    • If lim(an+1an)>1\lim(|\frac{a_{n+1}}{a_n}|) > 1, the series diverges
      • Example: For n2\sum n^2, lim((n+1)2n2)=lim(n2+2n+1n2)=1>1\lim(|\frac{(n+1)^2}{n^2}|) = \lim(\frac{n^2+2n+1}{n^2}) = 1 > 1, so the series diverges
    • If lim(an+1an)=1\lim(|\frac{a_{n+1}}{a_n}|) = 1, the test is inconclusive, and other tests should be used to determine convergence or divergence
      • Example: For 1n\sum \frac{1}{n}, lim(1/(n+1)1/n)=lim(nn+1)=1\lim(|\frac{1/(n+1)}{1/n}|) = \lim(\frac{n}{n+1}) = 1, so the Ratio Test is inconclusive (the series diverges by the p-series test)
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© 2024 Fiveable Inc. All rights reserved.
AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.

© 2024 Fiveable Inc. All rights reserved.
AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.
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