> 
> Prove $$\left[\dfrac{n+1}{2}\right]+\left[\dfrac{n+2}{4}\right]+\left[\dfrac{n+4}{8}\right]+\left[\dfrac{n+8}{16}\right] + \dots=n$$ 
>  where $[x]=\lfloor x\rfloor$
> 
> 
> 

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It was suggested that somehow I use the identity $[x]=\left[\dfrac x2\right]+\left[\dfrac{x+1}{2}\right]$$$$$After struggling for a while, I realised I wasn't getting anywhere using his hint, probably because I couldn't really understand how I was to use it. Instead I tried to use the Squeeze Theorem by rewriting the $nth$ term of the series (referred to later as S) as$$$$ $$t\_n=\left[\dfrac{n+2^k}{2^{k+1}}\right] \text{ where } 0\le k<\infty$$
$$\Rightarrow \dfrac{n+2^k}{2^{k+1}}-1<\left[\dfrac{n+2^k}{2^{k+1}}\right]\le \dfrac{n+2^k}{2^{k+1}}$$
$$$$
$$ \lim\_{k\to \infty}(k+1)\left(\dfrac{n+2^k}{2^{k+1}}-1\right)<S\le \lim\_{k\to \infty}(k+1)\left( \dfrac{n+2^k}{2^{k+1}}\right)$$

However these bounds are too loose as the limits diverge to $-\infty$ and $\infty$ respectively.
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Could somebody please show me how to prove the series is equal to $n$, either through the given hint, or through the selection of tighter bounds for the Squeeze Theorem? Many thanks!