MHB Emil's question at Yahoo Answers (Radius of convergence)

Click For Summary
The discussion addresses Emil's question about finding the radius of convergence for the series ∑((x-3)^n)/(n3^n) from n=1 to infinity. The ratio test is applied, leading to the limit L = |x-3|/3. The series converges absolutely if |x-3| < 3 and diverges if |x-3| > 3. Consequently, the radius of convergence is determined to be R = 3. This mathematical explanation clarifies Emil's confusion regarding the concept.
Mathematics news on Phys.org
Hello Emil,

Using the ratio test to the series $\displaystyle\sum_{n=1}^{\infty}\dfrac{(x-3)^n}{n3^n}$ we get: $$L=\displaystyle\lim_{n \to\infty}\left |\dfrac{(x-3)^{n+1}}{(n+1)3^{n+1}}\cdot\dfrac{n3^n}{(x-3)^n}\right |=\dfrac{|x-3|}{3}\displaystyle\lim_{n \to\infty}\frac{n}{n+1}=\dfrac{|x-3|}{3}$$ If $\left |{x-3}\right |/3<1$ (or equivalently $\left |{x-3}\right |<3$) the series is absolutely convergent. If $\left |{x-3}\right |/3>1$ (or equivalently $\left |{x-3}\right |>3$) the series is divergent. As a consequence, the radius of convergence is $R=3.$
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K