How am I supposed to prove this inequality? \tan(1)\leq\sum_{k=1}^\infty\frac{\sin(\frac{1}{k^2})}{\cos^2(\frac{1}{k+1})}

Bruce Partridge

Bruce Partridge

Answered question

2022-02-28

How am I supposed to prove this inequality?
tan(1)k=1sin(1k2)cos2(1k+1)

Answer & Explanation

Donald Erickson

Donald Erickson

Beginner2022-03-01Added 8 answers

If kN, k2 and x[0,12] then
ddxsin(xk)cos(x)=1kcos(xk)cos(x)sin(xk)sin(x)12kx2k0
so sin(xk)cos(x) is non-decreasing on this interval. In particular
sin(1k(k+1)}{cos(1k)}sin(1k2)cos(1k+1)
for all k2. This inequality also holds in fact for k=1 since sin(1)<sin(π2)=sin(2)
Now we get the following inequalities for all k1:
tan(1k)tan(1k+1)=sin(1k)cos(1k+1)cos(1k)sin(1k+1)cos(1k)cos(1k+1)
=sin(1k(k+1))cos(1k)cos(1k+1)sin(1k2)cos2(1k+1)
and summing from 1 to gives the required result.

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