\[log1+log2+log3...+log300\]
Is there any shortcut path to find the summation of the series?
SERIES
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sakibtanvir
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Phlembac Adib Hasan
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Re: SERIES
May be a way to find approximately : (but not so much helpful)
Surely the sum is equal to $log 300!$.Starling proved the following identity:\[log n!\approx \frac{log(2\pi n)}{2}+nlog \left (\frac{n}{e} \right)\]
Surely I could use it to prove $300!>100^{300}$.But I didn't, because I don't know the proof of this theorem.
Surely the sum is equal to $log 300!$.Starling proved the following identity:\[log n!\approx \frac{log(2\pi n)}{2}+nlog \left (\frac{n}{e} \right)\]
Surely I could use it to prove $300!>100^{300}$.But I didn't, because I don't know the proof of this theorem.
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