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January 29th, 2019, 10:43 AM   #1
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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.
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January 29th, 2019, 11:07 AM   #2
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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.
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January 29th, 2019, 11:22 AM   #3
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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.
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January 29th, 2019, 06:57 PM   #4
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Quote:
Originally Posted by idontknow View Post
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.
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July 6th, 2019, 04:48 AM   #5
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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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