Determine for what values of x the series converges.8.4.7 Petrovic

Question

Determine for what values of x the series converges.8.4.7 Petrovic

$$\sum_{n=1}^{\infty} \frac{x^n}{a^n + b^n}, a,b >0 $$

My trial:

I used the root test and I got that $C_{n} = (\frac{1}{a^n + b^n})^{1/n}|x|$, now taking the limit as $n \rightarrow \infty $ of $(\frac{1}{a^n + b^n})^{1/n}$ I managed to simplify it to $ \lim a^n (1 + (b/a)^{n})^{-1/n}$, but then I do not know what to do next, could anyone help me please?

Answer 1

assuming $a>b$ $$ \lim_{n\rightarrow \infty } \frac{x}{(a^n + b^n)^{1/n}}=\lim_{n\rightarrow \infty } \frac{x}{(a^n(1 + (b/a)^n))^{1/n}}= \lim_{n\rightarrow \infty } \frac{x}{a(1 + (b/a)^n)^{1/n}}= \frac{x}{a } $$

Great idea by user376343 in comments to use limit comparison test. I assumed x is positive , if x is negative then you can use leibniz alternating series

Answer 2

Completing my comment as an answer:

Due to the symmetry we can assume $\;a\geq b.\;$ We have $$0\leq \frac{|x|^n}{a^n+b^n}\leq \frac{|x|^n}{2a^n}.$$

The geometric series $\sum \frac{1}{2}\left(\frac {x}{a}\right)^n$ converges iff $-a<x<a,$ which is also the interval of convergence of the given series.