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PAGE 1

Sequence Convergence Analysis

Determine whether the following sequences are convergent or divergent.

  1. \[ \{a_n = 2n/(3n + 1)\} \] C
  2. \[ \{a_n = \cos n\pi\} \] D
  3. \[ \{a_n = n \sin(1/n)\} \] C

Sequences: \(\lim_{n \to \infty} a_n\) exists

  • A. (1) convergent (2) convergent (3) convergent
  • B. (1) divergent (2) convergent (3) convergent
  • C. (1) convergent (2) divergent (3) convergent
  • D. (1) convergent (2) convergent (3) divergent
  • E. (1) convergent (2) divergent (3) divergent

Step-by-Step Solutions

(1) Analysis

\[ \lim_{n \to \infty} \frac{2n}{3n+1} = \frac{2}{3} \]

(2) Analysis

The sequence values alternate: \(1, -1, 1, -1, 1, -1, \dots\)

no limit

A horizontal number line with tick marks and points at 1 and -1 showing sequence oscillation.

(3) Analysis

\[ \lim_{n \to \infty} n \sin\left(\frac{1}{n}\right) = \lim_{n \to \infty} \frac{\sin(1/n)}{1/n} \to \frac{0}{0} \]

Applying L'Hôpital's Rule:

\[ \lim_{n \to \infty} \frac{\cos(1/n) \cdot (-1/n^2)}{-1/n^2} = 1 \]
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Series Convergence Testing

Test the following series for convergence:

  1. \[ \sum_{n=1}^{\infty} \frac{1}{n + \sqrt{n}} \]
  2. \[ \sum_{n=1}^{\infty} \frac{2^n}{n + 3^n} \]
  3. \[ \sum_{n=1}^{\infty} \frac{5 - 2\sqrt{n}}{n^3} \]
  • A. I is divergent; II and III are convergent.
  • B. I and II are convergent; III is divergent.
  • C. I and III are divergent; II is convergent.
  • D. I, II and III are divergent.
  • E. I, II and III are convergent.

Analysis of Series (I)

What do these look like?

  • p-series?
  • geometric?
\[ (I) \quad \sum_{n=1}^{\infty} \frac{1}{n + \sqrt{n}} \]

When \(n\) is large, \(n > \sqrt{n}\), so:

\[ \frac{1}{n + \sqrt{n}} \approx \frac{1}{n} \]

Compare to \(\sum \frac{1}{n}\)?

But:

\[ \frac{1}{n + \sqrt{n}} \le \frac{1}{n} \]

\(\sum \frac{1}{n}\) diverges, so direct comparison doesn't work here.

Limit comparison?

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Let:

\[ b_n = \frac{1}{n} \quad \text{(known)} \]

\[ a_n = \frac{1}{n + \sqrt{n}} \]

\[ \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{1}{n + \sqrt{n}}}{\frac{1}{n}} = \lim_{n \to \infty} \frac{n}{n + \sqrt{n}} = 1 \]

So BOTH converge or BOTH diverge.

Here, \( \sum \frac{1}{n} \) diverges, we know \( \sum \frac{1}{n + \sqrt{n}} \) diverges.

(II) \( \sum_{n=1}^{\infty} \frac{2^n}{n + 3^n} \)

As \( n \to \infty \), \( n + 3^n \approx 3^n \)

So \[ \frac{2^n}{n + 3^n} \approx \frac{2^n}{3^n} = \left( \frac{2}{3} \right)^n \]

Convergent geometric series \( (r < 1) \)

Direct comparison ok:

\[ \frac{2^n}{n + 3^n} \le \frac{2^n}{3^n} \quad \text{because larger denominator} \]

So \( \sum \frac{2^n}{n + 3^n} \) must converge.

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(III) \( \sum_{n=1}^{\infty} \frac{5 - 2\sqrt{n}}{n^3} \)

As \( n \to \infty \):

\[ \frac{5 - 2\sqrt{n}}{n^3} \approx \frac{-2\sqrt{n}}{n^3} \approx \frac{-2}{n^{5/2}} \]

conv. p-series \( (p > 1) \)

Direct Comparison

\[ \frac{5 - 2\sqrt{n}}{n^3} \ge \frac{-2}{n^{5/2}} \]

finite or \( \infty \) not useful

Limit Comparison

\[ \lim_{n \to \infty} \frac{\frac{5 - 2\sqrt{n}}{n^3}}{\frac{-2\sqrt{n}}{n^3}} = \lim_{n \to \infty} \frac{5 - 2\sqrt{n}}{-2\sqrt{n}} = 1 \]

BOTH conv. or BOTH div.

So \( \sum \frac{5 - 2\sqrt{n}}{n^3} \) must conv.

Alternative approach:

\[ \sum_{n=1}^{\infty} \left( \frac{5}{n^3} - \frac{2\sqrt{n}}{n^3} \right) = \sum_{n=1}^{\infty} \frac{5}{n^3} - \sum_{n=1}^{\infty} \frac{2\sqrt{n}}{n^3} = \text{finite} \]

conv.

conv.

so conv.

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Alternating Series Estimation Theorem

The series \(\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}\) is convergent by the Alternating Series Test. According to the Alternating Series Estimation Theorem what is the smallest number of terms needed to find the sum of the series with error less than \(1/15\)?

  1. A. 1
  2. B. 4
  3. C. 5
  4. D. 2
  5. E. 3

|error| ≤ the magnitude of next term (first we throw away)

\[\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} = -1 + \frac{1}{4} - \frac{1}{9} + \frac{1}{16} - \frac{1}{25} + \dots\]

Analysis of Terms:

  • The first three terms are \(-1, \frac{1}{4}, -\frac{1}{9}\).
  • The fourth term is \(\frac{1}{16}\).
  • Since \(\frac{1}{16} < \frac{1}{15}\), this is the first term with magnitude less than the error threshold.
  • According to the theorem, we stop at the term before it.
  • Therefore, we need 3 terms.
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Alternative Method: Solving for n

We can also solve for \(n\) directly using the inequality for the magnitude of the next term:

\[\frac{1}{n^2} < \frac{1}{15}\]\[15 < n^2\]\[n^2 > 15\]\[n > \sqrt{15}\]

\(\sqrt{15}\) is bigger than 3 and less than 4.

So \(n=4\) is the first time the term magnitude \(\frac{1}{n^2} < \frac{1}{15}\).

So we sum up to the one before it → n = 3

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Convergence Analysis: Series with Variable c

For which values of c is \[ \sum_{n=1}^{\infty} \left( 1 + \frac{c^2}{n} \right)^n \] convergent?

  1. A. All values of c.
  2. B. \( |c| < 1 \).
  3. C. \( |c| < 2 \).
  4. D. \( |c| > 2 \).
  5. E. No values of c.

Analysis and Root Test

Let \( a_n = \left( 1 + \frac{c^2}{n} \right)^n \). Since the term has a power of \( n \), the root test looks good here.

Root Test:

\[ \lim_{n \to \infty} \sqrt[n]{|a_n|} \begin{cases} < 1 & \text{conv.} \\ = 1 & ? \\ > 1 & \text{div.} \end{cases} \]

Applying the root test to our series:

\[ \lim_{n \to \infty} \sqrt[n]{\left( 1 + \frac{c^2}{n} \right)^n} = \lim_{n \to \infty} \left( 1 + \frac{c^2}{n} \right) = 1 \quad ? \]

Divergence Test

Since the root test is inconclusive (limit equals 1), we check the limit of the terms themselves:

\[ \lim_{n \to \infty} \left( 1 + \frac{c^2}{n} \right)^n = e^{c^2} \neq 0 \]

The limit of the terms is \( e^{\text{something}} \) (specifically \( e^{c^2} \)), which is never 0. Therefore, the series fails the divergence test for all values of c. The correct answer is E.

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Convergence Analysis: Rational Function Series

Find all values of p such that the series \[ \sum_{k=1}^{\infty} \left( \frac{k^4 + 3k}{k^p + 2} \right)^{1/3} \] converges.

  1. A. \( p > 8 \)
  2. B. \( p > 6 \)
  3. C. \( p > 5 \)
  4. D. \( p > 4 \)
  5. E. \( p > 7 \)

Asymptotic Behavior

What does it look like as \( k \to \infty \)?

As \( k \to \infty \):

\[ \left( \frac{k^4 + 3k}{k^p + 2} \right)^{1/3} \approx \left( \frac{k^4}{k^p} \right)^{1/3} \]\[ \approx \left( \frac{1}{k^{p-4}} \right)^{1/3} \]\[ \approx \frac{1}{k^{(p-4)/3}} \]

Convergence Condition

The series behaves like a p-series. It is convergent if the exponent is greater than 1:

\[ \frac{p-4}{3} > 1 \]\[ p - 4 > 3 \]\[ p > 7 \]

Thus, the series converges for \( p > 7 \). The correct answer is E.

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Conditional Convergence of Series

8. Which of the following series converge conditionally?

  • A. \(\sum_{n=5}^{\infty} \frac{(-1)^n}{n^{1.1}}\)
  • B. \(\sum_{n=5}^{\infty} \frac{(-1)^n}{n^2-n+1}\)
  • C. \(\sum_{n=5}^{\infty} (-1)^n n\)
  • D. \(\sum_{n=5}^{\infty} (-1)^n e^{-n}\)
  • E. \(\sum_{n=5}^{\infty} \frac{(-1)^n}{\ln(\ln n)}\)

Definition:

Given \(\sum a_k\) converges but \(\sum |a_k|\) diverges.

Example:

\[\sum_{k \ge 1} \frac{(-1)^{k+1}}{k} \text{ conv.}\]\[\sum_{k \ge 1} \frac{1}{k} \text{ div.}\]

Analysis of Option A

\[\sum \frac{(-1)^n}{n^{1.1}} \text{ conv. because } \lim_{n \to \infty} \frac{1}{n^{1.1}} = 0 \text{ and } \frac{1}{n^{1.1}} \text{ is nonincreasing}\]
\[\sum \frac{1}{n^{1.1}} \text{ conv. so A. conv. absolutely}\]

Analysis of Option B

\[\sum \frac{(-1)^n}{n^2-n+1} \text{ looks like } \sum \frac{(-1)^n}{n^2} \text{ as } n \to \infty\]

\(\sum \frac{(-1)^n}{n^2-n+1}\) conv. by Alt. series test.

\(\sum \frac{1}{n^2}\) conv. p-series after taking absolute value.

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Analysis of Option C

\[\lim_{n \to \infty} (-1)^n n \neq 0 \text{ doesn't converge}\]

Analysis of Option D

\[\sum (-1)^n e^{-n} = \sum \frac{(-1)^n}{e^n} \text{ conv.? yes, by alt. series test}\]
\[\sum \frac{1}{e^n} \text{ conv.? yes}\]\[= \sum \left(\frac{1}{e}\right)^n \text{ geo. series w/ } r = \frac{1}{e} < 1\]

conv. absolutely

Analysis of Option E

\[\sum \frac{(-1)^n}{\ln(\ln(n))} \text{ conv. by alt. series test}\]
\[\sum \frac{1}{\ln(\ln(n))} \text{ conv.? No.}\]

Since the series converges but its absolute value diverges, it converges conditionally.

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Comparison of Logarithmic and Linear Growth

We begin by establishing the fundamental inequality between logarithmic and linear functions:

\[ \ln(n) < n \]

By applying the natural logarithm to both sides of the inequality, we obtain:

\[ \ln(\ln(n)) < \ln(n) \]
Coordinate graph with n on the x-axis and y on the vertical axis, showing y=n and y=ln(n) curves.

Reciprocal Inequalities

Taking the reciprocals of the established inequalities reverses the direction of the inequality signs:

\[ \frac{1}{\ln(\ln(n))} > \frac{1}{\ln(n)} > \frac{1}{n} \]

Series Divergence Analysis

Applying the summation operator to these terms, we compare the series:

\[ \sum \frac{1}{\ln(\ln(n))} > \sum \frac{1}{\ln(n)} > \sum \frac{1}{n} \]

Since the harmonic series \( \sum \frac{1}{n} \) is known to diverge (\( \to \infty \)), by the Comparison Test, the larger series must also diverge:

\( \sum \frac{1}{\ln(\ln(n))} \) is divergent (div.).

Conclusion:

So, \( \sum \frac{1}{\ln(\ln(n))} \) conv. conditionally.