Calculate the following infinite sum:

$$
\left[\frac{x+1}{2}\right]+\left[\frac{x+2}{4}\right]+\left[\frac{x+4}{8}\right]+\left[\frac{x+8}{16}\right]+\ldots
$$

Intermediate question: show that $[2 x]=[x]+[x+1 / 2]$.

## Solutions
We start by observing that this sum is finite. Indeed, if $2^{k}>x$ then $x+2^{k}<2^{k+1}$ and finally $\left[\frac{x+2^{k}}{2^{k+1}}\right]=0$. Let's now move on to the intermediate question. We simply make a case distinction: if $n \leq x < n+1/2$ then $[2x]=2n$ and $[x]+[x+1/2]=n+n$, and if $n+1/2 \leq x < n+1$ then $[2x]=2n+1$ and $[x]+[x+1]=n+(n+1)$. On the term of the sequence, this allows us to have

$$
\begin{gathered}
{\left[\frac{x+2^{k}}{2^{k+1}}\right]=\left[\frac{x}{2^{k+1}}+\frac{1}{2}\right]=\left[2 \frac{x}{2^{k+1}}\right]-\left[\frac{x}{2^{k+1}}\right]} \\
{\left[\frac{x+1}{2}\right]+\left[\frac{x+2}{4}\right]+\cdots=\left([x]-\left[\frac{x}{2}\right]\right)+\left(\left[\frac{x}{2}\right]-\left[\frac{x}{4}\right]\right)+\ldots}
\end{gathered}
$$

and the terms are telescoping, in the end only $[x]$ remains.

## 3 Exercises of the day

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## Thursday August 19