Flip some sum

Begin by grouping $f(x)$ : $f(x)=(1+\frac{1}{x}+\frac{1}{x^2}+\cdots)+(\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}+\cdots)+\cdots)$ By the formula for infinite geometric series, each inner series with initial term $a$ has sum $\frac{a}{1-\frac{1}{x}}=\frac{xa}{x-1}$ . Thus we have $f(x)=\frac{x}{x-1}+\frac{1}{x-1}+\frac{1}{x(x-1)}+\cdots$ However, this is an infinite geometric series with initial term $\frac{x}{x-1}$ and common ratio $\frac{1}{x}$ , so its sum is $\frac{\frac{x}{x-1}}{1-\frac{1}{x}}=\frac{x^2}{(x-1)^2}$ .

Using this formula, $f(-101)=\frac{101^2}{102^2}$ ; the reciprocal of this is $\frac{102^2}{101^2}$ .

By looking at the formula, we can easily see that the value of $x$ that gives this sum is $\boxed{102}$ .

We wish to simplify the above summation. Note that if we multiply the summation by $x$ we aquire $x*f(x)=x+2+\frac{3}{x}+\ldots$ . Subtracting the equations from one another we get that $f(x)=\frac{x^2}{(x-1)^2}$ . Plugging in $-101$ we find that $S=\frac{101^2}{102^2}$ . Thus, we solve $\frac{x^2}{(x-1)^2}=\frac{102^2}{101^2}$ . It is apparent that $x=102$ .