Solved by Henry K. Shin

Question:

A Semi-Harmomic Series

Prove that the series whose terms are sin^2(k)/k diverges.

Answer - Proof #1:

Because sin^2(k)/k = (1 - cos (2k))/(2k) = 1/(2k) - cos (2k)/(2k)
it suffices to show that the series whose terms are cos (2k)/(2k) converges.

To do this, I use abel's convergent test:

First it's obvious that 1/(2k) is positive, decreasing, and whose limit as k -> infinity is 0.
It suffices to show that a partial sum s_n whose terms are cos (2k) is bounded.
This is easy, since letting e^(2ki) = cos (2k) + i*sin (2k), we see that
|cos 2 + cos 4 + ... + cos (2n)| <= |(cos 2 + i*sin 2) + (cos 4 + i*sin 4) + ...
+ (cos (2n) + i*sin (2n))| = |e^(2i) + e^(4i) + ...
+ e^(2ni)| = |e^(2i) * (1 - e^(2ni))/(1 - e^(2i))| <= |(1 - e^(2ni))/(1 - e^(2i))| <= 2/|1 - e^(2i)| by triangular inequality.
 
Therefore the partial sum s_n is bounded, and thus the series w/ terms cos (2k)/(2k) converges.

-- Henry K. Shin

Answer - Proof #2:

Proof without Abel's Help:

Let A = sin^2(1)/1 + sin^2(2)/2 + sin^2(3)/3 + ... = Sigma(sin^2(k)/k, k =1 to inf) and 
let B = cos^2(1)/1 + cos^2(2)/2 + ... = Sigma(cos^2(k)/k, k = 1 to inf).

 
Then A + B = 1/1 + 1/2 + 1/3 + ... = Sigma(1/n, n = 1 to inf), a divergent series (harmonic series)


This reveals that one of A and B must diverge!

 
It suffices to show that then that A and B diverge in a similar "manner."


Now one would be tempted to find a bound for the term-to-term ratio
(sinÓ k/ k)/(cosÓ k / k) = tanÓ k.  However this is not easy to bind
with k an integer (not very promising either becauseÊfractional part of
(pi/2)*k, {(pi/2)*k}, is distributed uniformly over the interval [0,1))

 
Instead I regroup the terms of A and B:

 
A = [sin^2(1)/1 + sin^2(2)/2] + [sin^2(3)/3 + sin^2(4)/4)] + ... = Sigma(sin^2(2k-1)/(2k-1) + sin^2(2k)/(2k), k = 1 to infinity)

 
Let f(k) = 1/(2k-1) * sin^2(2k-1) + 1/(2k) * sin^2(2k) so that A = sigma(f(k), k = 1 to infinity).Ê

 
Similarly, B = Sigma(g(k), k = 1 to infinity) where g(k) = 1/(2k-1) * cos^2(2k-1) + 1/(2k) * cos^2(2k)

 
Now let h(x) = (sinÓ(x) + sinÓ (x+1))/(cosÓ(x) + cosÓ (x+1))

 
The inequalities 1/(2k-1) * (sinÓ (2k-1) + sinÓ (2k)) > 1/(2k-1) * sinÓ (2k-1) + 1/(2k) * sinÓ (2k)Ê= f(k) > 1/(2k) * (sinÓ (2k-1) + sinÓ (2k))
and
1/(2k-1) * (cosÓ (2k-1) + cosÓ (2k)) > 1/(2k-1) * cosÓ (2k-1) + 1/(2k) * cosÓ (2k) = g(k) > 1/(2k) * (cosÓ(2k-1) + cosÓ (2k))
hold.


 
Therefore
3M > (2k)/(2k-1) *Êh(2k-1) > f(k)/g(k) > (2k-1)/(2k) *h(2k-1) > (1/4)m
where M = maximum value of h(x) and m = minimum value of h(x).

 
It is clear that m and M are both positive.

Wherefore 3M * g(k) > f(k) > (1/4)m * g(k)
or
3M * B > A > (1/4)m * B

 
Therefore A and B both diverge. (if B diverges, then A > m*B so A diverges.
B cannot be convergent because this would imply that A is also convergent
according to M*B > A but A + B = harmonic series which is a divergent
series.)


QED

-- Henry K. Shin