Chapter 12: Problem 25
Find the interval of convergence. $$\sum(-1)^{k} \frac{k !}{k^{3}}(x-1)^{k}$$
Short Answer
Expert verified
The interval of convergence for the series \(\sum(-1)^{k} \frac{k !}{k^{3}}(x-1)^{k}\) is \(0 < x < 2\), which means the series converges for x in the open interval (0, 2).
Step by step solution
01
Write down the general term of the series
The general term of the given series is:
\(a_k = (-1)^k \frac{k!}{k^3}(x-1)^k\)
02
Apply the Ratio Test
Calculate the absolute value of the ratio of consecutive terms in the series:
\(|\frac{a_{k+1}}{a_k}| = \frac{(-1)^{k+1}\frac{(k+1)!}{(k+1)^3}(x-1)^{k+1}}{(-1)^k\frac{k!}{k^3}(x-1)^k}|\)
03
Simplify the expression
Simplify the ratio expression by canceling out terms and taking the absolute value:
\(|\frac{a_{k+1}}{a_k}| = \frac{(k+1)!}{(k+1)^3}\frac{k^3}{k!}\cdot|(x-1)| \\
= (k+1)\frac{k^3}{(k+1)^3}\cdot|(x-1)| \\
= \left(\frac{k}{k+1}\right)^2\cdot|(x-1)|\)
04
Calculate the limit
Take the limit of this expression as k approaches infinity:
\(\lim_{k\to\infty} \left(\frac{k}{k+1}\right)^2\cdot|(x-1)|
= \lim_{k\to\infty}\frac{k^2}{(k+1)^2}\cdot|(x-1)| \\
= |(x-1)|\lim_{k\to\infty}\frac{k^2}{(k+1)^2}\)
The last limit is equal to 1, so we have:
\(\lim_{k\to\infty} \left(\frac{k}{k+1}\right)^2\cdot|(x-1)| = |(x-1)|\)
05
Determine the convergence condition
Based on the Ratio Test, the given series converges if the limit calculated in Step 4 is less than 1, i.e.:
\(|(x-1)|< 1\)
06
Find the interval of convergence
Solve the inequality to find the interval of convergence:
\(-1 < (x-1) < 1 \\
0 < x < 2\)
Thus, the series converges for x in the open interval (0, 2).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Power Series
A power series is a type of infinite series where each term involves a power of a variable, often denoted as \(x\). In this context, the series is structured around powers of \((x-1)\). A power series typically takes the form:
- \( \sum a_k (x-c)^k \)
- Where \(a_k\) are coefficients and \(c\) is the center of the series.
Ratio Test
The ratio test is a powerful tool used to determine the convergence of a series. It is especially handy for infinite series with factorials and powers, like our power series. The test involves:
- Computing the ratio \( \frac{a_{k+1}}{a_k} \) of successive terms.
- Taking the absolute value of this ratio.
- Analyzing the behavior as \(k\) approaches infinity.
Convergence Condition
With the ratio test, the convergence condition arises from the limit comparison. For a series to converge, the limit of the absolute ratio must be less than 1 as \(k\) becomes very large. In mathematical terms:
- The series \( \sum a_k \) converges if \( \lim_{k\to\infty} |\frac{a_{k+1}}{a_k}| < 1 \).
Limit Analysis
Limit analysis involves evaluating how a function behaves as its variables approach certain values, often infinity or zero. In the context of the ratio test, we are interested in how the term \( \left(\frac{k}{k+1}\right)^2 \) behaves as \(k\) approaches infinity. Calculating this limit:
- \( \lim_{k\to\infty} \frac{k^2}{(k+1)^2} \)
- This simplifies to 1, since dividing both numerator and denominator by \(k^2\) yields \((\frac{k}{k+1})^2\).