Monday, 3 July 2017

sequences and series - Why does suminftyk=1fracsin(k)k=fracpi12?



Inspired by this question (and far more straightforward, I am guessing), Mathematica tells us that k=1sin(k)k converges to π12.



Presumably, this can be derived from the similarity of the Leibniz expansion of π 4k=1(1)k+12k1to the expansion of sin(x) as k=0(1)k(2k+1)!x2k+1, but I can't see how...



Could someone please explain how π12 is arrived at?


Answer




Here is one way, but it does not use the series you mention so much. I hope that's OK.



The series is:



sin(1)+sin(2)2+sin(3)3+



[ei+e2i2+e3i3+]



Let x=ei.




[x+x22+x33+]



differentiate:



[1+x+x2+x3+]



This is a geometric series, 11x



[11x]




Integrate:



[ln(x1)]=[ln(ei1)]



Now, suppose ln(ei1)=a+bi,



ei1=eaebi



cos(1)1+isin(1)=ea[cos(b)+isin(b)]




Equate real and imaginary parts:



cos(1)1=eacos(b)sin(1)=easin(b)



divide both:



cos(1)1sin(1)=easin(b)eacos(b)



cot(1/2)=tan(b)




b=tan1(cot(1/2))=12π2



But we need the negative of this, so finally:



π212


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