Three students, Eva, Alex, and Mary, solved the equation: $$ 1+\frac{3}{x}+\frac{9}{x^2}+\frac{27}{x^3}+\dots=\frac{x+1}{x} $$ where the left-hand side is an infinite geometric series. Each of them solved the equation in their own way.
Eva first multiplied both sides of the equation by $\frac{x}{3}$: $$ \frac{x}{3}+1+\frac{3}{x}+\frac{9}{x^2}+\frac{27}{x^3}+⋯=\frac{x+1}{3} $$ and then, by comparing both equations, she got: $$ \frac{x}{3}+\frac{x+1}{x}=\frac{x+1}{3} $$ Finally, she solved the equation above: $$ x^2+3x+3=x^2 + x\iff 2x=−3 \iff x=-\frac32 $$ Eva found that the equation has the solution $x=-\frac32$.
Alex used the formula for the sum of an infinite geometric series: $$ \frac{1}{1-\frac{3}{x}}=\frac{x+1}{x} $$ And by solving the above equation, he got: $$ \begin{align} \frac{1}{\frac{x-3}{x}}=\frac{x+1}{x}\cr \frac{x}{x-3}=\frac{x+1}{x}\cr x^2=x^2-2x−3\cr x=-\frac32 \end{align} $$ Alex also got the solution $x=-\frac32$.
Mary first multiplied both sides of the equation by $\frac{3}{x}$: $$ \frac3x+\frac9{x^2}+\frac{27}{x^3}+\dots =\frac{3x+3}{x^2} $$ and then by comparing both equations, she got: $$ \begin{align} \frac{x+1}{x}−1=\frac{3x+3}{x^2} \cr x^2+x−x^2=3x+3 \cr x=-\frac32 \end{align} $$ Mary found that the equation has the solution $x=-\frac32$.
Who proceeded correctly when solving the equation?
None of them
Mary
Eva
Alex
All three
The expression on the left side of the equation is an infinite geometric series with the common ratio $q=\frac3{x}$. The convergence of an infinite geometric series depends upon the value of its common ratio. The series converges when $|q|< 1$.
In our case, $$|q|< 1 \iff \left|\frac3{x}\right|< 1 \iff x \in(-\infty;-3) \cup (3;+\infty)$$
For $x=-\frac32$ , the series is divergent and therefore cannot converge to the value represented by the expression on the right side of the equation. Thus, the given equation has no solution.