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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 $\displaystyle n=\sum_{k=0}^\infty n_k 2^k$ $\displaystyle \sum_{j=1}^\infty \lfloor \frac{n}{2^j }\rfloor =\sum_{j=1}^\infty \sum_{k=j}^\infty n_k 2^{k-j} =\sum_{k=1}^\infty \sum_{j=1}^k n_k 2^{k-j} =\sum_{k=1}^\infty n_k (2^k-1)=n-\sum_{k=0}^\infty n_k$ Tags proof, series, sum Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post WWRtelescoping Complex Analysis 2 March 27th, 2014 12:08 AM SlamDunk Algebra 11 April 18th, 2012 11:21 AM geordief Calculus 2 January 8th, 2012 08:27 AM Weiler Calculus 10 January 2nd, 2012 06:00 PM wnvl Calculus 4 November 25th, 2011 12:44 PM

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