In $X=C[0,1]$ with $$\|f\| := \left(\int_0^1 |f(x)|^2 dx\right)^{\frac{1}{2}} $$ is $f_n (x) = x^n$, for $n=1,\dots, \infty$, a Cauchy sequence?
My attempt:
Suppose I have a big positive natural number $N$, and any $m , n > N$. Then I need to study $$ \|f_m - f_n\| = \|x^m - x^n\| = \sqrt{\int_0^1 (x^m - x^n)^2 dx}$$
I disregard the square root first. The inside integral can be calculated, and the value is $$\frac{1}{2m+1} + \frac{1}{2n+1} - \frac{2}{m+n+1}$$
What is next?
If $\varepsilon>0$, take $N\in\Bbb N$ such that $\frac1{2N+1}<\frac\varepsilon2$. Then, if $m,n\in\Bbb N$ are such that $m,n\geqslant N$,then\begin{align}\|f_m-f_n\|&=\frac1{2m+1}+\frac1{2n+1}-\frac2{m+n+1}\\&<\frac1{2m+1}+\frac1{2n+1}\\&\leqslant\frac2{2N+1}\\&<\varepsilon.\end{align}