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March 15th, 2017, 07:56 AM   #1
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What is the limit of this function?

Hello, I was playing around with some sums and noticed that (the weird brackets are supposed to be the floor function and all logs are in base two):

$\displaystyle \sum_{i=1}^{\left< \frac{x+1}{2}\right>}\left< \log (\frac{x}{2i-1})+1\right> = x$

So naturally I wanted to try (just without the floor):

$\displaystyle \lim_{\infty} \frac{\sum_{i=1}^{\left< \frac{x+1}{2}\right>}\log (\frac{x}{2i-1})+1}{x}$

For x = 10000 it equals 1.221347 and it seems it converges around that value. Anyone know where does that value come from or how can I calculate it?
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March 15th, 2017, 09:29 AM   #2
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Quote:
Originally Posted by Ryunaq View Post
Hello, I was playing around with some sums and noticed that (the weird brackets are supposed to be the floor function and all logs are in base two):

$\displaystyle \sum_{i=1}^{\left< \frac{x+1}{2}\right>}\left< \log (\frac{x}{2i-1})+1\right> = x$

So naturally I wanted to try (just without the floor):

$\displaystyle \lim_{\infty} \frac{\sum_{i=1}^{\left< \frac{x+1}{2}\right>}\log (\frac{x}{2i-1})+1}{x}$

For x = 10000 it equals 1.221347 and it seems it converges around that value. Anyone know where does that value come from or how can I calculate it?
Is it $x=a ; a\in \mathbb{N^+}$ or else ?
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March 15th, 2017, 09:58 AM   #3
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Oh right, I forgot, It's only for positive natural numbers.
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