Compute distance in $l^3$ of element $\{\frac{1}{n+1} \}$ from set $\{ \{x_n\} l^3 : x_{100}=4x_{101} \}$
I've started:
$ dist= inf_{\{x_n\}}\{ (\sum_{n=0}|\frac{1}{n+1}-x_n|^3)^{\frac{1}{3}} \}= inf_{\{x_n\}}\{ (\sum_{n=0, n!=100, n!=101}|\frac{1}{n+1}-x_n|^3 + |\frac{1}{101}-x_{100}|^3|+ |\frac{1}{102}-x_{101}|^3|)^{\frac{1}{3}} \}= inf_{\{x_n\}}\{ (\sum_{n=0, n!=100, n!=101}|\frac{1}{n+1}-x_n|^3 + |\frac{1}{101}-4x_{101}|^3|+ |\frac{1}{102}-x_{101}|^3|)^{\frac{1}{3}} \} $
But here I've stuck. I don't know if I can assume, that for $ n!=100 $ and $ n!=101 $ $x_n=\frac{1}{n+1} $ to get those sum equals zero, and compute minimum of function $ |\frac{1}{101}-4x_{101}|^3|+ |\frac{1}{102}-x_{101}|^3$ ? Is my thought right? Or maybe I've just started from wrong side?
Thank you for any sugestion