How to evaluate $\lim_{N\to\infty}\sup_{x\in(1,\infty)}\big|\sum_{n=N+1}^{\infty}\frac{1}{1+x^n}\big|$?

Question

How to evaluate the following limit

$$\lim_{N\to\infty}\sup_{x\in(1,\infty)}\big|\sum_{n=N+1}^{\infty}\frac{1}{1+x^n}\big|\ ? $$

My hunch is that this limit should be zero, but I am unable to prove it .

Answer 1

First, let's calculate $$\lim_{x\to 1^+}\big|\sum_{n=N+1}^{\infty}\frac{1}{1+x^n}\big|$$Given arbitrary $M\in \Bbb N$ and $M>N$, let $1<x<2^{1\over M}$. Therefore $$1<x^{n}<2^{n\over M}\le 2\quad,\quad N< n\le M$$therefore $${1\over 3}\le {1\over 1+x^n}<{1\over 2}\quad ,\quad N< n\le M$$so we can write$$\left|\sum_{n=N+1}^{\infty}\frac{1}{1+x^n}\right|{>\left|\sum_{n=N+1}^{M}\frac{1}{1+x^n}\right|\\=\sum_{n=N+1}^{M}\frac{1}{1+x^n}\\\ge\sum_{n=N+1}^{M}\frac{1}{3}\\={M-N\over 3}\quad,\quad 1<x<2^{1\over M}}$$since $M$ can be chosen arbitrarily large, then so can be $\left|\sum_{n=N+1}^{\infty}\frac{1}{1+x^n}\right|$ for any amount of $N$ and therefore the final answer is $\infty$.

Answer 2

Hint: For each $n,$ what is $\sup_{x\in (1,\infty)}\dfrac{1}{1+x^n}?$