Part 2: Sequences, Series, Power Series

Sequences: For each sequence, determine whether it converges or diverges. If it converges, find the limit.

a) $a_n = \frac {3^n}{n!}$

• $\lim_{n\to\infty} a_n = 0$

  • Factorials grow faster than exponentials

b) $a_n = \frac n{3n+2}$

• $\frac {n}{3n+2} \cdot \frac {\frac 1n}{\frac 1n} \Rightarrow \frac 1{3 + \frac 2n}$

• $\lim_{n\to\infty} a_n = \frac 13$

• Converges to $\frac 13$

c) $a_n = \frac {3n}{n²+4n+5}$

• Since there is a larger exponent in the denominator, $a_n$ converges to 0

d) $a_n = \frac {\cos n}{n}$

• Since $\lim_{n\to\infty} \cos n \le 1$

  • $\lim_{n\to\infty} \frac {\cos n }{n} \le \frac 1n = 0$

e) $a_n = \sqrt {\frac {n+3}{16n+3}}$

• $(\frac {n+3}{16n+3})^{\frac 12}$

• $(\frac {1 + \frac 3n} {16 + \frac 3n } )^{\frac 12}$

• $(\frac 1{16})^{\frac 12}$

• $\frac 1{4}$

f) $a_n = \frac {n!}{(n+2)!}$

• $(n+2)! = (n+2)(n+1)n!$

• $\frac {n!}{(n+2)!} \Rightarrow \frac {1}{(n+2)(n+1)}$

Converges to 0

Determine whether each series converges or diverges. Use any appropriate convergence test.

a) $\sum_{n=1}^{\infty} \frac 1{n^4}$

• Use P-Series (converges when p > 1 in the form $\frac {1}{n^p}$)

• $p = 4$, therefore the series converges

b) $\sum_{n=1}^{\infty} \frac {\ln n}n$

• Use the integral test

• $\int_1^{\infty} \frac {\ln x}x dx$

  • $u = \ln x$

  • $du = \frac 1x dx$

• $\int_1^{\infty} u du$

• $[\frac {u²}{2}]_1^{\infty}$

• $\frac 12(\ln² |\infty| - \ln² |1|)$

  • Series abso-fucking-lutely diverges

c) $\sum_{n=1}^{\infty} \frac 1{1+n²}$

• Use the comparison test

• \frac 1{1+n²} < \frac 1{n²}

• $\frac 1{n²}$ converges, so $\frac 1{1+n²}$ converges

d) $\sum_{n=1}^{\infty} \frac {n}{4n³-1}$

• Use the limit comparison test

• $\frac {n}{4n³-1} \cdot \frac {1n}{1n} \Rightarrow \frac {1}{4n²}$

• $b_n = \frac 1{n²}$

• $a_n = \frac {n}{4n³-1}$

• $\frac {a_n}{b_n} = \frac {n³}{4n³-1} \cdot \frac {\frac 1{n³}}{\frac 1{n³}} = \frac 1{4 - \frac 1{n³}} = \frac 14 \ne 0$

• Since $\frac 14$ is a positive non-infinite number, $a_n$ has the same behavior as $b_n$, so both converge

e) $\sum_{n=2}^{\infty} \frac {n^4}{n^5+1}$

• $\frac {n^4}{n^5 - 1} \cdot \frac {\frac 1{n^4}}{\frac 1{n^4}}= \frac 1n$

• $b_n = \frac 1n$

• $a_n = \frac {n^4}{n^5+1}$

• $\frac {n^4 \cdot n}{n^5 + 1} = \frac {n^5}{n^5+1} = 1$

• Since $1 \ne 0$, $a_n$ has the same behavior as $b_n$, and $b_n$ diverges, so $a_n$ diverges

f) $\sum_{n=2}^{\infty} \frac {n^4}{n^5-1}$

• Same steps as the previous one since the limit is the same at the end, so diverges

g) $\sum_{n=5}^{\infty} \frac {1}{\sqrt n -2}$

• $a_n =\frac 1{n^{\frac 12} - 2}$

• $b_n = \frac 1{n^{\frac 12}}$

• $\frac {n^{\frac 12}} {n^{\frac 12} - 2} = 1$

• Since $1 \ne 0$, this $a_n$ has the same behavior as $b_n$, so both diverge.

h) $\sum_{n=1}^{\infty} \frac {5n²}{\sqrt {3n^5+n}}$

• $a_n = \frac {5n²}{\sqrt {3n^5 + n}}$

• $b_n = \frac {1}{\sqrt n}$

• $\frac {5n²\sqrt n} {\sqrt{3n^5+n}} = \frac {5n^{\frac 52}}{\sqrt {3n^5+n}} = \frac {5}{\sqrt 3}$

• Since $\frac 5{\sqrt 3} \ne 0$, $a_n$ and $b_n$ have the same behavior, so both diverge

i) $\sum_{n=1}^{\infty} \frac n{(n+1)^{\frac 52}}$

• $\Rightarrow \frac n{n^{\frac 5n}} \cdot \frac {\frac 1n}{\frac 1n} = \frac {1}{n^{\frac 32}}$

• $a_n = \frac {n}{(n+1)^{\frac 52}}$

• $b_n=\frac {1}{n^{\frac 32}}$

• $\frac {n \cdot n^{\frac 32}}{(n+1)^{\frac 52}}$

• $\frac {n^{\frac 52}}{(n+1)^\frac {5}{2}} = 1$

• Since $1 \ne 0$, $a_n$ has the same behavior as $b_n$, so they both converge

j) Determine whether the series $\sum_{n=1}^{\infty} (-1)^n \frac 1n$ converges absolutely, conditionally, or diverges

Series: Determine whether each series converges or diverges. Use any appropriate convergence test

a) $\sum_{n=1}^\infty \frac {(-1)^n}{n³}$

• $(-1)^n \frac 1{n³}$

  • Is it decreasing? Yes

  • Is the limit zero? Yes

• Converges absolutely

b) $\sum_{n=1}^\infty \frac {\sin n}{n²}$

• Comparison test:

  • \frac {\sin n}{n²} < \frac {1}{n²}

  • $\frac 1{n²}$ is absolutely convergent

• $\frac {\sin n}{n²}$ absolutely convergent

c) $\sum_{n=1}^{\infty} \frac {3^n}{n!}$

• Ratio Test:

  • $L = \lim_{n\to\infty} \frac {\frac {3^{n+1}}{(n+1)!}}{\frac {3^n}{n!}}$

  • $L = \lim_{n\to\infty} \frac {3^{n+1}n!} {(n+1)!3^n}$

  • $L = \lim_{n\to\infty} \frac {3^n \cdot 3n!}{(n+1)!3^n}$

  • $L = \lim_{n\to\infty} \frac {3n!}{(n+1)!}$

  • $L = \lim_{n\to\infty} \frac {3n!}{(n+1)n!}$

  • L = \lim_{n\to\infty} \frac {3}{n+1} = 0 < 1

• This converges absolutely

d) $\sum_{n=1}^\infty \frac {n³}{5^n}$

• Ratio test:

  • $L = \lim_{n\to\infty} \frac {\frac {(n+1)³}{5^{n+1}}} {\frac {n³} {5^n}}$

  • $L = \lim_{n\to\infty} \frac {(n+1)³5^n} {5^{n+1}n³}$

  • $L = \lim_{n\to\infty} \frac {(n+1)³}{5n³}$

  • L = \lim_{n\to\infty} \frac 15 (\frac {n+1}{n})³ = \frac 15 < 1

• Converges absolutely

e) $\sum_{n=1}^\infty \frac {(-2)^n}{n!}$

• Absolute Convergence Test:

  • $\sum_{n=1}^\infty \frac {2^n}{n!}$

  • Ratio Test:

    • $L=\lim_{n\to\infty} \frac {\frac {2^{n+1}} {(n+1)!}} {\frac {2^n}{n!} }$

    • $L = \lim_{n\to\infty} \frac {2^{n+1}n!}{(n+1)!(2^n)}$

    • $L = \lim_{n\to\infty} \frac {2}{n+1} = 0$

• Converges Absolutely

Power Series: For each, find the radius and interval of convergence. Test endpoints when appropriate

a) $\sum_{n=0}^\infty c_n (x-3)^n$ $c_n = \frac 1{n+1}$

• Rewrite as $\sum_{n=0}^\infty \frac {(x-3)^n} {n+1}$

• Ratio Test:

  • $L = \lim_{n\to\infty} \frac {\frac {(x-3)^{n+1}}{n+2}}{\frac {(x-3)^n}{n+1}}$

  • $L = \lim_{n\to\infty} |\frac {(x-3)^{n+1}(n+1)}{(n+2)(x-3)^n}|$

  • $L = \lim_{n\to\infty} |x-3||\frac {n+1} {n+2}|$

    • The limit as $n\to\infty$ is 1

  • $L = |x-3|$

• Radius of convergence:

  • For the series to converge, we require L < 1

    • |x-3| < 1

  • The radius of convergence is $R = 1$. The inequality defines the open interval of convergence

    • 2<x<4 

• Endpoints:

  • By replacing x with 2 and 4, we can see that it is convergent when 2 is plugged in and divergent when 4 is plugged in, so $[2, 4)$

b) $\sum_{n=1}^{\infty} \frac {x^n}{n2^n}$

• Ratio Test:

  • $L = \lim_{n\to\infty} \frac {\frac {x^{n+1}} {(n+1)2^{n+1}} } {\frac {x^n} {n2^n} }$

  • $L = \lim_{n\to\infty} \frac {x^{n+ 1}n2^n} {(n+1)2^{n+1}x^n}$

  • $L = \lim_{n\to\infty} |\frac x2| \frac n{n+1} = \frac x2$

  • $L = \frac x2$

• Radius of convergence:

  • For the series to converge, we require L < 1

    • \frac x2 < 1 \Rightarrow x < 2

  • The radius of convergence is $R=2$

    • - 2 < x < 2

• Endpoints

  • $[-2,2)$

c) $\sum_{n=0}^\infty \frac {nx^n}{5^n}$

• Ratio Test:

  • $L = \lim_{n\to\infty} \frac {\frac {(n+1)x^{n+1}} {5^{n+1}}} {\frac {nx^n} {5^n}}$

  • $L = \lim_{n\to\infty} \frac {(n+1)x^{n+1}5^n} {5^{n+1}nx^n}$

  • $L = \lim_{n\to\infty} \frac {(n+1)x}{5n} = \frac x5$

• Radius of convergence:

  • \frac x5 < 1 \Rightarrow x < 5

  • The radius of convergence is $R = 5$

    • -5 < x < 5

• Endpoints:

  • (-5 < x < 5)

d) $\sum_{n=0}^{\infty} \frac {(x+2)^n}{3^n}$

• Geometric series:

  • $(\frac {x+2}{3})^n$

  • |x+2| < 3 \Rightarrow -5 < x < 1

• Radius of convergence is $R = 1$

• Endpoints are $(-5,1)$

e) $\sum_{n=2}^\infty \frac {(x-1)^n} {n\ln n}$

• Ratio test:

  • $L = \lim_{n\to\infty} \frac {\frac {(x-1)^{n+1}} {(n+1)\ln (n+1)}} {\frac {(x-1)^n} {n\ln n}}$

  • $L = \lim_{n\to\infty} \frac {(x-1)^{n+1}n\ln n} {(n+1)\ln(n+1)(x-1)^n}$

  • $L = \lim_{n\to\infty} \frac {(x-1)\ln(n)n} {(n+1)\ln (n+1)}$

  • $L = \lim_{n\to\infty} \frac {n}{n+1} \cdot \frac {\ln n}{\ln (n + 1)} \cdot |x-1| = |x-1|$

• Radius of convergence:

  • |x-1| < 1 \Rightarrow 0 < x <2

  • $R = 1$

• Endpoints:

  • $\sum_{n=2}^\infty \frac {(0-1)^n}{n\ln n} \Rightarrow$ Converges

  • $\sum_{n=2}^\infty \frac {1^n}{n\ln n} \Rightarrow$ Diverges

  • $[0,2)$

f) $\sum_{n=0}^\infty n²(x-4)^n$

• Ratio test:

  • $L = \lim_{n\to\infty} \frac {(n+1)²(x-4)^{n+1}} {n²(x-4)^n}$

  • $L = \lim_{n\to\infty} (\frac {n+1}{n})² |x-4| = |x-4|$

• Radius of convergence:

  • |x-4| < 1

  • Radius of convergence is $R = 1$

  • 3 < x < 5

• Endpoints:

  • $(3,5)$

g) $\sum_{n=1}^\infty \frac {(x-5)^n}{n!}$

• Ratio test:

  • $L = \lim_{n\to\infty} \frac {\frac {(x-5)^{n+1}} {(n+1)!} } {\frac {(x-5)^n} {n!}}$

  • $L = \lim_{n\to\infty} \frac {(x-5)^{n+1}n!} {(n+1)!(x-5)^n}$

  • $L = \lim_{n\to\infty} \frac {(x-5)} {(n+1)} = 0$

• Always convergent

• $R = \infty$ $(-\infty, \infty)$

a) Write a power series for $\frac 2{3+x}$ and determine its interval of convergence

• Rewrite:

  • $\frac 2{3+x} = \frac 23 \cdot \frac {1}{1+ \frac x3}$

• Use geometric expansion:

  • $\frac 1{1+u} = \sum_{n=0}^\infty (-u)^n$, |u| < 1

• With $u = \frac x3$

  • $\frac {2}{3+x} = \frac 23 \sum_{n=0}^\infty (-1)^n (\frac x3)^n$

  • $\frac 2{3+x}= \sum_{n=0}^\infty (-1)^n \frac {2x^n}{3^{n+1}}$, |x| < 3

b) Write a power series for $\ln(1-x)$ and determine its interval of convergence

• Start from the geometric series:

  • $\frac {1}{1-t} = \sum_{n=0}^\infty t^n$, |t| < 1

• Integrate both sides from 0 to x

  • $\int_0^x \frac 1{1+t} dt = -\ln |1 - x|$

  • \int_0^x \sum_{

c) Write a power series for \frac x{(1-x)²}$</p><p></p><ul><li><p>Start from geometric:</p><ul><li><p>$\frac 1{1-x} = \sum_{n=0}^\infty x^n$</p></li></ul></li><li><p>Differentiate</p><ul><li><p>$\frac {1}{(1-x)²} = \sum_{n=0}^\infty nx^{n-1}$</p></li></ul></li><li><p>Multiply by x</p><ul><li><p>$\frac x{(1-x)²} = \sum_{n=0}^{\infty} nx^n$</p></li></ul></li></ul><p></p><ul><li><p>Final:</p><ul><li><p>$\sum_{n=0}^\infty nx^n$,$|x| < 1$</p></li></ul></li></ul><p></p><p>d) Write a power series for$e^{2x}$and find its radius of convergence</p><p></p><ul><li><p>Use exponential series:</p><ul><li><p>$e^{2x} = \sum_{n=0}^\infty \frac {(2x)^n}{n!} = \sum_{n=0}^\infty \frac {2^n}{n!}x^n$</p></li></ul></li><li><p>Radius of convergence$R = \infty$</p><ul><li><p>$e^{2x} = \sum_{n=0}^\infty \frac {2^n}{n!}x^n$,$R = \infty$</p></li></ul></li></ul><p></p><p>e) Write a power series for$e^{x³}$and find its radius of convergence</p><p></p><ul><li><p>$e^x = \sum_{n=0}^\infty \frac {x^n}{n!}$</p></li><li><p>$e^{x^3} = \sum_{n=0}^\infty \frac {{x³}^n}{n!}$,$R = \infty$</p></li></ul><p></p><p>f) Write the Maclaurin series for$\sin x$and its interval of convergence</p><p></p><ul><li><p>$\sin x = \sum_{n=0}^{\infty} (-1)^n \frac {x^{2n+1}}{(2n+1)!}$</p></li><li><p>$R = \infty$,$-\infty < x < \infty$</p></li></ul><p></p><p>g) Write the Maclaurin series for$\cos x $and its interval of convergence</p><p></p><ul><li><p>$\cos x = \sum_{n=0}^\infty (-1)^n \frac {x^{2n}}{(2n)!}$</p></li><li><p>$R = \infty$,$-\infty < x < \infty$</p></li></ul><p></p><p>h) Write the geometric series for$\frac 1 {1-x}$</p><p></p><ul><li><p>$\frac 1{1-x} =\sum_{n=0}^\infty x^n$</p></li><li><p>$|x| < 1$</p></li></ul><p></p><p>i) Find a power series for$\frac x{1-x}$</p><p></p><ul><li><p>Multiply geometric series by x:</p><ul><li><p>$\frac x{1-x} = x\sum_{n=0}^\infty x^n = \sum_{n=0}^\infty x^{n+1}$,$|x| < 1$</p></li></ul></li></ul><p></p><p>j) Find a power series for$\frac x{1+x}$</p><p></p><ul><li><p>$\sum_{n=0}^\infty = (-1)^nx^{n}$,$|x| < 1$</p></li></ul><p></p><p>k) Write the Maclaurin series for$\sin(3x)$</p><p></p><ul><li><p>$\sum_{n=0}^\infty (-1)^n \frac {(3x)^{2n+1}}{(2n+1)!}$</p></li></ul><p></p><p>l) Find the degree-3 Taylor polynomial for$\sqrt x$at$a=1$</p><p></p><ul><li><p>$f(x) = x^\frac 12$</p></li></ul><p></p><ul><li><p>$f^0(x) = x^{\frac 12}$</p></li><li><p>$f^1(x) = \frac 12 x^{-\frac 12}$</p></li><li><p>$f²(x) = -\frac 14 x^{-\frac 32}$</p></li><li><p>$f³(x) = \frac 38 x^{-\frac 52}$</p></li></ul><p></p><ul><li><p>$f^0(1) = 1^\frac 12 = 1$</p></li><li><p>$f^1(1) = \frac 12 (1)^{-\frac 12} = \frac 12$</p></li><li><p>$f²(1) = -\frac 14$</p></li><li><p>$f³(1) = \frac 38$</p></li></ul><p></p><ul><li><p>$T_3(x) = \frac {1}{0!}(x-1)^0 + \frac {\frac 12}{1!}(x-1)^1 + \frac {-\frac 14}{2!}(x-1)² + \frac {\frac 38}{3!}(x-1)³$</p></li><li><p>$T_3(x) = 1+ \frac 12 (x-1) - \frac 18 (x-1)² + \frac 1{16} (x-1)³$</p></li></ul><p></p><p>m) Find the degree-4 Maclaurin polynomial for$\ln(1+x)$</p><p></p><ul><li><p>$\ln(1+x) = \sum_{n=1}^\infty (-1)^{n+1} \frac {x^n}{n}$</p></li></ul><p></p><ul><li><p>$x - \frac {x²}{2} + \frac {x³}3 - \frac {x^4}{4}$$