Royboy01's blog

By Royboy01, history, 2 months ago, In English

Sorry if its already been answered.

 
 
 
 
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2 months ago, # |
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We can approximate the sum of this infinite harmonic sequence with the natural log function (Think of it like integration instead of addition), and the error is less than $$$1$$$ (It's called the Euler-Mascheroni constant)

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2 months ago, # |
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lets say it is smaller than 1 + 1/2 + 1/2 + 1/4 + 1/4 + 1/4 + 1/4 + 1/8 + ... (we have 1/2^i 2^i times) and (1/2^i) * 2^i = 1 so we have n * log(n)

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2 months ago, # |
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Good job bruh

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2 months ago, # |
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Bibek..rocks

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2 months ago, # |
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We can apply integral test for it 1+/2+1/3+...+1/n < integral of (1/x)dx with lower limit 1 and upper limit n. So the sum is < log(n)