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 $$\left|\frac{x^2(1-(\cos x)^k)}{1-\cos x}-\frac{x^2}{1-\cos x}\right| =\left|\frac{x^2}{1-\cos x}\right|$$

  • $\begingroup$ Weierstrass M-Test is straightforward. The series is uniformly convergent. $\endgroup$ – Crostul Apr 8 '17 at 7:57
  • $\begingroup$ I can't read the Chinese text, but the identity you wrote is wrong: check the subtraction. $\endgroup$ – user228113 Apr 8 '17 at 8:28
  • $\begingroup$ That should be an inequality, not an equality: $|\cos^k x| \le 1$, but it is not always $= 1$. $\endgroup$ – Paul Sinclair Apr 8 '17 at 14:16

Define $f(x):=\sum_{k=0}^\infty x^2(\cos x)^k$ and observe that

$$\sum_{k=0}^\infty x^2(\cos x)^k=x^2\sum_{k=0}^\infty (\cos x)^k=\frac{x^2}{1-\cos x},\quad\text{for }x\in(0,1]$$


$$\lim_{x\to 0^+}f(x)=\lim_{x\to 0^+}\frac{x^2}{1-\cos x}=\lim_{x\to 0^+}\frac{2x}{\sin x}=\lim_{x\to 0^+}\frac{2}{\cos x}=2\neq f(0)=0$$

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 $(0,1]$).


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