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Excellent
thanks! :)
Too many methods are never enough.
right! MORE METHODS! :)
I found a longer way and got the wrong answer! Yay! Close but no cigar.
awesome! I had a feeling someone might try it :)
@@owl3math I'm so close. I have your answer but with -2ln2 +ln3. I'm just scratching my head trying to work out where the rest of the ln's have got to.
It’s those kind of errors that can be annoying and hard to find
Not the fastest, but a nice study. I enjoyed it ; )
Right thanks Mike!
S=1/6•sum[n=1,♾️](1/(n+1/3)-1/(n+1/2))S=1/6•(¥(3/2)-¥(4/3))¥(x+1)=1/x+¥(x)S=1/6•(-1+¥(1/2)-¥(1/3))¥(1/2)=-ř-2ln(2)¥(1/3)+¥(2/3)=-2ř-3ln(3)¥(1/3)-¥(2/3)=-pi/sqrt(3)¥(1/3)=-ř-3/2•ln(3)-pi/2sqrt(3)S=-1/3•ln(2)+1/4•ln(3)+pi/12sqrt(3)-1/6
Nice thanks 🙏
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Excellent
thanks! :)
Too many methods are never enough.
right! MORE METHODS! :)
I found a longer way and got the wrong answer! Yay! Close but no cigar.
awesome! I had a feeling someone might try it :)
@@owl3math I'm so close. I have your answer but with -2ln2 +ln3. I'm just scratching my head trying to work out where the rest of the ln's have got to.
It’s those kind of errors that can be annoying and hard to find
Not the fastest, but a nice study. I enjoyed it ; )
Right thanks Mike!
S=1/6•sum[n=1,♾️](1/(n+1/3)-1/(n+1/2))
S=1/6•(¥(3/2)-¥(4/3))
¥(x+1)=1/x+¥(x)
S=1/6•(-1+¥(1/2)-¥(1/3))
¥(1/2)=-ř-2ln(2)
¥(1/3)+¥(2/3)=-2ř-3ln(3)
¥(1/3)-¥(2/3)=-pi/sqrt(3)
¥(1/3)=-ř-3/2•ln(3)-pi/2sqrt(3)
S=-1/3•ln(2)+1/4•ln(3)+pi/12sqrt(3)-1/6
Nice thanks 🙏