MHB Joanne 's question at Yahoo Answers (Interval of convergence)

  • Thread starter Thread starter Fernando Revilla
  • Start date Start date
  • Tags Tags
    Convergence
AI Thread Summary
The series in question is $\sum_{n=2}^{\infty}\frac{(-1)^n}{(n-1)2^n}(x-1)^n$. Using the ratio test, it converges absolutely for $|x-1|<2$, leading to the interval of convergence $x\in (-1,3)$. At the endpoints, $x=-1$ results in divergence, while $x=3$ leads to conditional convergence. Therefore, the series converges if and only if $x\in(-1,3]$. This analysis provides a clear understanding of the interval of convergence for the series.
Mathematics news on Phys.org
Hello Joanne,

The series is $\displaystyle\sum_{n=2}^{\infty}\dfrac{(-1)^n}{(n-1)2^n}(x-1)^n.$ Using the ratio test: $$\begin{aligned}L&=\lim_{n\to \infty}\left|\frac{u_{n+1}}{u_n}\right|\\&=\lim_{n\to \infty}\left|\dfrac{(-1)^{n+1}(x-1)^{n+1}}{n2^{n+1}}\cdot\frac{(n-1)2^n}{(-1)^n(x-1)^n}\right|\\&=\lim_{n\to \infty}\left|\dfrac{n-1}{2n}(x-1)\right|\\&=\frac{|x-1|}{2}<1\\&\Leftrightarrow |x-1|<2\\& \Leftrightarrow x\in (-1,3)\end{aligned}$$

So, the series is absolutely convergent if $|x-1|<2$ and divergent if $|x-1|>2$. If $|x-1|=2$ (i.e. $x=-1$ or $x=3$) we have

$(a)\;x=-1$. The series is $\displaystyle\sum_{n=2}^{\infty}\dfrac{(-1)^n}{(n-1)2^n}(-2)^n=\displaystyle\sum_{n=2}^{\infty}\dfrac{1}{n-1}.$ Using the limit comparison test we easily verify that the series is divergent.

$(b)\;x=3$. The series is $\displaystyle\sum_{n=2}^{\infty}\dfrac{(-1)^n}{(n-1)2^n}2^n=\displaystyle\sum_{n=2}^{\infty}\dfrac{(-1)^n}{n-1}.$ Using the Lebniz alteranting series criterion we easily verify that the series is conditionally convergent.

As a consequence the given series is convergent iff $x\in(-1,3]$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top