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Magazine Chapter 11: Problem 44
Find the interval of convergence of the series. $$ \sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{2}}(x-1)^{n} $$
Short Answer
Expert verified
The interval of convergence is [0, 2).
Step by step solution
01
Identify the Power Series
The given series is \( \sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{2}}(x-1)^{n} \). This is a power series centered at 1, with general term \( a_n = \frac{1}{n(\ln n)^{2}} \) and the powers of \( (x-1)^n \). We need to find the interval of convergence.
02
Use the Ratio Test
Apply the ratio test to determine the interval where the series converges. Consider the expression:\[\lim_{{n \to \infty}} \left| \frac{a_{n+1} (x-1)^{n+1}}{a_n (x-1)^n} \right| = \lim_{{n \to \infty}} \left| \frac{1}{(n+1)(\ln (n+1))^{2}} \cdot \frac{n(\ln n)^{2}}{1} \cdot (x-1) \right|\]
03
Simplify the Limit Expression
Simplify the expression for the limit:\[= \lim_{{n \to \infty}} \left| (x-1) \cdot \frac{n}{n+1} \cdot \left(\frac{\ln n}{\ln (n+1)}\right)^2 \right|.\]As \( n \) approaches infinity, both \( \frac{n}{n+1} \to 1 \) and \( \left(\frac{\ln n}{\ln (n+1)}\right)^2 \to 1 \). Therefore, the limit simplifies to:\[|x - 1|\]
04
Determine the Convergence Interval
For the series to converge, the ratio test requires:\[|x-1| < 1\]This implies that \( 0 < x < 2 \). However, the endpoints need to be checked separately to determine if the series converges at these points.
05
Check Endpoints: x = 0
At \( x = 0 \), the series becomes \( \sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{2}}(-1)^{n} \). Check for convergence using the Alternating Series Test. Since \( \frac{1}{n(\ln n)^{2}} \) is decreasing and approaches 0, the series converges at \( x = 0 \).
06
Check Endpoints: x = 2
At \( x = 2 \), the series becomes \( \sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{2}} \). This is a p-series with \( p = 1 \) multiplied by \( (\ln n)^{-2} \), and does not converge due to comparison with the harmonic series \( \sum 1/n \). Thus, it diverges at \( x = 2 \).
07
Conclusion on Interval of Convergence
Considering the radius of convergence and checking endpoints, the interval of convergence is \( [0, 2) \). The series converges for values of \( x \) within this interval, including 0 but not 2.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Power Series
A power series is a series of the form \( \sum_{n=0}^{\infty} c_n (x-a)^n \), where \( c_n \) are coefficients and \( a \) is the center of the series. It is essentially a polynomial with infinitely many terms. In the given example, we have the power series centered at \( 1 \), with the general form:
- \( a_n = \frac{1}{n(\ln n)^{2}} \)
- Powers of \( (x-1)^n \)
Ratio Test
The ratio test is a method used to determine the absolute convergence of a series. For a series \( \sum a_n \), the ratio test states that it converges absolutely if:\[ \lim_{{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right| < 1 \]In the solution, applying the ratio test to the series:
- We examine \( \lim_{{n \to \infty}} \left| \frac{a_{n+1} (x-1)^{n+1}}{a_n (x-1)^n} \right| \)
- This simplifies to \( |x - 1| \) after considering large \( n \)
Alternating Series Test
The alternating series test helps in determining the convergence of series where the terms alternate in sign. For the series \( \sum (-1)^n b_n \), it converges if:
- The terms \( b_n \) are decreasing
- The terms approach zero as \( n \to \infty \)
Harmonic Series
The harmonic series is a well-known divergent series of the form \( \sum_{n=1}^{\infty} \frac{1}{n} \). It is crucial in the convergence analysis of series. When analyzing the endpoint \( x = 2 \), the series becomes analogous to:\( \sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{2}} \).This is a modification of the harmonic series, adjusted by \( (\ln n)^{-2} \). However, the harmonic series itself diverges, so by comparison, this particular transformed series also diverges at \( x = 2 \). This comparison underscores why the given series does not converge at this endpoint.
