The problem is stated as:
Show that $\sum _{k=1}^{\infty }\frac{\left(-1\right)^kx^k}{k}$ converges uniformly on [0,1]
My attempt:
So, since we have an alternating series, applying Weierstrass M - test in this case, will not help since we will have that $\sup_{x\in[0,1]} u_k(x) = 1/k$ for $u_k(x) := (-1)^k x^k / k$. But $\sum_{k=1}^{\infty} 1/k$, diverges.
Instead, we check uniform convergence applying the definition of uniform convergence directly:
$\sup_{x\in[0,1]}|\sum _{k=n}^{\infty }\frac{\left(-1\right)^kx^k}{k}| \leq \sup_{x\in[0,1]}\sum _{k=n}^{\infty }|\frac{\left(-1\right)^kx^k}{k}| = \sup_{x\in[0,1]}\sum _{k=n}^{\infty }|\frac{x^k}{k}| = \sum _{k=n}^{\infty }\frac{1}{k}$
But this diverges too! In my textbook however, they somehow went from $\sup_{x\in[0,1]}|\sum _{k=n}^{\infty }\frac{\left(-1\right)^kx^k}{k}| \leq 1/n \rightarrow 0, n\rightarrow \infty$. However, I can't see at all how that step is possible. I'd be glad if someone could clarify this for me.
Thanks in advance!