Need help with some Calculus II power series problems?

Question

I am trying to find the interval of convergence for two power series right now.

They are:

A. $\frac{1+2n}{1+3n}x^n$ where $n$ starts at $0$ and approaches infinity

B. $\frac{1+2^n}{1+3^n}x^n$ where $n$ starts at $0$ and approaches infinity

Thanks in advance for your help with these problems!

Answer 1

HINT

For the first by ratio test

$$\left|\frac{(1+2(n+1))(x^{n+1})}{ (1+3(n+1))}\frac{ (1+3n)}{(1+2n)(x^n)}\right|=|x|\frac{(2n+3)(3n+1)}{(3n+4)(2n+1)}$$

For the second by ratio test

$$\left|\frac{(1+2^{n+1})(x^{n+1})}{ (1+3^{n+1})}\frac{ (1+3^n)}{(1+2^n)(x^n)}\right|=|x|\frac{(2\cdot2^n+1)(3^n+1)}{(3\cdot3^n+1)(2^n+1)}$$

From these in both cases we can find the condition of convercence for $|x|<R$ then we need to check for the cases $|x|=R$.

Answer 2

Hints: Cauchy-Hadamard formula and

$$\begin{align*}&\sqrt[n]{\frac{1+2n}{1+3n}}=\frac{\sqrt[n]{1+2n}}{\sqrt[n]{1+3n}}\\{}\\ &\sqrt[n]{\frac{1+2^n}{1+3^n}}=\frac23\sqrt[n]{\frac{1+\frac1{2^n}}{1+\frac1{3^n}}}\end{align*}$$