Finding the interval of convergence of the infinite series $\sum_{n=1}^{\infty}\frac{2^n + x^n}{1+3^n x^n}$ Exercise problem 3.2.1(c) in Problems in Mathematical Analysis, by Kaczor and Nowak asks to find where the following infinite series converges pointwise:
$$\sum_{n=1}^{\infty} \frac{2^n + x^n}{1+3^n x^n}$$
Fix $x = x_0$. We can write:
\begin{align*}
\frac{2^n + x_{0}^{n}}{1+3^n x_{0}^{n}} &\leq \frac{2^n + x_0^n}{3^n x_0^n} = \left(\frac{2}{3x_0}\right)^n + \left(\frac{1}{3}\right)^n
\end{align*}
The term on the right hand side converges for all $|x_0| > \frac{2}{3}$. So, by the comparison test, the original series converges for $|x|>2/3$.
However, consider the interval $(\frac{2}{3},1)$. If we pick the point $x_0$ from this interval, then we can write:
$$\lim \frac{2^n + x_0^n}{1+3^n x_0^n} = \frac{\lim 2^n + \lim x_0^n}{1+\lim 3^n x_0^n}= \frac{\lim 2^n + \lim x_0^n}{1 + \lim 3^n \cdot \lim x_0^n}=\lim 2^n$$
which does not exist. So, by the $n$th term test, the series is divergent in this interval.
We can do this, because $\lim (a_n/b_n) = a/b$ provided $b \neq 0$.
This contradicts the conclusion obtained earlier.
Do you have a hint/clue to proceed with this exercise problem, without revealing the entire solution (proof)?
 A: If $x_0\in(\frac{2}{3},1)$ then $0\le\lim \frac{2^n + x_0^n}{1+3^n x_0^n} = 
\lim \frac{2^n}{1+3^n x_0^n} +\lim \frac{x_0^n}{1+3^n x_0^n} \le 
\lim \frac{2^n}{3^n x_0^n} +\lim \frac{x_0^n}{3^n x_0^n} = 
\lim(\frac{2/3}{x_0})^n +\lim \frac{1}{3^n} = 0+0
=0.$
The way you take limit in one part of your expression, and leave another limit to be taken later, is not correct. It seems that what you are suggesting (with extra details inserted by me according to my interpretation of what you omitted) is that $\lim \frac{2^n + x_0^n}{1+3^n x_0^n} = \frac{\lim 2^n + \lim x_0^n}{1 + \lim 3^n \cdot \lim x_0^n}=\frac{\lim 2^n + 0}{1 + \lim3^n\cdot0}=\frac{\lim 2^n}{1 + 0}=\lim 2^n.$
But if the above was correct then I could similarly prove that $\lim1=0$. Indeed $\lim 1 = \lim 5^n\cdot \lim(\frac15)^n=\lim 5^n\cdot0=0.$ Or perhaps I do not follow what you are suggesting, but you should look at the steps more carefully yourself, try to justify each, and inserting more details as necessary, which hopefully would indicate which step was not justified.
