Review

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Test 3 Review

1 Sample Test 3A

(problem 1) Determine whether the infinite series converges or diverges: $\displaystyle \sum _{n=1}^\infty \frac {n+1}{2n+3}$

Use the Test for Divergence

Ratio of lead coeff: $\displaystyle \lim _{n \to \infty } a_n = \frac 12 \neq 0$

(problem 2) Use the Integral Test to determine whether the infinite series converges or diverges (if it applies): $\displaystyle \sum _{n=2}^\infty \frac {1}{n\ln (n)}$ .

Show the derivative is negative for $x$ sufficiently large

Use u-sub to get the anti-derivative

Write the improper integral using a limit

The integral diverges

See problem 2a in section 3.5

(problem 3) Determine whether the infinite series converges or diverges: $\displaystyle \sum _{n=1}^\infty \frac {\sin ^2(n)}{n^3}$

Use the Direct Comparison Test

$\displaystyle 0 \leq \sin ^2(n) \leq 1$ for all $n$

Compare to $p$ -series with $p = 3 > 1 \implies$ convergence

(problem 4) Determine whether the infinite series converges or diverges: $\displaystyle \sum _{n=1}^\infty \frac {n^2 + 5n + 12}{n^3 + 6n^2 + 2n}$

Use the Limit Comparison Test

Compare with the divergent Harmonic Series

$\displaystyle \lim _{n \to \infty } \frac {a_n}{b_n} = 1$ and $0 < 1 < \infty$

(problem 5) Determine whether the infinite series converges absolutely, converges conditionally or diverges: $\displaystyle \sum _{n=0}^\infty (-1)^n \frac {n^2 + 2}{n^3 + 3}$

Use the Alternating Series Test to see that the series converges

Use the Limit Comparison Test (with the Harmonic Series) to see that the non-alternating counterpart diverges

The alternating series converges conditionally

(problem 6) Determine whether the infinite series converges or diverges: $\displaystyle \sum _{n=1}^\infty (-1)^n\frac {n^2}{3^n}$

Use the Ratio Test

$L = 1/3 < 1$ so the series converges absolutely (this is an alternating series)

$\frac {3^n}{3^{n+1}} = \frac 13$

(problem 7) Determine whether the infinite series converges or diverges: $\displaystyle \sum _{n=1}^\infty \left (\frac {n+ 2}{3n+1}\right )^n$

Use the Root Test

Ratio of lead coeff: $L = 1/3 < 1$ so the series converges

(problem 8) Find the interval of convergence of the power series (be sure to check the endpoints): $\displaystyle \sum _{n=1}^\infty \frac {n(x-2)^n}{5^n}$

Use the Ratio Test

The endpoints are $-3$ and $7$

The series diverges at both endpoints by the test for divergence

(problem 9) Find a power series representation for the function and include the interval of convergence in your answer: $\displaystyle f(x) = \frac {x^2}{3 + x}$

Factor $x^2$ from the numerator and $3$ from the denominator

Use the formula: $\displaystyle \frac {1}{1-x} = \sum _{n = 0}^\infty x^n, \; |x| <1$
with $-x/3$ in place of $x$

Answer: $\displaystyle \frac {x^2}{3 + x} = \sum _{n = 0}^\infty (-1)^n \frac {x^{n+2}}{3^{n+1}}, \; |x| <3$

2 Sample Test 3B

(problem 1) Determine whether the infinite series converges or diverges: $\displaystyle \sum _{n=1}^\infty \sqrt [n] 2$

Use the Test for Divergence

$\displaystyle \lim _{n \to \infty } a_n = 2^0 = 1 \neq 0$

(problem 2) Use the Integral Test to determine whether the infinite series converges or diverges (if it applies): $\displaystyle \sum _{n=1}^\infty \frac {1}{1 + n^2}$ .

Show the derivative is negative for $x$ sufficiently large

The anti-derivative is the inverse tangent function

Write the improper integral using a limit

The integral converges

See example 1 in section 3.5

(problem 3) Determine whether the infinite series converges or diverges: $\displaystyle \sum _{n=1}^\infty \frac {1}{\sqrt {n^2 + 1}}$

Use the Limit Comparison Test with the Harmonic Series

(problem 4) Determine whether the infinite series converges or diverges: $\displaystyle \sum _{n=1}^\infty \frac {\cos ^4(n)}{n\sqrt {n}}$

Use the Direct Comparison Test

$\displaystyle 0 \leq \cos ^4(n) \leq 1$ for all $n$

Since $n\sqrt n = n^{3/2}$ , compare to $p$ -series with $p = 3/2 > 1 \implies$ convergence

(problem 5) Determine whether the infinite series converges absolutely, converges conditionally or diverges: $\displaystyle \sum _{n=1}^\infty (-1)^{n+1} \frac {n^3}{n^4 + 2}$

Use the Alternating Series Test to see that the series converges

Use the derivative to show that the terms are decreasing

Use the Limit Comparison Test (with the Harmonic Series) to see that the non-alternating counterpart diverges

The alternating series converges conditionally

(problem 6) Determine whether the infinite series converges or diverges: $\displaystyle \sum _{n=1}^\infty \frac {n^5 10^{n+1}}{(2n)!}$

Use the Ratio Test

$L = 0 < 1$ so the series converges

(problem 7) Determine whether the infinite series converges or diverges: $\displaystyle \sum _{n=0}^\infty (-1)^n\left (\frac {n^2 + 2n+ 4}{4n^2 + 5}\right )^n$

Use the Root Test

Ratio of lead coeff: $L = 1/4 < 1$ so the series converges absolutely (this is an alternating series)

(problem 8) Find the interval of convergence of the power series (be sure to check the endpoints): $\displaystyle \sum _{n=1}^\infty \frac {(x + 1)^n}{n 3^n}$

Use the Ratio Test

The endpoints are $-4$ and $2$

The series converges at $x = -4$ (alternating Harmonic) and diverges at $x = 2$ (Harmonic)

(problem 9) Find a power series representation for the function and include the interval of convergence in your answer: $\displaystyle f(x) = \frac {4x}{1 - 4x^2}$

Factor $4x$ from the numerator

Use the formula: $\displaystyle \frac {1}{1-x} = \sum _{n = 0}^\infty x^n, \; |x| <1$
with $-4x^2$ in place of $x$

Answer: $\displaystyle \frac {4x}{1-4x^2} = \sum _{n = 0}^\infty 4^{n+1}x^{2n+1} , \; |x| <\frac 12$

2024-09-27 13:59:14