Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (c)

MATH101 April 2017

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Question 01 (c)
For which real numbers $x$ does the power series $\sum _{n=0}^{\infty }{\frac {(2x-3)^{n}}{n+1}}$ converge?

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Consider the ratio test, the p-series test, and the alternating series test.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. First, apply the ratio test for the given series $\sum _{n=0}^{\infty }a_{n}=\sum _{n=0}^{\infty }{\frac {(2x-3)^{n}}{n+1}}$. Since $\lim _{n\to \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|=\lim _{n\to \infty }\left|{\frac {\frac {(2x-3)^{n+1}}{n+2}}{\frac {(2x-3)^{n}}{n+1}}}\right|=\lim _{n\to \infty }|2x-3|\left|{\frac {n+1}{n+2}}\right|=|2x-3|,$ by the ratio test, the series converges for $x$ satisfying $|2x-3|<1$. (i.e., $x\in (1,2)$) and diverges for $x$ satisfying $|2x-3|>1$ (i.e., $x\in (-\infty ,1)\cup (2,\infty )$), When $x=2$, the given series diverges by the p-series test; $\sum _{n=0}^{\infty }{\frac {(2\cdot 2-3)^{n}}{n+1}}=\sum _{n=0}^{\infty }{\frac {1}{n+1}}=\sum _{n=1}^{\infty }{\frac {1}{n}}=+\infty$. On the other hand, when $x=1$, we can see that the given series is an alternating series; $\sum _{n=0}^{\infty }{\frac {(2\cdot 1-3)^{n}}{n+1}}=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{n+1}}$. Since ${\frac {1}{n+1}}$ decreasing and converges to $0$ as $n$ goes to infinity, by the alternating series test, it converges. To summarize, the given series converges when $\color {blue}x\in [1,2)$.