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Problem Of The Week #400 Jan 5th, 2020

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
Hi MHB,

I will give the community another week to attempt at last week's high school POTW. I welcome anyone of you who are interested in this problem to give it one more try and I am looking forward to receiving your submission!(Happy)
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
Solution from other:

Summing just the $k=0$ and $k=1$ terms from the identity \(\displaystyle \ln 2=\sum_{k=0}^\infty \dfrac{2}{2k+1}\left(\dfrac{7}{31^{2k+1}}+\dfrac{3}{161^{2k+1}}+\dfrac{5}{49^{2k+1}}\right)\) gives

$\ln 2>\dfrac{29558488681560}{42643891494953}\\ \ln2>0.693147\\ (\ln2)^5>(0.693147)^5 \\ (\ln 2)^5>0.160002 \\ (\ln 2)^5>\dfrac{4}{25}\\ \ln2>\left(\dfrac{2}{5}\right)^{\frac{2}{5}}$
 
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