Find the interval of convergence of the series [math: ∑n=0∞(−1)n+1n10n(x−2)n] \displaystyle\sum\limits_{n=0}^ \infty \frac{(-1)^{n+1}}{n 10^n} (x-2)^n .

Question

BlackTom AI · Accepted Answer

First, I restate the problem in my own words: we are asked for the interval of convergence of the power series sum from n=0 to infinity of [(-1)^{n+1} / (n 10^n)] (x-2)^n. The general term is a_n = (-1)^{n+1} (x-2)^n / (n 10^n). The usual way to determine the interval is to examine the radius of convergence via the absolute value of the terms and then check the endpoints separately.

- Step 1: Determine the radius of convergence using the root/ratio idea on the absolute value. Consider the absolute series sum |x-2|^n / (n 10^n). This behaves like sum [ (|x-2|/10)^n ] / n. This is a known p-series-like form: for r = |x-2|/10, the series sum r^n / n converges if and only if r < 1 and diverges if r > 1. Therefore, we obtain the condition |x-2|/10 < 1, i.e., |x-2| < 10. This gives a candidate interval (2-10, 2+10) = (-8, 12) for absolute convergence, hence for convergence in the interior.

- Step 2: Check the endpoints x = -8 and x = 12 separately, because the ratio/root test is inconclusive there.
  - At x = 12: substitute into the series. We get (x-2)^n = 10^n, so the nth term is (-1)^{n+1} 10^n / (n 10^n) = (-1)^{n+1} / n. This is the alternating harmonic series, which converges (conditionally) to ln 2. So the series converges at x = 12.
  - At x = -8: substitute into the series. Then (x-2)^n = (-10)^n, so the nth term is (-1)^{n+1} (-10)^n / (n 10^n) = (-1)^{n+1} (-1)^n / n = (-1)^{2n+1} / n = -1/n. This is the negative of the harmonic series, which diverges. So the series diverges at x = -8.

- Step 3: Compile the interval. Since the interior part (-8, 12) is convergent, and the right endpoint x = 12 yields convergence while the left endpoint x = -8 yields divergence, the interval of convergence is (-8, 12].

Note on missing options: the provided data shows the answer options field as null, so I cannot comment on individual distractors. The analysis above is self-contained and shows why the interval (-8, 12] is correct. The key ideas are the radius of convergence determined by the comparison to sum r^n/n and the endpoint tests for x = 12 and x = -8.

Find the interval of convergence of the series [math: ∑n=0∞(−1)n+1n10n(x−2)n] \displaystyle\sum\limits_{n=0}^ \infty \frac{(-1)^{n+1}}{n 10^n} (x-2)^n .单项选择题

类似问题

Suppose the power series \(\displaystyle \sum_{n=1}^{\infty}a_nx^n\) converges at \(x=4\) and \(x=-3\), at which of the following values of \(x\) is the series definitely convergent (select one or more)?

What type interval of convergence does the power series [math: x+2x2+3x3+⋯]x+2x^{2}+3x^{3}+\cdots have?

What is the interval of convergence of the power series [math: ∑n=1∞xnn⋅5n?]\sum_{n=1}^{\infty}\frac{x^{n}}{n\cdot 5^{n}}?

What type interval of convergence does the power series \(x+2x^{2}+3x^{3}+\cdots\) have?

What is the interval of convergence of the power series \[\sum_{n=1}^{\infty}\frac{x^{n}}{n\cdot 5^{n}}?\]

Find the interval of convergence of the series [math: ∑n=1∞n!2n(x−5)n] \displaystyle\sum\limits_{n=1}^ \infty \frac{n!}{2^n} (x-5)^n .

Question at position 8 An excess of growth hormone (GH) before puberty can result in the development ofgigantism.Graves' disease.acromegaly.pituitary dwarfism.Clear my selection

更多留学生实用工具

智能学习助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单