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Algebra Difficulty 5.3 AIME, harder Find the answer

Find the sum of the infinite series 1+2(11998)+3(11998)2+4(11998)3+…1+2\left(\frac{1}{1998}\right)+3\left(\frac{1}{1998}\right)^{2}+4\left(\frac{1}{1998}\right)^{3}+\ldots

A number or a short expression. Spacing and $ signs are ignored.

Solution

We can rewrite the sum as (1+11998+(11998)2+…)+(11998+(11998)2+(11998)3+…)+((11998)2+(11998)3+…)+…\left(1+\frac{1}{1998}+\left(\frac{1}{1998}\right)^{2}+\ldots\right)+\left(\frac{1}{1998}+\left(\frac{1}{1998}\right)^{2}+\left(\frac{1}{1998}\right)^{3}+\ldots\right)+\left(\left(\frac{1}{1998}\right)^{2}+\left(\frac{1}{1998}\right)^{3}+\ldots\right)+\ldots Evaluating each of the infinite sums gives 11−11998+119981−11998+(11998)21−11998+…=19981997⋅(1+11998+(11998)2+…)=19981997⋅(1+11998+(11998)2+…)\frac{1}{1-\frac{1}{1998}}+\frac{\frac{1}{1998}}{1-\frac{1}{1998}}+\frac{\left(\frac{1}{1998}\right)^{2}}{1-\frac{1}{1998}}+\ldots=\frac{1998}{1997} \cdot\left(1+\frac{1}{1998}+\left(\frac{1}{1998}\right)^{2}+\ldots\right)=\frac{1998}{1997} \cdot\left(1+\frac{1}{1998}+\left(\frac{1}{1998}\right)^{2}+\ldots\right), which is equal to (19981997)2\left(\frac{1998}{1997}\right)^{2}, or 39920043988009\frac{3992004}{3988009}, as desired.

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.