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1. Find the sum of the series ∑n=1∞(−1)n34n\sum_{n=1}^{\infty}(-1)^{n} \frac{3}{4^{n}} 2. ...
May 31, 2024
Solution by Steps
step 1
Recognize that this is an alternating geometric series with the first term a=−34 a = -\frac{3}{4} and common ratio r=−14 r = -\frac{1}{4}
step 2
The sum of an infinite geometric series ∑n=0∞arn \sum_{n=0}^{\infty} ar^n is given by a1−r \frac{a}{1-r} if |r| < 1
step 3
Substitute a=−34 a = -\frac{3}{4} and r=−14 r = -\frac{1}{4} into the formula: −341−(−14)=−341+14=−3454=−35 \frac{-\frac{3}{4}}{1 - (-\frac{1}{4})} = \frac{-\frac{3}{4}}{1 + \frac{1}{4}} = \frac{-\frac{3}{4}}{\frac{5}{4}} = -\frac{3}{5}
Answer
The sum of the series is −35 -\frac{3}{5} .
Key Concept
Geometric Series Sum Formula
Explanation
The sum of an infinite geometric series can be found using the formula a1−r \frac{a}{1-r} where a a is the first term and r r is the common ratio.
Question 2: Does the series converge absolutely, converge conditionally, or diverge? ∑n=1∞ln⁡nn3 \sum_{n=1}^{\infty} \frac{\ln n}{n^{3}}
step 1
To determine the convergence, apply the comparison test. Compare ln⁡nn3 \frac{\ln n}{n^3} with 1n2 \frac{1}{n^2}
step 2
Note that ln⁡n \ln n grows slower than any polynomial, so ln⁡nn3≤nn3=1n2 \frac{\ln n}{n^3} \leq \frac{n}{n^3} = \frac{1}{n^2} for large n n
step 3
Since ∑n=1∞1n2 \sum_{n=1}^{\infty} \frac{1}{n^2} is a convergent p-series with p = 2 > 1 , by the comparison test, ∑n=1∞ln⁡nn3 \sum_{n=1}^{\infty} \frac{\ln n}{n^3} also converges
step 4
To check for absolute convergence, consider ∑n=1∞∣ln⁡nn3∣=∑n=1∞ln⁡nn3 \sum_{n=1}^{\infty} \left| \frac{\ln n}{n^3} \right| = \sum_{n=1}^{\infty} \frac{\ln n}{n^3} , which we have already shown to converge
Answer
The series converges absolutely.
Key Concept
Comparison Test for Series Convergence
Explanation
The comparison test helps determine the convergence of a series by comparing it to a known convergent or divergent series.
Question 3: What is the sum of the series? π−π33!+π55!−π77!⋯ \pi - \frac{\pi^3}{3!} + \frac{\pi^5}{5!} - \frac{\pi^7}{7!} \cdots
step 1
Recognize that this series is the Taylor series expansion of sin⁡x \sin x evaluated at x=π x = \pi
step 2
The Taylor series for sin⁡x \sin x is ∑n=0∞(−1)nx2n+1(2n+1)! \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}
step 3
Substitute x=π x = \pi into the series: sin⁡π=0 \sin \pi = 0
Answer
The sum of the series is 0 0 .
Key Concept
Taylor Series Expansion
Explanation
The Taylor series expansion of a function at a point provides an infinite series representation of the function.
Question 4: Find Taylor series at x=0 x=0 for the function e−x2 e^{-x^2}
step 1
The Taylor series for eu e^u at u=0 u = 0 is ∑n=0∞unn! \sum_{n=0}^{\infty} \frac{u^n}{n!}
step 2
Substitute u=−x2 u = -x^2 into the series: e−x2=∑n=0∞(−x2)nn! e^{-x^2} = \sum_{n=0}^{\infty} \frac{(-x^2)^n}{n!}
step 3
Simplify the series: e−x2=∑n=0∞(−1)nx2nn! e^{-x^2} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!}
Answer
The Taylor series for e−x2 e^{-x^2} at x=0 x=0 is ∑n=0∞(−1)nx2nn! \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!} .
Key Concept
Taylor Series for Exponential Functions
Explanation
The Taylor series for eu e^u can be adapted for functions of the form ef(x) e^{f(x)} by substituting f(x) f(x) into the series.
Question 5: Find the first four nonzero terms of the function's Taylor series at x=a x=a for f(x)=1x f(x)=\frac{1}{x}
step 1
The Taylor series for f(x) f(x) at x=a x = a is ∑n=0∞f(n)(a)n!(x−a)n \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n
step 2
Compute the derivatives of f(x)=1x f(x) = \frac{1}{x} : f(x)=x−1 f(x) = x^{-1} , f′(x)=−x−2 f'(x) = -x^{-2} , f′′(x)=2x−3 f''(x) = 2x^{-3} , f′′′(x)=−6x−4 f'''(x) = -6x^{-4}
step 3
Evaluate the derivatives at x=a x = a : f(a)=1a f(a) = \frac{1}{a} , f′(a)=−1a2 f'(a) = -\frac{1}{a^2} , f′′(a)=2a3 f''(a) = \frac{2}{a^3} , f′′′(a)=−6a4 f'''(a) = -\frac{6}{a^4}
step 4
Substitute into the Taylor series formula: 1a−(x−a)a2+(x−a)2a3−(x−a)3a4 \frac{1}{a} - \frac{(x-a)}{a^2} + \frac{(x-a)^2}{a^3} - \frac{(x-a)^3}{a^4}
Answer
The first four nonzero terms of the Taylor series for 1x \frac{1}{x} at x=a x=a are 1a−(x−a)a2+(x−a)2a3−(x−a)3a4 \frac{1}{a} - \frac{(x-a)}{a^2} + \frac{(x-a)^2}{a^3} - \frac{(x-a)^3}{a^4} .
Key Concept
Taylor Series for Rational Functions
Explanation
The Taylor series for a function at a point provides an approximation of the function using its derivatives at that point.
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