I want to show $$\sum_{n=0}^\infty x^n$$$$x\in(-1,1)$$ does not converge uniformly using the negation of Cauchy's Criterion for uniform convergence of series of functions.
Cauchy's Criterion states that $\sum f_n$ converges uniformly iff $$\forall \epsilon>0\text{ }\exists l\in N\text{ }\forall x\in D\text{ }\forall m,n\in N \text{ }(m>n\ge l\Rightarrow |f_{n+1}(x)+...+f_{m}(x)|<\epsilon)$$
Hence the negation would be:$$\exists \epsilon>0\text{ }\forall l\in N\text{ }\exists x\in D\text{ }\exists m,n\in N \text{ }(m>n\ge l\land |f_{n+1}(x)+...+f_{m}(x)|\ge\epsilon)$$
For the question, I have taken $\epsilon=\frac{1}{4}$ and $\forall l\in N$ let $ x_l=\frac{1}{2^\frac{1}{l+1}}$ and $m=l+1,n=l$.
Hence,$$|f_{n+1}(x)+...+f_{m}(x)|=|f_{l+1}(x)|=|\frac{1}{2}|\ge\epsilon)$$
I know that the sum is a geometric sum and it could be shown not to converge uniformly using the fact that $\forall n\in N$ the partial sums of $ x^n$ are bounded on (-1,1).
However, using Cauchy's criterion is the above method correct?