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 January 29th, 2019, 10:43 AM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 742 Thanks: 98 Series sum proof How to prove it or disprove it? Is it true? $\displaystyle \sum_{j=1}^{\infty} \lfloor \frac{n}{2^j }\rfloor=n+1 \;$, $\displaystyle n,j \in \mathbb{N}$. Last edited by skipjack; January 29th, 2019 at 11:18 PM.
 January 29th, 2019, 11:07 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,590 Thanks: 1434 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.
 January 29th, 2019, 11:22 AM #3 Senior Member   Joined: Dec 2015 From: somewhere Posts: 742 Thanks: 98 Something about odd and even numbers. Got it by inspection; it is not true. Last edited by skipjack; January 29th, 2019 at 11:18 PM.
January 29th, 2019, 06:57 PM   #4
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Quote:
 Originally Posted by idontknow 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.

Last edited by skipjack; January 29th, 2019 at 11:20 PM.

 July 6th, 2019, 04:48 AM #5 Newbie     Joined: Jun 2016 From: Hong Kong Posts: 25 Thanks: 2 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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