Determine whether the series converges uniformly? Determine whether the series converges uniformly?
$$\sum\limits_{k=0}^{\infty} \ x^2 \ (\cos x)^k$$ on $[0,1]$.
I tried to find the upper bound of the series and got the following
$$\left|\frac{x^2(1-(\cos x)^k)}{1-\cos x}-\frac{x^2}{1-\cos x}\right|
=\left|\frac{x^2}{1-\cos x}\right|$$

Define $f(x):=\sum_{k=0}^\infty x^2(\cos x)^k$ and observe that 
$$\sum_{k=0}^\infty x^2(\cos x)^k=x^2\sum_{k=0}^\infty (\cos x)^k=\frac{x^2}{1-\cos x},\quad\text{for }x\in(0,1]$$
Hence
$$\lim_{x\to 0^+}f(x)=\lim_{x\to 0^+}\frac{x^2}{1-\cos x}=\lim_{x\to 0^+}\frac{2x}{\sin x}=\lim_{x\to 0^+}\frac{2}{\cos x}=2\neq f(0)=0$$
Then if $f$ would be uniformly convergent it would be continuous in $[0,1]$ (because the functions $x^2(\cos x)^k$ are continuous), hence $f$ does not converges uniformly in $[0,1]$. 
However we have that for $x\in[\alpha,1]$ for any $\alpha\in(0,1]$
$$0\le f(x)\le\sum_{k=0}^\infty(\cos \alpha)^k<\infty$$
Then $f$ converges locally uniformly in $(0,1]$ (what implies that $f$ is continuous in $(0,1]$).
