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January 29th, 2019, 10:43 AM  #1 
Senior Member Joined: Dec 2015 From: somewhere Posts: 600 Thanks: 87  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: 2,531 Thanks: 1390 
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: 600 Thanks: 87 
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 
Senior Member Joined: Sep 2016 From: USA Posts: 635 Thanks: 401 Math Focus: Dynamical systems, analytic function theory, numerics  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 $\displaystyle n=\sum_{k=0}^\infty n_k 2^k$ $\displaystyle \sum_{j=1}^\infty \lfloor \frac{n}{2^j }\rfloor =\sum_{j=1}^\infty \sum_{k=j}^\infty n_k 2^{kj} =\sum_{k=1}^\infty \sum_{j=1}^k n_k 2^{kj} =\sum_{k=1}^\infty n_k (2^k1)=n\sum_{k=0}^\infty n_k$ 

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