Prove $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}x^k$ converges uniformly on $[0,1]$.

Question

Prove $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}x^k$ converges uniformly on $[0,1]$.

I don't think this series is uniformly convergent on $[0,1]$. If I let $x=0$, then $s_n=\sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{k}(0)^k \to 0$ as $n\to\infty$. If I let $x=1,$ then $s_n=\sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{k} \to \ln2$ as $n\to\infty$. So, the series is not uniformly convergent since for different values of $x$, I get different limits.

Is my reasoning valid? If not, how can you prove the series is uniformly convergent using the alternating series test?

Answer 1

The series converges uniformly on $[0,1]$ by the Dirichlet test. Here is an argument that parallels the proof of this theorem applied to this specific example.

With $a_k = (-1)^{k+1}$, $S_k = \sum_{j=1}^k a_k,$ and $f_k(x) = x^k/k$ summation by parts gives

$$\left|\sum_{k=n+1}^m a_k f_k(x)\right| = \left|f_m(x) S_m - f_{n+1}(x) S_n - \sum_{k= n+1}^{m-1} (f_k(x) - f_{k+1}(x))S_k\right| \\ \leqslant f_m(x) |S_m| + f_{n+1}(x) |S_n| + \sum_{k= n+1}^{m-1} |(f_k(x) - f_{k+1}(x))||S_k|. $$

Since $|S_k| \leqslant 1$ is uniformly bounded and $f_k(x)$ is decreasing we have

$$\left|\sum_{k=n+1}^m a_k f_k(x)\right| \leqslant 2f_{n+1}(x) = \frac{2 x^{n+1}}{n+1} \leqslant \frac{2}{n+1} \to 0,$$

and the series converges uniformly by the Cauchy criterion for $x \in [0,1]$.

Answer 2

If you test for absolute convergence and apply D'Alembert's Ratio test then you will find that in the end the limit would be equal to x. If x = 1 the Ratio test shall fail . Then you will have to apply Raabe's Test .

For x < 1 or x> 1 you will find out whether the series is convergent or not. But for x=1 you will have apply other test.

You can apply the ratio test to check.

But your reasoning looks right for checking whether convergent for all values of x or not.

You can test for absolute convergence or apply Lebiniz's test .