I need the ratio of x to y in this equation:

sum [n*(x/10)^n/10^(n-2)] == y from n=1 to ∞

Thank you!

  • 4
    $\begingroup$ Is this about Mathamatica the software, or is it about math? Should we then forward the question to Mathematics? $\endgroup$ – Johu Sep 19 '18 at 23:23

Something like this maybe?

Solve[Sum[n (x/10)^n/10^(n-2),{n,1,\[Infinity]}]==y/.x->a y,a]/.a->x/y

{{x/y->(100 (50+y-10 Sqrt[25+y]))/y^2},{x/y->(100 (50+y+10 Sqrt[25+y]))/y^2}}

x/Sum[n (x/10)^n/10^(n - 2), {n, \[Infinity]}]
(-100 + x)^2/10000

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