How would one attack a sum like this:
$$\lfloor 1 \cdot \sqrt{2} \rfloor + \lfloor 2 \cdot \sqrt{2} \rfloor + \lfloor 3 \cdot \sqrt{2} \rfloor + ... + \lfloor n \cdot \sqrt{2} \rfloor$$
or simply:
$$\sum_{i=1}^n\lfloor i \sqrt{2} \rfloor$$
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I dont know if this would be useful here $$\lfloor i\sqrt 2\rfloor=i+\lfloor i{\sqrt 2}\rfloor$$ because $\sqrt2\approx 1.4$. If I remember correctly this case is discussed in Concrete math of Graham and Knuth. – Masacroso Jun 02 '17 at 13:18
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HINT:
Since the sum $1+2+3+4 \cdots$ or $\sum_{n=0}^\infty n$ diverges according to Wikipedia, by the convergence test this sum is larger than the sum of all natural numbers, so it also diverges.
Therefore, the series doesn't approach a general limit, so the sum has to be expressed in terms of $i$ only.
Toby Mak
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