Converging Power Series: Finding the Radius of Convergence for (3x+4)^n/n

cokezero
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1. The radius of convergence of the power series the sum n=1 to infinity of (3x+4)^n / n is

a 0
b 1/3
c 2/3
d 3/4
e 4/3

2. the sum n=1 to infinity of (3x+4)^n / n



3. no idea

do the ration test to get abs value 3x+4 < 1 ?
 
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The ratio test is indeed the way to go. It's always a good place to start when the nth term of the series involves n! or a constant to the nth power.
 
i know but what do i do to get from the ratio test to the radius of convergence?
 
The ratio test says that \sum_{n=1}^\infty a_n is abs. convergent if

\lim_{n \rightarrow \infty} \frac{|a_{n+1}|}{|a_n|} &lt; 1,

and divergent if the limit is greater than 1 (assuming in both cases that the limit exists, of course).

So find out for what x your series converges using that test.
 
yes, so i have

lim n --> oo ((3x + 4)^(n+1)/ (n +1)) * (n/ (3x +4)^n)

which simplifies to lim (3x + 4) (n/ (n+1))

so is it abs value (3x+4) < 1 if it converges? but i don't think i have this right b/c none of the answer choices fit to make this statement true.

a) 0
b) 1/3
c) 2/3
d) 3/4
e) 4/3
 
You haven't finished yet. You need x, not 3x+4. Saying that |3x+4|< 1 means -1< 3x+4< 1. Now what interval does x lie in? What is the length of that interval? Of course, the "radius" of convergence is half the length of the interval of convergence.
 
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...

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