Does $f_n(x) = \sum_{k=0}^n x^k$ converge uniformly where f maps from $[0, 3/4]$ to reals

Question

Does $f_n(x) = \sum_{k=0}^n x^k$ converge uniformly where f maps from $[0, 3/4]$ to reals?

I found the pointwise limit to be $\frac{1}{1-x} $for $ x \in (0, 3/4]$ and $0$ for $x=0$. I do not know how to proceed

Answer 1

Observe by the geometric series summation formula that $$ f_n(x)=\frac{1-x^{n+1}}{1-x}, $$ and also let $f(x)=1/(1-x)$ denote the limit function. Then $$ |f(x)-f_n(x)|=\frac{|x|^{n+1}}{|1-x|}. $$ To establish uniform convergence, we must show that for every $\epsilon>0$ there exists an $N$ such that for all $n>N$, we have $|f(x)-f_n(x)|<\epsilon$ - where the bounds do not depend on the value of $x$. In this case, observe from $1-x\geq \tfrac14$ and $x\leq \tfrac34$ that we obtain the uniform bound $$ |f(x)-f_n(x)|\leq 4 \left(\frac{3}{4}\right)^{n+1}, $$ allowing you to choose the appropriate $N$ given $\epsilon$, as required.