$14 \cdot 41$ Try for any positive integer $n$, to calculate the sum $\sum_{k=0}^{\infty}\left[\frac{n+2^{k}}{2^{k+1}}\right]$.
(10th International Mathematical Olympiad, 1968)
【Solution】First, prove an identity:
For all real numbers $x$, we have
$$
\left[x+\frac{1}{2}\right]=[2 x]-[x] \text {. }
$$

To this end, let $x=[x]+\alpha$, then $0 \leqslant \alpha<1$.
Thus, we have
$$
\begin{aligned}
{\left[x+\frac{1}{2}\right]=} & {\left[[x]+\alpha+\frac{1}{2}\right]=[x]+\left[\alpha+\frac{1}{2}\right], } \\
{[2 x]-[x] } & =[2[x]+2 \alpha]-[[x]+\alpha] \\
& =[x]+[2 \alpha]-[\alpha] .
\end{aligned}
$$

Comparing (2) and (3), we see that (1) is equivalent to
$$
\left[\alpha+\frac{1}{2}\right]=[2 \alpha]-[\alpha], \quad 0 \leqslant \alpha<1 .
$$

When $0 \leqslant \alpha<\frac{1}{2}$, $\frac{1}{2} \leqslant \alpha+\frac{1}{2}<1, \quad 0 \leqslant 2 \alpha<1$, thus
$$
\left[\alpha+\frac{1}{2}\right]=0,[2 \alpha]-[\alpha]=0 \text {. }
$$

The equation $\quad\left[\alpha+\frac{1}{2}\right]=[2 \alpha]-[\alpha]$ holds.
When $\frac{1}{2} \leqslant \alpha<1$, $1 \leqslant \alpha+\frac{1}{2}<\frac{3}{2}, 1 \leqslant 2 \alpha<2$, thus
$$
\left[\alpha+\frac{1}{2}\right]=1,[2 \alpha]-[\alpha]=1 \text {. }
$$

The equation $\quad\left[\alpha+\frac{1}{2}\right]=[2 \alpha]-[\alpha]$ holds.
Therefore, (1) holds.
From (1), we get
$$
\begin{aligned}
& {\left[\frac{n+2^{k}}{2^{k+1}}\right] } \\
= & {\left[\frac{n}{2^{k+1}}+\frac{1}{2}\right] } \\
= & {\left[2 \cdot \frac{n}{2^{k+1}}\right]-\left[\frac{n}{2^{k+1}}\right] } \\
= & {\left[\frac{n}{2^{k}}\right]-\left[\frac{n}{2^{k+1}}\right] . }
\end{aligned}
$$

Thus, we have
$$
\begin{aligned}
& \sum_{k=0}^{\infty}\left[\frac{n+2^{k}}{2^{k+1}}\right] \\
= & \left([n]-\left[\frac{n}{2}\right]\right)+\left(\left[\frac{n}{2}\right]-\left[\frac{n}{2^{2}}\right]\right)+\left(\left[\frac{n}{2^{2}}\right]-\left[\frac{n}{2^{3}}\right]\right)+\cdots
\end{aligned}
$$

Since for a fixed natural number $n$, when $k$ is sufficiently large, $-n<2^{k},\left[\frac{n}{2^{k}}\right]=0$. Therefore, the above expression is
$$
\sum_{k=0}^{\infty}\left[\frac{n+2^{k}}{2^{k+1}}\right]=[n]=n \text {. }
$$