Series sum proof

Dec 2015
1,084
169
Earth
How to prove it or disprove it? Is it true?
\(\displaystyle \sum_{j=1}^{\infty} \lfloor \frac{n}{2^j }\rfloor=n+1 \; \), \(\displaystyle n,j \in \mathbb{N}\).
 
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romsek

Math Team
Sep 2015
2,960
1,674
USA
Your questions are all over the map.

Can I ask what exactly you ask them for? They don't seem to be related to a few classes. Are you a tutor getting answers here? I would be fine with that if you were up front about it.
 
Dec 2015
1,084
169
Earth
Something about odd and even numbers.
Got it by inspection; it is not true.
 
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SDK

Sep 2016
801
544
USA
How to prove it or disprove it? Is it true?
\(\displaystyle \sum_{j=1}^{\infty} \lfloor \frac{n}{2^j }\rfloor=n+1 \; \), \(\displaystyle n,j \in \mathbb{N}\).
Take $n = 1$, then for all $j \in \mathbb{N}$ you have $\frac{n}{2^j} < 1$ so the sum on the left hand side is equal to 0.
 
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Jun 2016
25
2
Hong Kong
Let $\displaystyle n=\sum_{k=0}^\infty n_k 2^k$

$\displaystyle \sum_{j=1}^\infty \lfloor \frac{n}{2^j }\rfloor
=\sum_{j=1}^\infty \sum_{k=j}^\infty n_k 2^{k-j}
=\sum_{k=1}^\infty \sum_{j=1}^k n_k 2^{k-j}
=\sum_{k=1}^\infty n_k (2^k-1)=n-\sum_{k=0}^\infty n_k$
 
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