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