Controlling power series by a function

Question

I was working on a problem where the following power series was considered $$ \sum_{n=0}^{\infty} \dfrac{x^n}{n+2^n} $$ I found the interval of convergence: $-2<x<2$. Then they as to prove that $$ \sum_{n=0}^{\infty} \dfrac{x^n}{n+2^n} \leqslant \dfrac{x}{2-x}, \forall 0\leqslant x <2. $$ Does anyone can help on how to prove such inequality, or let me know how I can start with it.

Answer 1

For $0\le x \lt2$ $$ \sum_{n=0}^{\infty} \dfrac{x^n}{n+2^n} \le \sum_{n=0}^{\infty} \dfrac{x^n}{2^n} =\frac{1}{1-\frac{x}{2}} =\frac{2}{2-x}$$ using the sum of the terms of geometric series.

But the inequality of the question is not correct.

Answer 2

What you think about that?? $$ \sum_{n=0}^{\infty} \dfrac{x^n}{n+2^n} \leqslant \sum_{n=0}^{\infty} \dfrac{x^n}{2^n} =\sum_{n=0}^{\infty} \left(\dfrac{x}{2}\right)^n = \dfrac{1}{1-\frac{x}{2}} $$

Answer 3

The inequality seems to be wrong.

But we can do the following.

$$\sum_{n=0}^{\infty} \dfrac{x^n}{n+2^n}\leq\sum_{n=0}^{\infty} \left[\dfrac{x}{2}\right]^n = \dfrac{1}{1-\dfrac{x}{2}}=\dfrac{2}{2-x}.$$

At the same time we can write

$$\sum_{n=0}^{\infty} \dfrac{x^n}{n+2^n}\geq\sum_{n=0}^{\infty} \dfrac{x^n}{2^n+2^n} = \dfrac{1}{2}\sum_{n=1}^{\infty}\left[\dfrac{x}{2}\right]^n=\dfrac{1}{2-x}.$$

Hence, we can conclude

$$\dfrac{1}{2-x} \leq \sum_{n=0}^{\infty} \dfrac{x^n}{n+2^n}\leq \dfrac{2}{2-x}$$