What is the Interval of Convergence for the Series x^n/(n3^n)?

  • Thread starter Thread starter ganondorf29
  • Start date Start date
  • Tags Tags
    Convergence Series
ganondorf29
Messages
49
Reaction score
0

Homework Statement


Find the radius of convergence and the interval of convergence for
<br /> \sum_{n=0}^\infty \frac{x^n}{n3^n}<br /> <br />

Homework Equations


The Attempt at a Solution


Ok, so I first applied the ratio test.
<br /> \lim_{n\rightarrow\infty}<br /> <br /> |<br /> \frac{x^{n+1}}{(n+1)3^(n+1)} /<br /> \frac{x^n}{n3^n}<br /> |<br />

After some cancellations I got

<br /> L = |x/3|<br /> <br />Does this mean that the interval of convergence is from -3<x<3 ?
 
Physics news on Phys.org
Yes.
 
You have to check the endpoints of your interval separately to see if the series converges there since the ratio test doesn't give any information when the limit is 1
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

Similar threads

Replies
5
Views
2K
Replies
3
Views
1K
Replies
2
Views
2K
Replies
1
Views
1K
Replies
6
Views
2K
Replies
4
Views
2K
Replies
4
Views
1K
Replies
3
Views
2K
Back
Top