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1989 USAMO Problems/Problem 1

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Problem

For each positive integer n, let \begin{align*}S_n &= 1 + \frac 12 + \frac 13 + \cdots + \frac 1n \\T_n &= S_1 + S_2 + S_3 + \cdots + S_n \\U_n &=... Find, with proof, integers 0 < a,\ b,\ c,\ d < 1000000 such that T_{1988} = a S_{1989} - b and U_{1988} = c S_{1989} - d.

Solution

We note that for all integers n \ge 2, \begin{align*}T_{n-1} &= 1 + \left(1 + \frac 12\right) + \left(1 + \frac 12 + \frac 13\right) + \ldots + \left(1 + \frac ...

It then follows that \begin{align*}U_{n-1} &= \sum_{i=2}^{n} \frac{T_{i-1}}{i} = \sum_{i=2}^{n}\ (S_{i} - 1) = T_{n-1} + S_n - (n-1) - S_1 \\&...

If we let n=1989, we see that (a,b,c,d) = (1989,1989,1990, 2\cdot 1989) is a suitable solution. \blacksquare

Notice that it is also possible to use induction to prove the equations relating T_n and U_n with S_n.


Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See also

1989 USAMO (Problems)
Preceded by
First question 		1 • 2 • 3 • 4 • 5 Followed by
Problem 2
All USAMO Problems and Solutions
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