2019 Putnam Problem B2

On Saturday, December 7, 2019, I took the 2019 Putnam competition. The Putnam competition is the premier undergraduate math competition in the United States. It has two sessions of six problems each with 3 hours allotted for each session. The problems are all proof-based and are graded on a 0-10 scale, with 10 points being given only for complete proofs. This exam is infamous for having a median score of 0-1 points out of a total of 120. This was my first time taking the Putnam, so my goal was simply to get a positive score.

According to AoPS, discussion is now allowed on the 2019 Putnam.

Putnam Problem B2 concerns evaluating a complex limit. The question reads:

2019 Putnam Problem B2:
For all n>=1, let a_n = sum_(k=1)^(n-1) (sin((2k-1)pi/(2n)))/(cos^2((k-1)pi/(2n))cos^2(kpi/(2n))).
Determine lim_(n->inf) a_n/n^3.

I apologize if it is difficult to read the problem above. The link I provided has LaTeX for this problem. This problem was particularly satisfying because it gave me a chance to put an obscure trig identity I learned in my Physics class to use. Here’s my solution:

Solution #1: Telescoping L’Hopital Approach

We can recall that cos(x)+cos(y) = 2*cos((x+y)/2)*cos((y-x)/2) and that cos(x)-cos(y) = 2*sin((x+y)/2)*sin((y-x)/2). It follows that cos^2(x) – cos^2(y) = 4*cos((x+y)/2)*sin((x+y)/2)*cos((y-x)/2)*sin((y-x)/2) = sin(x+y)*sin(y-x). It follows that 1/(cos^2((k-1)pi/(2n))) – 1/(cos^2(kpi/(2n))) = sin((2k-1)pi/(2n))*sin(pi/2n)/(cos^2((k-1)pi/(2n))*cos^2(kpi/(2n))). With this substitution, it is possible to telescope the series to get that the given series a_n is equivalent to (1 – 1/(cos^2((n-1)pi/(2n)))/(sin(pi/2n)). At this point, we can use L’Hopital’s Rule to quickly find the limit as n approaches infinity of a_n/(n^3). We get that the answer is 8/pi^3 as desired.

I was pretty satisfied when this worked. I have almost never used these trig identities before, so seeing them simplify such a complex series was pretty great.

Thanks for reading! See you tomorrow.