# Need help with some Calculus II power series problems? [closed]

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!

## closed as off-topic by Leucippus, user296602, Xam, José Carlos Santos, Parcly TaxelMar 10 '18 at 23:23

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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 $|x|=R$.

• Thanks for the helpful information. It appears from your computations that as n approaches infinity, problem one will converge to |x|(2/3)(3/2) and will end up with an interval of (-1, 1). It also appears that the second problem will converge to |x|(2/3) and end up with an interval of |2x/3| < 1 or (-3/2, 3/2). Am I figuring correctly? Thanks! – Stephen Lanford Mar 10 '18 at 17:15
• @StephenLanford Yes it is and what about |x|=1? – gimusi Mar 10 '18 at 17:19
• My apologies, what exactly are you asking? – Stephen Lanford Mar 10 '18 at 17:24
• @StephenLanford usually by these criteria we can find the value for convergence $|x|<R$ (un less $R=\infty$) then we need always to check also the cases $|x|=R$ when $R\neq \infty$ since these test are not conclusive. – gimusi Mar 10 '18 at 17:27
• Are you saying I need to plug in my values of -1, 1 and -3/2, 3/2 into my original functions and see whether they converge or diverge? – Stephen Lanford Mar 10 '18 at 17:30

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*}