1989 USAMO Problems/Problem 1

Problem

For each positive integer $n$ , let $\begin{align*} S_n &= 1 + \frac 12 + \frac 13 + \cdots + \frac 1n \\ T_n &= S_1 + S_2 + S_3 + \cdots + S_n \\ U_n &= \frac{T_1}{2} + \frac{T_2}{3} + \frac{T_3}{4} + \cdots + \frac{T_n}{n+1}. \end{align*}$ 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 $n \ge 2$ , $\begin{align*} T_{n-1} &= 1 + \left(1 + \frac 12\right) + \left(1 + \frac 12 + \frac 13\right) + \ldots + \left(1 + \frac 12 + \frac 13 + \ldots + \frac 1{n-1}\right) \\ &= \sum_{i=1}^{n-1} \left(\frac {n-i}i\right) = n\left(\sum_{i=1}^{n-1} \frac{1}{i}\right) - (n-1) = n\left(\sum_{i=1}^{n} \frac{1}{i}\right) - n \\ &= n \cdot S_{n} - n . \end{align*}$

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

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.

1989 USAMO (Problems • Resources)
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1989 USAMO Problems/Problem 1

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