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What are the differences between the Direct Comparison Test and the Limit Comparison Test?

Direct Comparison: Requires direct inequality (anbna_n \le b_n). Limit Comparison: Compares the limit of the ratio of terms. Direct Comparison: More straightforward when applicable. Limit Comparison: Useful when direct inequality is hard to establish.

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What are the differences between the Direct Comparison Test and the Limit Comparison Test?

Direct Comparison: Requires direct inequality (anbna_n \le b_n). Limit Comparison: Compares the limit of the ratio of terms. Direct Comparison: More straightforward when applicable. Limit Comparison: Useful when direct inequality is hard to establish.

Compare and contrast p-series and geometric series.

p-series: Form 1np\sum \frac{1}{n^p}, converges if p>1p > 1. Geometric series: Form arn\sum ar^n, converges if r<1|r| < 1. Both: Useful for comparison tests. p-series: Depends on exponent p. Geometric series: Depends on common ratio r.

Compare Direct Comparison Test when an<bna_n < b_n and bn\sum b_n converges vs. when an\sum a_n diverges.

bn\sum b_n converges: Implies an\sum a_n converges. an\sum a_n diverges: Implies bn\sum b_n diverges. The first shows convergence, the second shows divergence.

Compare Limit Comparison Test when the limit is a finite positive number vs. when the limit is zero or infinity.

Finite positive number: Both series behave alike (both converge or both diverge). Zero or infinity: The test is inconclusive and a different comparison series must be found.

Compare using bn=1npb_n = \frac{1}{n^p} with p>1p>1 and p1p \le 1 in comparison tests.

p>1p>1: bn\sum b_n converges, useful for showing convergence of a smaller series. p1p \le 1: bn\sum b_n diverges, useful for showing divergence of a larger series.

What are the similarities and differences between Direct Comparison Test and Integral Test?

Direct Comparison: Compares series to another series. Integral Test: Relates series convergence to integral convergence. Both: Used to determine convergence/divergence. Integral Test: Requires function to be continuous, positive, and decreasing.

Compare the conditions for convergence of a geometric series with r=0.5 and r=2.

r=0.5: Converges because |0.5| < 1. r=2: Diverges because |2| > 1. Convergence depends on the absolute value of r being less than 1.

Compare using Direct Comparison with ana_n and bnb_n where ana_n is always greater than bnb_n.

If an>bna_n > b_n and an\sum a_n converges, the test is inconclusive. If an>bna_n > b_n and bn\sum b_n diverges, then an\sum a_n diverges.

Compare the usefulness of comparison tests for series with alternating signs vs. series with only positive terms.

Comparison tests are typically used for series with positive terms. Alternating series often require the Alternating Series Test.

Compare the difficulty of applying the Direct Comparison Test when the inequality is obvious vs. when it requires manipulation.

Obvious inequality: Application is straightforward. Requires manipulation: More complex, requires algebraic skills to establish the inequality.

How to determine if n=11n2+1\sum_{n=1}^{\infty} \frac{1}{n^2+1} converges or diverges using the Direct Comparison Test?

  1. Recognize 1n2+1<1n2\frac{1}{n^2+1} < \frac{1}{n^2}. 2. Know that n=11n2\sum_{n=1}^{\infty} \frac{1}{n^2} converges (p-series with p=2 > 1). 3. Conclude that n=11n2+1\sum_{n=1}^{\infty} \frac{1}{n^2+1} converges by the Direct Comparison Test.

How to determine if n=1nn32\sum_{n=1}^{\infty} \frac{n}{n^3-2} converges or diverges using the Limit Comparison Test?

  1. Choose bn=nn3=1n2b_n = \frac{n}{n^3} = \frac{1}{n^2}. 2. Evaluate limnanbn=limnnn321n2=1\lim_{n\to\infty} \frac{a_n}{b_n} = \lim_{n\to\infty} \frac{\frac{n}{n^3-2}}{\frac{1}{n^2}} = 1. 3. Since the limit is finite and positive, and n=11n2\sum_{n=1}^{\infty} \frac{1}{n^2} converges, conclude that n=1nn32\sum_{n=1}^{\infty} \frac{n}{n^3-2} converges.

How to choose a comparison series bnb_n for n=12n+1n2+n+1\sum_{n=1}^{\infty} \frac{2n+1}{n^2+n+1}?

  1. Focus on the dominant terms: 2n in the numerator and n2n^2 in the denominator. 2. Form bn=2nn2=2n=1nb_n = \frac{2n}{n^2} = \frac{2}{n} = \frac{1}{n}.

How to determine if n=11n1\sum_{n=1}^{\infty} \frac{1}{\sqrt{n} - 1} diverges?

  1. Compare to 1n\frac{1}{\sqrt{n}}. 2. Note that 1n1>1n\frac{1}{\sqrt{n}-1} > \frac{1}{\sqrt{n}}. 3. Recognize that n=11n\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} diverges (p-series with p=1/2 < 1). 4. Conclude that n=11n1\sum_{n=1}^{\infty} \frac{1}{\sqrt{n} - 1} diverges by the Direct Comparison Test.

How to handle a series with a sine function in the numerator, such as n=1sin(n)n2\sum_{n=1}^{\infty} \frac{\sin(n)}{n^2}?

  1. Use the fact that 1sin(n)1-1 \le \sin(n) \le 1. 2. Compare to n=11n2\sum_{n=1}^{\infty} \frac{1}{n^2}. 3. Since sin(n)n21n2\left|\frac{\sin(n)}{n^2}\right| \le \frac{1}{n^2} and n=11n2\sum_{n=1}^{\infty} \frac{1}{n^2} converges, conclude that n=1sin(n)n2\sum_{n=1}^{\infty} \frac{\sin(n)}{n^2} converges absolutely by the Direct Comparison Test.

Given n=112n+n\sum_{n=1}^{\infty} \frac{1}{2^n + n}, how do you select a suitable bnb_n?

  1. Notice that 2^n grows faster than nn. 2. Choose bn=12nb_n = \frac{1}{2^n}. 3. Use the Direct Comparison Test since 12n+n<12n\frac{1}{2^n + n} < \frac{1}{2^n}.

How to determine if n=21nln(n)\sum_{n=2}^{\infty} \frac{1}{n\ln(n)} diverges?

  1. Recognize that this is not directly comparable to a p-series or geometric series. 2. Consider the Integral Test (not a comparison test, but relevant). 3. Since 21xln(x)dx\int_2^{\infty} \frac{1}{x\ln(x)} dx diverges, conclude that n=21nln(n)\sum_{n=2}^{\infty} \frac{1}{n\ln(n)} diverges.

How to determine if n=13n4n2n\sum_{n=1}^{\infty} \frac{3^n}{4^n - 2^n} converges?

  1. Compare to bn=3n4n=(34)nb_n = \frac{3^n}{4^n} = (\frac{3}{4})^n. 2. Use the Limit Comparison Test. 3. Since n=1(34)n\sum_{n=1}^{\infty} (\frac{3}{4})^n converges (geometric series with |r| < 1), conclude that n=13n4n2n\sum_{n=1}^{\infty} \frac{3^n}{4^n - 2^n} converges.

How do you know when to use the Direct Comparison Test vs. the Limit Comparison Test?

Direct Comparison Test: when you can easily show an<bna_n < b_n or an>bna_n > b_n. Limit Comparison Test: when it's difficult to find a direct inequality, but the limit of the ratio is easy to compute.

What is the first step in determining convergence/divergence using comparison tests?

Identify a suitable comparison series (bnb_n) with known convergence/divergence behavior.

Define Direct Comparison Test.

Compares a series to another known series to determine convergence/divergence. If 0 le a_n le b_n and bn\sum b_n converges, then an\sum a_n converges. If an\sum a_n diverges, then bn\sum b_n diverges.

Define Limit Comparison Test.

Compares the limit of the ratio of two series terms. If limnanbn=c\lim_{n\to\infty} \frac{a_n}{b_n} = c, where 0 < c < \infty, then both series either converge or diverge.

Define Convergence.

A series converges if the sequence of its partial sums approaches a finite limit.

Define Divergence.

A series diverges if the sequence of its partial sums does not approach a finite limit.

Define p-series.

A series of the form n=11np\sum_{n=1}^{\infty} \frac{1}{n^p}, where pp is a real number.

Define Geometric Series.

A series of the form n=0arn\sum_{n=0}^{\infty} ar^n, where aa is a constant and rr is the common ratio.

What is a series?

The sum of the terms of a sequence.

Define ana_n and bnb_n in the context of comparison tests.

ana_n and bnb_n are the terms of the two series being compared. They must be non-negative for comparison tests to be valid.

What does it mean for a limit to be 'finite'?

A finite limit is a real number (not infinity).

Define 'end behavior' in the context of series.

How the terms of a series behave as nn approaches infinity.