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๐Ÿƒ๐Ÿฝโ€โ™€๏ธโ€โžก๏ธIntro to Mathematical Analysis Unit 2 Review

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2.4 Limit Theorems for Sequences

2.4 Limit Theorems for Sequences

Written by the Fiveable Content Team โ€ข Last updated August 2025
Written by the Fiveable Content Team โ€ข Last updated August 2025
๐Ÿƒ๐Ÿฝโ€โ™€๏ธโ€โžก๏ธIntro to Mathematical Analysis
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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

  • The Sum Rule states that if limโก(an)=L\lim(a_n) = L and limโก(bn)=M\lim(b_n) = M, then limโก(an+bn)=L+M\lim(a_n + b_n) = L + M
    • Example: If limโก(an)=3\lim(a_n) = 3 and limโก(bn)=5\lim(b_n) = 5, then limโก(an+bn)=3+5=8\lim(a_n + b_n) = 3 + 5 = 8
  • The Difference Rule states that if limโก(an)=L\lim(a_n) = L and limโก(bn)=M\lim(b_n) = M, then limโก(anโˆ’bn)=Lโˆ’M\lim(a_n - b_n) = L - M
    • Example: If limโก(an)=7\lim(a_n) = 7 and limโก(bn)=2\lim(b_n) = 2, then limโก(anโˆ’bn)=7โˆ’2=5\lim(a_n - b_n) = 7 - 2 = 5
  • The Constant Multiple Rule states that if limโก(an)=L\lim(a_n) = L and cc is a constant, then limโก(cโˆ—an)=cโˆ—L\lim(c * a_n) = c * L
    • Example: If limโก(an)=4\lim(a_n) = 4 and c=3c = 3, then limโก(3โˆ—an)=3โˆ—4=12\lim(3 * a_n) = 3 * 4 = 12
  • The Product Rule states that if limโก(an)=L\lim(a_n) = L and limโก(bn)=M\lim(b_n) = M, then limโก(anโˆ—bn)=Lโˆ—M\lim(a_n * b_n) = L * M
    • Example: If limโก(an)=2\lim(a_n) = 2 and limโก(bn)=6\lim(b_n) = 6, then limโก(anโˆ—bn)=2โˆ—6=12\lim(a_n * b_n) = 2 * 6 = 12

Advanced Algebraic Limit Theorems

  • The Quotient Rule states that if limโก(an)=L\lim(a_n) = L and limโก(bn)=M\lim(b_n) = M, where Mโ‰ 0M \neq 0, then limโก(an/bn)=L/M\lim(a_n / b_n) = L / M
    • Example: If limโก(an)=10\lim(a_n) = 10 and limโก(bn)=2\lim(b_n) = 2, then limโก(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 limโก(bn)=0\lim(b_n) = 0, the limit of the quotient sequence may not exist or may require further investigation using other techniques (L'Hรดpital's Rule)
  • The Power Rule states that if limโก(an)=L\lim(a_n) = L and pp is a real number, then limโก(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 limโก(an)=4\lim(a_n) = 4 and p=1/2p = 1/2, then limโก(an1/2)=41/2=2\lim(a_n^{1/2}) = 4^{1/2} = 2

Applying Limit Theorems to Sequences

Fundamental Algebraic Limit Theorems, How Do You Calculate a Limit Algebraically? โ€“ Math FAQ

Evaluating Limits Using Algebraic Limit Theorems

  • To evaluate the limit of a sequence using the Algebraic Limit Theorems, first identify the individual limits of the component sequences
    • Example: To find limโก(3anโˆ’2bn)\lim(3a_n - 2b_n), identify limโก(an)\lim(a_n) and limโก(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 limโก(an)=5\lim(a_n) = 5 and limโก(bn)=3\lim(b_n) = 3, then limโก(3anโˆ’2bn)=3โ‹…limโก(an)โˆ’2โ‹…limโก(bn)=3โ‹…5โˆ’2โ‹…3=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 limโก(an)=4\lim(a_n) = 4 and limโก(bn)=0\lim(b_n) = 0, then limโก(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=(โˆ’1)na_n = (-1)^n and bn=(โˆ’1)n+1b_n = (-1)^{n+1}, then limโก(an)\lim(a_n) and limโก(bn)\lim(b_n) do not exist, but limโก(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 limโก(2an+3bncnโˆ’dn)\lim(\frac{2a_n + 3b_n}{c_n - d_n}), apply the Sum, Constant Multiple, and Quotient Rules

Monotone Convergence Theorem

Fundamental Algebraic Limit Theorems, Finding Limits: Properties of Limits ยท Precalculus

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 anโ‰คan+1a_n \leq a_{n+1} for all nโˆˆNn \in \mathbb{N}, and non-increasing if anโ‰ฅan+1a_n \geq a_{n+1} for all nโˆˆNn \in \mathbb{N}
    • Example: The sequence (1โˆ’1n)(1 - \frac{1}{n}) is non-decreasing because 1โˆ’1nโ‰ค1โˆ’1n+11 - \frac{1}{n} \leq 1 - \frac{1}{n+1} for all nโˆˆNn \in \mathbb{N}
  • A sequence is bounded if there exist real numbers mm and MM such that mโ‰คanโ‰คMm \leq a_n \leq M for all nโˆˆNn \in \mathbb{N}
    • Example: The sequence (1n)(\frac{1}{n}) is bounded because 0โ‰ค1nโ‰ค10 \leq \frac{1}{n} \leq 1 for all nโˆˆNn \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+1โ‰คn+1n+2\frac{n}{n+1} \leq \frac{n+1}{n+2} for all nโˆˆNn \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 0โ‰คnn+1โ‰ค10 \leq \frac{n}{n+1} \leq 1 for all nโˆˆNn \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 0โ‰คanโ‰คbn0 \leq a_n \leq b_n for all nโ‰ฅNn \geq N (some NโˆˆNN \in \mathbb{N}) and limโก(bn)=0\lim(b_n) = 0, then limโก(an)=0\lim(a_n) = 0 (Squeeze Theorem)
    • Example: To show that limโก(sinโกnn)=0\lim(\frac{\sin n}{n}) = 0, note that 0โ‰คโˆฃsinโกnnโˆฃโ‰ค1n0 \leq |\frac{\sin n}{n}| \leq \frac{1}{n} for all nโ‰ฅ1n \geq 1 and limโก(1n)=0\lim(\frac{1}{n}) = 0
  • If anโ‰คbna_n \leq b_n for all nโ‰ฅNn \geq N and โˆ‘bn\sum b_n converges, then โˆ‘an\sum a_n converges
    • Example: To show that โˆ‘1n2\sum \frac{1}{n^2} converges, compare it to โˆ‘1n(nโˆ’1)\sum \frac{1}{n(n-1)}, which converges by the p-series test
  • If anโ‰ฅbnโ‰ฅ0a_n \geq b_n \geq 0 for all nโ‰ฅNn \geq N and โˆ‘bn\sum b_n diverges, then โˆ‘an\sum a_n diverges
    • Example: To show that โˆ‘1n\sum \frac{1}{\sqrt{n}} diverges, compare it to โˆ‘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 โˆ‘an\sum a_n by examining the limit of the ratio of consecutive terms, limโก(โˆฃan+1anโˆฃ)\lim(|\frac{a_{n+1}}{a_n}|)
    • If limโก(โˆฃan+1anโˆฃ)<1\lim(|\frac{a_{n+1}}{a_n}|) < 1, the series converges absolutely
      • Example: For โˆ‘2nn!\sum \frac{2^n}{n!}, limโก(โˆฃ2n+1/(n+1)!2n/n!โˆฃ)=limโก(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 limโก(โˆฃan+1anโˆฃ)>1\lim(|\frac{a_{n+1}}{a_n}|) > 1, the series diverges
      • Example: For โˆ‘n2\sum n^2, limโก(โˆฃ(n+1)2n2โˆฃ)=limโก(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 limโก(โˆฃan+1anโˆฃ)=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 โˆ‘1n\sum \frac{1}{n}, limโก(โˆฃ1/(n+1)1/nโˆฃ)=limโก(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)