Inspired by this question (and far more straightforward, I am guessing), Mathematica tells us that ∞∑k=1sin(k)k converges to π−12.
Presumably, this can be derived from the similarity of the Leibniz expansion of π 4∞∑k=1(−1)k+12k−1to 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 π−12 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+⋅⋅⋅
ℑ[ei+e2i2+e3i3+⋅⋅⋅]
Let x=ei.
ℑ[x+x22+x33+⋅⋅⋅]
differentiate:
ℑ[1+x+x2+x3+⋅⋅⋅]
This is a geometric series, 11−x
ℑ[11−x]
Integrate:
−ℑ[ln(x−1)]=−ℑ[ln(ei−1)]
Now, suppose ln(ei−1)=a+bi,
ei−1=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=tan−1(cot(1/2))=12−π2
But we need the negative of this, so finally:
π2−12
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