Consider the sequence of functions $f_n(x)=\chi_{(0,1)}\min(2^{n/4},x^{-1/4})$ for $n=1,2,3,...$. I want to prove this sequence is Cauchy with respect to the metric $$d(f,g)=\bigg(\int\limits_{0}^1|f(x)-g(x)|^2dx\bigg)^{1/2}.$$
Here is my work so far:
We note that for $\{f_n\}$ as defined $$d(f_n,f_{n+1})^2=\int\limits_{0}^{2^{-n}}|f_{n+1}-f_{n}|^2dx\leq\int\limits_{0}^{2^{-n}}|2^{n/4}|^2dx=2^{-n/2}.$$ So $d(f_n,f_{n+1})\leq 2^{-n/4}$. (Note that we are only taking the integral between $0$ and $2^{-n}$ since this is where the difference of the two functions will be non-zero.) We want to prove that for $\epsilon >0$ we have that there exists $N\in\mathbb{N}$ so $d(f_m,f_n)<\epsilon$ for $n,m\geq N$ where $n>m$ without loss of generality. But we have $$d(f_m,f_n)\leq d(f_m,f_{m+1})+d(f_{m+1},f_{m+2})+...+d(f_{n-1},f_n)\leq 2^{-m/4}+2^{-(m+1)/4}+...+2^{-(n-1)/4}\leq 2^{-m/4}+2^{-m/4}+...+2^{-m/4}=2^{-m/4}\cdot(n-m),$$ by the triangle inequality. But from here, I'm having trouble finding the desired $N$. Any help would be appreciated.