1989 USAMO Problems/Problem 1

Problem

For each positive integer $n$ , let

$S_n = 1 + \frac 12 + \frac 13 + \cdots + \frac 1n$

$T_n = S_1 + S_2 + S_3 + \cdots + S_n$

$U_n = \frac{T_1}{2} + \frac{T_2}{3} + \frac{T_3}{4} + \cdots + \frac{T_n}{n+1}$ .

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

If we re-group the terms of $T_{n-1}$ ,

$\begin{align*} T_{n-1} &= \left(1\right) + \left(1 + \frac 12\right) + \left(1 + \frac 12 + \frac 13\right) + \ldots + \left(1 + \frac 12 + \frac 13 + \ldots + \frac 1{n-1}\right) \\ &= \sum_{i=1}^{n-1} \left(\frac {n-i}i\right) = n\left(\sum_{i=1}^{n-1} \frac{1}{i}\right) - (n-1) = n\left(\sum_{i=1}^{n} \frac{1}{i}\right) - n \\ &= n \cdot S_{n} - n\end{align*}$

Thus, for $n = 1989$ , $T_{1988} = 1989 S_{1989} - 1989 \Longrightarrow \boxed{a = b = 1989}$ .

For the second part, applying this result gives

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

For $n = 1989$ , we get that $\boxed{c = 1990, d = 2 \cdot 1989 = 3978}$ .

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

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