For function $f(x)=\sum_{n=1}^{\infty}\frac{n[nx]}{4^n}$, prove or disprove that $\int_{-1}^1f(x)dx$ is a rational number.

Question

Let $f(x)=\sum_{n=1}^{\infty}\frac{n[nx]}{4^n},$ where $[a] $ stands for the integer part of $a\in\mathbb{R}$.

$\mathbf{1.}$ Show that the series is uniformly convergent on $[-1,1]$. Show that the function $f(x)$ is Riemann integrable on $[-1,1].$

$\mathbf{2.}$ Prove or disprove that $\int_{-1}^1f(x)dx$ is a rational number.

I think, I'm almost good for the first part. However, I'm stuck on the second part. Here is my thoughts.

$\mathbf{Part 1) }$ As $x\in[-1,1]$, we have $|[nx]|\leq n$ for all $n\in\mathbb{N}$. Hence $|\sum_{n=1}^{\infty}\frac{n[nx]}{4^n}|\leq \sum_{n=1}^{\infty}\frac{n^2}{4^n}$. By M-Weierstrass, we get $f(x)$ is uniformly convergent.

Also, we know that if Riemann integrable sequence $f_n$ converges to $f$ over a bounded interval $I$, then $f $ is also integrable.

Now define $f_N(x):=\sum_{n=1}^N\frac{n[nx]}{4^n}.$ We claim that the only discontinuities of $f_N$ are the rational numbers.

Suppose $x\in\mathbb{Q}$, is in the form $x=\frac{p}{q}$, where $p, \text{and}\:\:q\neq 0$ are inetgers. For this $x$. there is inetger $N$ sufficiency large ($N\geq q$), such that $f_N(x)$ is discontinuous at $x$. Note that the function $\frac{q[qx]}{4^q}$ is discontinuous at $x=\frac{p}{q}$.

Now if $x_0\notin \mathbb{Q}, $ so for all $N\in \mathbb{N}$, the function, $f_N(x)$ is continuous at $x_0$.

By what we have seen, the function $f_N$ has countably many discontinuities on $[-1,1]$, so it's Riemann integrable, and as $f_N\to f$ uniformly, $f$ is Riemann integrable.

$\mathbf{Part 2) }$ Define $g_n$ as follow.

$$g_N(x)=\int_{-1}^1f_N(x)dx=\int_{-1}^1\sum_{n=1}^N\frac{n[nx]}{4^n}=\sum_{n=1}^N\int_{-1}^1\frac{n[nx]}{4^n}. $$

Since $f_N(x)$ for each $x\in[-1,1]$ is rational number, by above we get $g_N(x)$ is also a rational number, so we have a sequence of rational numbers $g_N(x)$ which is convergent uniformly to $\int_{-1}^1f(x)dx$. But I don't know how to show $\int_{-1}^1f(x)dx$ is a rational number. Thank you.

Answer 1

\begin{align} \int_{-1}^1f(x)dx&=\sum_{n=1}^{+\infty}\frac{n}{4^n}\int_{-1}^1[nx]dx\\ &=\sum_{n=1}^{+\infty}\frac{1}{4^n}\int_{-n}^n[t]dt\\ &=\sum_{n=1}^{+\infty}\frac{1}{4^n}\sum_{k=-n}^{n-1}k\\ &=-\sum_{n=1}^{+\infty}\frac{n}{4^n}\\ &=-{4\over9} \end{align}