Prove that $\sum_{n=1}^\infty \dfrac{(-1)^n}{n} x^n$ converges uniformely on $[0,1]$.
Using the Leibiz criteria, I could show, that the series converges for $x\in[0,1]$. But how can I show the unifom convergence? I know about the Weierstrass M-test but that doesn't apply here...
The Dini Theorem: If each $f_n:[0,1]\to \Bbb R$ is continuous and if $(f_n(x))_n$ converges monotonically to $f(x)$ for each $x\in [0,1],$ and if $f$ is continuous, then $f_n\to f$ uniformly on $[0,1].$
Apply this, first to $(g_n)_n,$ where $g_n(x)=\sum_{j=1}^{2n}(-1)^jx^j/j,$ and second to $(h_n)_n$ where $h_n(x)=\sum_{j=1}^{2n-1}(-1)^j x^j/j.$
The series in the Q converges to $f(x)=-\ln (1+x),$ which is continuous on $[0,1].$
Proof of the Dini Theorem. Given $\epsilon >0:$
For each $x\in [0,1]$ take $n_x$ such that $0\leq |f(x)-f_{n_x}|\leq\epsilon /3.$ Then take $a_x>0$ such that $(y\in [0,1] \land |y-x|<a_x)\implies |f_{n_x}(y)-f_{n_x}(x)|\leq\epsilon /3.$ And take $b_x>0$ such that $(y\in [0,1]\land |y-x|<b_x)\implies |f(y)-f(x)|\leq\epsilon /3.$
Let $c_x=\min (a_x,b_x).$
Now $\{(-c_x+x,c_x+x):x\in [0,1]\}$ is an open cover of $[0,1] ,$ and $[0,1]$ is compact . So there exists a finite non-empty $S\subset [0,1]$ such that $\cup_{x\in S}(-c_x+x,c_x+x)\supset [0,1].$
Let $M=\max \{n_x:x\in S\}.$
Now for $y\in [0,1]$ take $x\in S$ such that $y\in (-c_x+x,c_x+x).$ For any $m\geq M$ we have $$|f_m(y)-f(y)|\leq |f_{n_x}(y)-f(y)|\leq$$ $$\leq |f_{n_x}(y)-f_{n_x}(x)| +|f_{n_x}(x)-f(x)|+|f(x)-f(y)|\leq$$ $$\leq \epsilon /3+\epsilon /3 +\epsilon /3=\epsilon.$$ Therefore $\sup_{m\geq M}\sup_{y\in [0,1]}|f(y)-f_m(y)|\leq \epsilon.$
Remarks: The Dini Theorem holds if the domain is any compact space (not just for $[0,1].$) Compactness is used to ensure that $M$ exists, because $S$ is finite. Monotonicity of each sequence $(f_n(y))_n$ is used to ensure that $m\geq M\geq n_x\implies |f_m(y)-f(y)|\leq |f_{n_x}(y)-f(y)|.$
Further remarks. In the Dini Theorem we cannot omit the continuity of $f.$ For example if $f_n(x)=x^n$ for $x\in [0,1].$ And we cannot omit the monotonicity of each sequence $(f_n(x))_n$. For example if $f_n:[0,1]\to [0,n]$ where $f_n(x)=0$ when $x\in [0,1/3n]\cup [1/n,1]$ and $f_n(1/2n)=n.$ Interestingly, in this example each sequence $(f_n(x))_n$ is eventually monotonic but the convergence is not uniform.