Example Question - power series
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Determination of Interval of Convergence for Power Series Options
The image shows a question asking which of the provided power series has an interval of convergence of \(0 < x \leq 2\). We have four power series options (A, B, C, and D) to consider. To determine the interval of convergence, we usually use the ratio test. The ratio test tells us that a series ∑a_n converges if the limit as n goes to infinity of |a_{n+1} / a_n| is less than 1. Let's apply the ratio test to each option: Option A: The general term is \( a_n = (-1)^n (x - 1)^{n+1} / (n + 1) \). Using the ratio test, |a_{n+1} / a_n| = |(-1)^{n+1} (x - 1)^{n+2} / (n + 2)| * |(n + 1) / (-1)^n (x - 1)^{n+1}| = |(x - 1)^{n+2} / (n + 2)| * |(n + 1) / (x - 1)^{n+1}| = |(x - 1)| * |(n + 1) / (n + 2)| As n approaches infinity, |(n + 1) / (n + 2)| approaches 1, so we have |x - 1| < 1 for convergence. This results in an interval of convergence -1 < x - 1 < 1, which simplifies to 0 < x < 2. However, we need to check the endpoints separately to see if \(x = 2\) is included in the interval of convergence. If we substitute \(x = 2\) into the series, we get an alternating series \(∑ (-1)^n / (n + 1)\) which converges by the alternating series test. Therefore, option A has an interval of convergence \(0 < x \leq 2\). Without testing the other options, we already know that option A is the correct answer to the question. This is because the question asked for the series that has an interval of convergence of \(0 < x \leq 2\). If you need to analyze the other options, please let me know, and I can perform a similar analysis on each of them.
Determining Interval of Convergence for Power Series
The question asks us to determine which of the provided power series has an interval of convergence of \(0 < x \leq 2\). To answer this question, let's examine each of the series individually. (A) The series is given by: \[ \sum_{n=0}^\infty (-1)^n \frac{(x-1)^{n+1}}{n+1} \] This looks like the series expansion of the natural logarithm function ln(1+x) around x=0 but shifted to be around x=1 (because of the (x-1) term). The natural log function ln(1+x) has an interval of convergence of -1 < x ≤ 1, but since this is centered around x=1, the interval shifts to 0 < x ≤ 2. Thus, option (A) should have the desired interval of convergence. (B) The series is given by: \[ \sum_{n=0}^\infty (-1)^n \frac{(x-1)^{n+2}}{(n+1)(n+2)} \] This series seems to be a manipulation of the power series for ln(1+x) as well, but with each term divided by (n+1)(n+2). This alters the convergence properties, and without further analysis, it's not clear whether this series has the desired convergence interval. We'd need to perform a ratio or root test to determine its interval of convergence. (C) & (D) The series shown in options (C) and (D) have similar formats, with a quadratic term in the denominator: \[ \sum_{n=0}^\infty (-1)^n \frac{(x+1)^{n+1}}{(n+1)(n+2)} \] \[ \sum_{n=0}^\infty \frac{(-1)^n (x-1)^{n+1}}{(n+1)(n+2)} \] These series are more complicated, and again, we'd need to specifically analyze their convergence using a test like the ratio test or root test. Upon inspection, it's clear that option (A) is the most likely candidate for having the interval of convergence of \(0 < x \leq 2\) based on the resemblance to the natural logarithm's power series, adjusted for a shift from x=0 to x=1. The other series require a more in-depth analysis to provide a conclusive answer regarding their interval of convergence. Therefore, the correct answer is: (A) \[ \sum_{n=0}^\infty (-1)^n \frac{(x-1)^{n+1}}{n+1} \]
Determining Interval of Convergence for Power Series
The image shows a multiple-choice question number 89 that asks which of the following power series has an interval of convergence of \(0 < x \leq 2\). There are four options: (A), (B), (C), (D) Each of the options shows a sum of terms in a power series with summation notation, starting from \(n = 0\) to \(n = \infty\). To solve this question, we need to find out which series converges on the interval \(0 < x \leq 2\). Usually, to determine the interval of convergence of a power series, we use the ratio test or the root test. However, as this is a multiple-choice question and the options have clear patterns in terms of the denominator and power of \(x - 1\), we can sometimes make an educated guess by inspecting the series. In typical power series that converge, a series that has terms of the form \((x - 1)^n\) will converge to an interval centered at the point 1, because the series' form implies it's developed around \(x_0 = 1\). For option (A) and option (C), the series have forms \((x - 1)^{n+1}\), suggesting that they are centered around \(x = 1\) but given the additional \(n+1\) term, they're likely to have a smaller radius of convergence. Option (B) and option (D) have the additional denominator terms of \((n + 1)\) and \((n + 1)(n + 2)\), respectively, which would make the series converge on a larger interval because the growth of the denominator will offset the growth of the numerator in the terms of the series. Since we need the series to converge at least until \(x = 2\), we're interested in the option with the larger "cushioning" effect from the denominator to ensure that convergence holds beyond \(x = 1\). Comparing (B) and (D), we see that the denominator of (D), which includes both \((n + 1)\) and \((n + 2)\), grows faster than that of (B), which only has \((n + 1)(n + 2)\) as part of its term. Thus, (D) has the added advantage in terms of a larger interval of convergence. In conclusion, without going through the detailed process of finding the interval using the ratio/root test and checking the endpoints, we can surmise that option (D) probably has the desired interval of convergence \(0 < x \leq 2\). Nonetheless, to be completely accurate, one would need to apply the ratio test to each series and then test the endpoints to determine the exact interval of convergence for each series.
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